how to solve a estimate problem

Estimation (Introduction)

Estimation is a skill for life..

As you walk around and live your life, imagine if you could easily estimate:

• how much a bill will be,
• which item is the best value for money
• the size, areas and angles of things

Also, it would be great if you could quickly guess how many people are in a room, how many cars in the street, how many boxes on the shelf, or even how many seagulls on the beach.

We are not talking exact answers here, but answers that are good enough for your life.

Exact vs Estimate

In mathematics we often stress getting an exact answer.

But in everyday life a few cents will not make much difference ... you should focus on the dollars!

Estimation is ...

... finding a number that is close enough to the right answer.

• You are not trying to get the exact right answer
• What you want is something that is good enough (usually in a hurry!)

Estimation can save you money . Always do a quick estimation of how much you should pay:

Example: you want to buy five magazines that cost $1.95 each. When you go to buy them the cost is $12.25. Is that right?

"five at $1.95 each is about 5 times 2, or about $10"

so $12.25 seems too much!

Ask to have the total checked.

Estimation can save you time (when the calculation does not have to be exact):

Example: you want to plant a row of flowers. The row is 58.3cm long. The plants should be 6cm apart. How many do you need?

"58.3 is nearly 60, and 60 divided by 6 is 10, so 10 plants should be enough."

Estimation can save you from making mistakes with your calculator :

Example: you are calculating 107 times 56, and the calculator shows this:

Is that right?

"107 times 56 is a bit more more than 100 times 50, which is 5000 "

Ooops! you must have typed something wrong ...

... in fact you pressed 17×56 (you left out the zero), and without estimating you could have made a really big mistake!

In fact don't trust calculators or computers ...

Computers don't understand what you are doing.

So use your brain power to double-check everything

And one last important reason:

Fun, and Good for your Brain

It is actually good fun to do estimating because it keeps your mind active. As you walk around see how good you are at estimating answers before they come up.

• At the supermarket try to add up everything yourself, and then compare it to the final bill
• Estimate how long it will take you to get home
• When shopping, see which is the better bargain by estimating the price per quantity

Example: You are making invitation cards. It took you 3 minutes and 20 seconds to make one card, but you need to do 15 more ... how long will it take? Think:

"... it took a little more than 3 minutes for one card, and 3 × 15 is 45...

... add a bit more ... maybe 50 minutes and I will be done."

And then see how close your estimate was when you finish.

Skill and Practice

Estimating takes skill ... but how do you get this skill?

Lots of Practice .

So we made some Estimation Games for you to practice with!

Practice is the best way to become good at estimating, because you need to develop your own "strategies", as different numbers need different methods:

• 550 + 298: 298 is nearly 300 so an estimate is 550+300 = 850
• 550 + 248: 50+48 is nearly 100 so an estimate is 500+200 +100 = 800

In one case it seemed easy to change one number and then add.

In the other case I added the hundreds together and then increased the result by 100

So the first step is to think "how do I go about making this estimate?"

Tips and Tricks

After doing some practice, read our page on Estimation Tips and Tricks

Estimating Counts, Lengths and More

Estimation is not always about doing calculations.

It is important for you to be able to estimate how many things you can see, or how long something is or how big something is.

Example: How many objects are here:

You could count them, but what if you don't need an exact answer, just one that is close enough? Can you guess well?

Read more on our page on Visual Estimation .

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How to Estimate a Math Problem

•  / 

Estimation is a great way to check your work and solve math problems quickly. Once you learn to round numbers successfully, you can do much more complex math equations in your head. Here are some tips and tricks for mastering the art of rounding and estimation.

How to Estimate

Rounding numbers.

When you estimate math problems, start by rounding off the numbers. Rounding makes math problems easier to estimate. If you're working with decimals, you might want to round to the nearest whole number. If you have a number with a decimal value of 0.5 or higher, you round up (5.7 rounds to 6, for instance). The opposite is also true - numbers with decimal values lower than 0.5 are rounded down. For instance, 4.3 rounds down to 4.

Whole Numbers

If you're working with whole numbers, you'll probably want to round to the nearest ten or hundred, depending on how close you need your estimation to be. Naturally, close rounding will make your estimation more accurate. For example, if you round 4,557 to 4,560 (the nearest ten), your estimation will be more accurate than if you round to 4,600 (the nearest hundred).

Sample Estimation Question

Once you've rounded the numbers, you can estimate the answer in your head. As an example, we'll estimate the math problem 51 x 69. This would be a hard problem to do mentally, so to make it easier, you'll start by rounding to the nearest ten. You can round 51 to 50 and 69 to 70. Then, in your head, multiply 5 x 7 = 35 and add two zeros to your answer. The final, estimated answer is 50 x 70 = 3,500.

Now you know that 51 x 69 is about 3,500. By checking it on a calculator, you'll see that 51 x 69 = 3,519. That's only 19 off, which isn't bad for a quick estimation!

Practice Problems

Using the skills you've learned, try to estimate the following math problems. You can start by rounding to the nearest whole number that is easy enough to work with (for instance, 20 instead of 19 and 50 instead of 48).

• 78 ÷ 41

Although there are really no 'wrong' answers to these problems, since you're estimating, the answers below are the ones you should have if you've followed the steps explained above. The important thing is that your estimated answers are in the ballpark of the correct answers, which you can find using a calculator. It's also important that you can explain how you arrived at your estimated answers.

• If you round 49 to 50 and 99 to 100, you'll get an estimated answer of 150.
• Start by rounding 153 to 150 and 33 to 30. After you subtract, you should get an estimated answer of 120.
• Round 56 to 60 and 81 to 80, and then multiply. Your estimated answer will be 4,800.
• After you've rounded 67 to 70 and 42 to 40, you'll multiply to get the estimated answer of 2,800.
• Begin by rounding 78 to 80 and 41 to 40. Then, divide to get an estimated answer of 2.

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In order to access this I need to be confident with:

Here is everything you need to know about estimation for. You’ll learn the definition of estimation and how to use various estimation strategies to make calculations quicker and easier.

Students will first learn about estimation in 3rd grade math as part of their work with operations and algebraic thinking. They will expand upon that knowledge as they progress through elementary math.

What is estimation?

Estimation is when you use approximate values in a calculation to give an approximate answer rather than an exact answer.

Estimation helps to make calculations quicker and easier.

Let’s look at some estimation strategies.

Common Core State Standards

How does this relate to 4th grade math and 5th grade math?

• Grade 3 – Operations and Algebraic Thinking (3.OA.D.8) Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
• Grade 4 – Number and Operations – Base Ten (4.NBT.A.3) Use place value understanding to round multi-digit whole numbers to any place.
• Grade 5 – Number and Operations – Base Ten (5.NBT.A.4) Use place value understanding to round decimals to any place

How to estimate calculations

In order to estimate using compatible numbers.

Round each number so that they are compatible.

Create the new calculation using the compatible numbers.

Do the calculation using the compatible numbers to give an approximate answer, using the ‘approximately equal to’ symbol \textbf{≈}

In order to estimate using rounding.

Round each number in the calculation the nearest whole number, nearest ten, nearest hundred, or nearest thousand.

Create the calculation using the rounded numbers.

Do the calculation using the rounded numbers to give an approximate answer, using the ‘approximately equal to’ symbol \textbf{≈}

In order to estimate a  real world scenario using rounding or compatible numbers.

Read the problem and decide which operation(s) to use.

Create an equation.

Use rounding or compatible numbers to write a new equation.

Do the calculation(s) using the rounded numbers of the compatible numbers to give an approximate answer, using the ‘approximately equal to’ symbol \textbf{≈}

Label the answer.

[FREE] Rounding Numbers Check for Understanding (Grade 3 to 5)

Use this quiz to check your grade 3 to 5 students’ understanding of rounding numbers. 10+ questions with answers covering a range of 3rd, 4th, and 5th grade rounding numbers topics to identify areas of strength and support!

Estimation examples

Example 1: estimate the product using compatible numbers.

Estimate the value of 228 \times 19 by using compatible numbers.

228 is close to 220.

19 is close to 20

2 Create the new calculation using the compatible numbers.

220 \times 20 =

This calculation is much easier to do than 228\times 19.

3 Do the calculation(s) using the rounded numbers of the compatible numbers to give an approximate answer, using the ‘approximately equal to’ symbol \textbf{≈}

220\times20=4400

228\times19=4,332

228\times19\approx4,400

Example 2: estimate quotients using compatible numbers

Estimate the value of 2774\div73

2774 is close to 2800.

73 is close to 70.

2800\div70=40

2774\div73=38

2774\div73\approx40

Example 3: estimate the product of decimal numbers by rounding to the nearest whole number

Estimate the value by rounding to the nearest whole number.

6.9\times 9.6

6.9 rounded to the nearest whole number is 7. \;\; 9 > 5 so round up by adding 1 to the rounded digit. 6 + 1 = 7

9.6 rounded to the nearest whole number is 10. \;\; 6 > 5, so round up by adding 1 to the rounded digit. 9 + 1 = 10

This calculation is much easier to do than 6.9\times9.6

7\times10=70

6.9\times9.6=66.24

6.9\times9.6\approx70

Example 4: estimate the quotient of decimal numbers using rounding

783.12\div2.4

783.12 round to the nearest ten. 3 < 5, so round down by leaving the rounded digit alone. The number will round to 780.

2.4 round to the nearest whole number. 4 < 5, so round down by leaving the rounded digit alone. The number will round to 2.

780\div2=390

783.12\div 2.4 = 326.3

783.12\div2.4\approx390

Example 5: word problem using estimation

Benny has saved \$127.54. Sandy has saved 2.5 times as much money as Benny.

Estimate how much money Sandy has saved.

Multiplication will be used to solve this problem.

127.54\times2.5= \; ?

Rounding strategy will be used.

127.54 round to the nearest ten. 7 > 5, so round up by adding one to the rounding digit. 2 + 1 = 3, so it will round to 130.

2.5 rounded to the nearest whole number is 3. \; 5 = 5, so round up by adding one to the rounded digit.

130\times3=?

130\times3=390

127.54\times 2.5 = 318.85

127.54\times2.5\approx390

Sandy saved approximately \$390.

Example 6: word problem using estimation

Janna has \$21.16. She wants to buy the large candy bars for her party. The large candy bars cost \$1.84. Estimate how many large candy bars she can buy.

Division is the operation that will be used.

21.16\div 1.84=

21.16 round to the nearest ten. 1 < 5, so round down by leaving the rounded digit alone. The number will round to be 20.

1.84 round to the nearest whole number. 8 > 5, so round up by adding 1 to the rounded digit. 1 + 1 = 2. The number will round to be 2.

20\div 2 = 10

21.16\div1.84 = 11.5

21.16\div1.84\approx 10

Janna can buy approximately 10 large candy bars.

Teaching tips for Estimation

• Reinforcing number sense is essential to estimation strategies. Use number lines and manipulatives to help students build a deep understanding of numbers and how they relate to one another.
• Have students check their answers for reasonableness by estimating products, estimating quotients, estimating sums and estimating differences.
• Estimating strategies are used in real-life constantly. Use real-life learning activities in class so that students can see the importance of estimating skills.
• When teaching the strategy of compatible numbers, emphasize to students that they estimated pair of numbers must work well together. Rounding strategies such as rounding to the nearest whole number might not give a pair of compatible numbers.
• Demonstrate to students that the closer the rounded numbers are to the actual number the better estimate they’ll get.
• Although worksheets with estimation math problems help students practice the strategies, having students do real-life problems is more meaningful.

Our favorite mistakes

• It is common to find the actual calculation rather than use estimation Estimation strategies are used to make the calculation quick and easily done mentally. 5.8\times2.3 is not easy to do quickly. So using the estimation strategy of rounding 5.8 to 6 and 2.3 to 2 will make the calculation quick and easy. 6\times 2 = 12. The actual answer is 13.34.
• Students have difficulty understanding estimation or applying an estimation strategy. Students have trouble estimating “how many” because they have weak number sense. Number sense helps you to understand how numbers relate to one another. Reinforcing number lines and visual models helps students to make sense of the numbers.
• Students think they have to use one particular estimation strategy. Estimation strategies exist to help students make calculations quick and easy. There is NOT one right strategy to estimate. Students should use the strategy they are most comfortable with.

Related lessons on rounding numbers

• Rounding numbers
• Rounding decimals
• Place value
• Significant figures

Practice Estimation questions

1. Use compatible numbers to estimate the value of 22\times 19?

22 is close to 20

19 is close to 20 20\times20 = 400 22\times 19\approx 400

The actual answer is 418 which is close to 400.

2. What are the appropriate compatible numbers to estimate the value of 219\div74

Compatible numbers are a pair of numbers that work well together.

In this case, estimating 219 to be 210 and estimating 74 to be 70 will make a pair of compatible numbers. 210\div30 is easy and quick to calculate.

The compatible numbers are 210 and 70.

3. Estimate the value of 7.2 \times 98 using rounding.

7.2 rounded to the nearest whole number is 7. \;\; 2 < 5 , so round down by leaving the rounding digit alone.

98 rounded to the nearest hundred is 100. \;\; 8 > 5, so round up by adding 1 to the rounded digit.

4. Estimate the value of 389\div81 using rounding.

389 round to the nearest hundred. 8 > 5, so round up by adding one to the rounded digit. 3 + 1 = 4, so 400.

81 round to the nearest ten. 1 < 5, so round down by leaving the rounding digit alone.

5. Jenny has saved \$114.65 in her savings account. Brandon has saved 1.5 times the amount of money. Estimate the amount of money Brandon has saved.

You will estimate the value of 114.65\times 1.5.

Round \$114.65 to the nearest ten. 4 < 5 so leave the rounded digit alone, so it will round to be \$110. \; 1.5 round to the nearest whole number. 5 = 5, so round up by adding one to the rounded digit. 1 + 1 = 2.

6. There are 678 boxes in 11 storage spaces. Estimate how many boxes there are in 1 storage space by rounding the values to the nearest ten.

\approx{57} boxes

\approx{68} boxes

\approx{64} boxes

\approx{58} boxes

You will estimate the value of 678\div11.

Round 678 to the nearest ten. 8 > 5, round up by adding 1 to the rounded digit. 7 + 1 = 8, so the rounded number is 680.

Round 11 to the nearest ten. 1 < 5, leave the rounded digit alone, so the rounded number is 10.

Estimation FAQs

Rounding is taking a number and rounding it to a specific place, for example rounding to the nearest tens place, nearest hundreds place, nearest hundredths place, etc.. Compatible numbers are numbers that work well with each other.

There is no right or wrong way to estimate. In many cases it depends on the actual problem.

Round to the place that will make the calculations the easiest to do.

The closer your estimated answer is to the exact answer, the better the estimate.

The next lessons are

• Factors and multiples
• Fractions operations

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When we estimate , we’re making an educated guess . We’re using what we already know to make decisions about our data.

We estimate the answers to Math problems when our answers don’t have to be exact. For example, if you want to download mp3 songs from the Internet, you may need to purchase a gift card and add the points to your account before buying the music. If you want to buy six songs and each song costs between $.99 and $1.29 each, you can estimate the total cost of the songs to predict about how much money you’ll need.

This situation is an example of making an educated guess . You used what you already knew about the cost of the songs to guess the total cost so you could purchase the gift card. Estimating the cost helps you buy a gift card with enough money on it to pay for the downloads without leaving too much extra money in your account.

In order to estimate , we need to know how to round numbers . Below is an explanation of how to round numbers using the Common Method.

When we round numbers, we either increase a number by 1 or leave it the same to make mental math easier. It also helps us to make some numbers easier to understand when we’re comparing sets of numbers.

One important thing to remember is that when we round numbers, our answers will not be as accurate as they would have if we had kept our original numbers.

Here’s how the Common Method of rounding works:

Step 1. Decide what number you want to keep. For example, if we want to know how many books we can buy for $15 or less and we’re comparing the prices of three books whose costs are $5.99, $4.98, and $7.99, we might decide to round to the nearest dollar. In that case, we’ll keep the 5, 4, and 7.

Step 2. Look at the number to the right of the number you decided to keep . If the number is a 5 or above, round the number you’re keeping up by adding 1. If the number is a 4 or less, round the number by keeping it the same.

In our book example, we round the prices to $6.00, $5.00, and $8.00 because the numbers to the right of the numbers we want to keep are above 5. That means we can buy the $6.00 and $5.00 books for under $15.00.

We round numbers when we estimate – just like in our book example above.

We wanted to know which books we could buy for less than $15, but we didn’t need to know the exact total cost. Estimation was close enough to give us the information we needed.

Estimation also helps us check our answers in Math. For example, if we add 1,540 to 575, we can round these two numbers to make checking our work easier. We could round the first number to 1,550 and the second to 580. These numbers are easier to add in our heads. When we add these numbers we get 2,130. If the answer we have is 2,800, we know we’ve gone wrong somewhere and we need to find our mistake.

You can also estimate decimal numbers. Use the two step Common Method of rounding. Decide what place (example: tenths, hundredths, etc.) you want to round your number to and use the number to the right of it and round up by adding 1 or round down by keeping it the same . Here are some examples of rounding to different decimal places (the number we need to round is in blue):

Exercise 1. Use the Common Method to round each of the blue numbers below. Type your answer in the space given.

Answer: 360. The number to the right of the 5 is 6. Since it’s above a 5, we round the 5 up by adding 1. That gives us 360.

Answer: 2,800. The number to the right of the 7 is 5. Since it’s a 5 or above, we round the 7 up by adding 1. That gives us 2,800.

Answer: 500. The number to the right of the 5 is less than 5, so we round down by leaving it the same. That gives us 500.

Exercise 2. Read the situations below. Then, decide whether we can use estimation. Type “yes” in the box if we can estimate. Type “no” in the box if we can’t. Remember, when we estimate, our answers will not be as accurate as if we’d used the exact numbers.

Answer: Yes . We don’t need to know exactly how much flour is left to make our decision. We can subtract 4 cups of flour from the 12 – 1 / 2 we had when we bought the bag and see that there are 8 – ½ cups of flour left. Then, we can multiply 3 by 2 and see that we’ll need 6 cups of flour for the bake sale cakes. Our measurement doesn’t need to be exact. As long as we know we have more than 6 cups of flour left, we know we don’t need to buy anymore.

Answer: No . The auditorium manager needs to know exactly how many “reserved” tickets are left for the report. If estimation is used, the board of directors may advertise that there are 100 “reserved” tickets when there are only 50! Customers may get angry if they learn that there aren’t enough tickets for their group.

• Probability math
• The population of endangered species is changing at a rate of P′(t) = 30 − 20t.
• 3x - 7y = 4 and -6x + y = 5

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Course: 3rd grade   >   Unit 8

2-step estimation word problems.

• (Choice A)   5 ‍   bags A 5 ‍   bags
• (Choice B)   16 ‍   bags B 16 ‍   bags
• (Choice C)   24 ‍   bags C 24 ‍   bags
• (Choice D)   35 ‍   bags D 35 ‍   bags

Estimation in Maths

Have you ever observed the headlines of any news channels closely?

Let's look at a couple of examples. 

• The US drew with Canada in a football match watched by 51,000 spectators in the stadium and 40 million TV viewers worldwide.
• Approximately, 2000 people were killed and more than 50,000 injured in a cyclonic storm in the coastal areas.
• Over 13 million passengers are carried over the 63,000-kilometer route of railways every day.

Can we firmly say that the numbers quoted in the news are the exact match to the number of people in these scenarios? 

No! We know that these are not the exact numbers.

The word "approximately" indicates that the number (of people) was close to these given numbers.

Clearly, 51,000 could be 50,800 or 51,300, but definitely not 70,000

In the same way, 13 million passengers could indicate more than 12 million but less than 14 million people, but definitely not 20 million.

The quantities given in the examples above are not exact counts but are estimated to give an idea of the quantity.

This indicates that we round numbers to make them easier to work with or to express a number to an acceptable level of accuracy. 

This natural and quick process of assessing data is called estimation.

Let's learn how to estimate and solve problems involving estimation.

Table of Contents  

What does estimation mean.

Estimation of a number is a reasonable guess of the actual value to make calculations easier and realistic.

Estimation means approximating a quantity to the required accuracy.

This is obtained by rounding off the numbers involved in the calculation and getting a quick and rough answer. 

How Do You Calculate an Estimate?

We subconsciously involve estimation in all our day-to-day activities.

We typically estimate numbers to a certain level of accuracy.

Numbers are rounded to a particular place value to make them as close to a whole number as possible.

The General Rule of Estimation

In all cases of rounding, we need to do the following:

• Find the digit in the place we want to round to.
• If the digit to the right is 0-4 i.e., 0, 1, 2, 3, 4: we leave the digit alone (round down).
• If the digit to the right is 5-9 i.e., 5, 6, 7, 8, 9: we increase the digit by 1 (round up).
• Finally, we remove any extra digits that have not been rounded off. 

Estimating to Different Place Values

Let's assume that you want to estimate the total cost of several different items in your grocery list or that there are a few gift items you want to buy.

How would you roughly calculate the amount of money you'd need?

You would round the prices to the nearest tens place because it involves money and it becomes easier to calculate when you round off in this way.

If you are working with extremely large numbers, you might want to round the number to the nearest one million or one billion.

Did we use the same rounding off techniques in all these situations?

There are no hard and fast rules when you want to estimate the outcomes of numbers.

The procedure varies with the degree of accuracy required as well as how quickly we need the estimate.

Example :  We need to estimate \(854917.632\)

(a) to the nearest whole number.

(b) to the nearest thousand.

(c) to the nearest tenth.

(a) To estimate \(854917.632\) to the nearest whole number, we need to check the number in the tenth place. 

The digit in the tenths place is 6, so we need to round up.

Hence, we increase the digit in the units place by 1

Then we get the number 854917 + 1 = 854918 

\(\therefore\) The estimation of \(854917.632\) is \(854918\)

(b) To estimate \(854917.632\) to the nearest thousand, we need to check the number in the hundreds place.

4 is in the thousands place. The digit to its right is 9

So, we round up 4 i.e., increase the digit in the thousands place by 1

Then we get the number 854000 + 1000 = 855000 

(c) \(854917.632\) to the nearest tenth: we need to check the number in the hundredths place.

The digit in the hundredths place is 3

So, we round down 6 i.e., we leave the digit alone.

Estimating Counts, Lengths, and More.

Let us see some real-life scenarios involving estimation, such as:

• counting numbers
• measuring the length, breadth, or other dimensions of a place or figure
• totaling the amount in a bill etc.

Solved Examples

Tia and John collect marbles.

Tia has collected 335 and John has collected 442 marbles.

Choose a better estimate to find the number of marbles they own together. 

Here, a better estimate can be obtained by rounding off the digits to the nearest tens.

To get an estimate of the sum, round off all the numbers to the same place value, then add them.

Number of marbles Tia has collected = 335 rounded off to 340

Number of marbles John has collected = 442 rounded off to 440

\(\therefore\) Total number of marbles = 340 + 440 = 780

A train terminal has 12 platforms.

Each platform has about 115 trains departing each day.

Find an estimate of the number of trains leaving the terminal every day.

This is an estimation problem involving multiplication.

The rule is to estimate all the numbers that we need to multiply to the nearest round number. 

The total number of trains = Number of platforms \(\times\) Number of trains leaving each platform

 \begin{eqnarray}\text{Total number of trains} &=& 12 \times 115\\ \text{Estimated value}&=&10 \times120\end{eqnarray}

In a town, there are 1,759 families.

An average of two children from each family will attend a welfare program organized at various schools.

Each school can accommodate 320 children.

Calculate the minimum number of school avenues that are needed in the region.

The total number of families = 1759 rounded off to 1800

The capacity of each school = 320 rounded off to 300

The final estimated answer =  \( 1800\div 300 = 6\)

Ron is planning a major yard renovation.

He wants to reseed his entire lawn. 

Here is an overhead view of his yard.

Help Ron find the estimate of the area of his yard.

The dimensions of the yard are 12.3 ft and 14.5 ft.

The rounded off dimensions will be 12 ft and 15 ft.

The yard is in the form of a rectangle.

\begin{eqnarray}           \text{Area of rectangle} &=&\text {Length} \times\text {Breadth}\\ \text{Area of rectangle} &=& 12\hspace{0.2cm}\text {ft} \times 15\hspace{0.2cm}\text {ft}\\ &=&180\hspace{0.25cm} \text {sq.ft}     \end{eqnarray}

Estimate the value of the following sum by first rounding each value to the nearest hundredth:

123.616283 + 5.123 + 6.183478

Let us first estimate each number and then we will add the numbers to get the required estimate.  

Estimate the area and perimeter of a circle with a radius of 4.25 cm.

When to Use Estimates?

Imagine you organize a carnival in your town.

The first thing you do is find out roughly how many guests will visit the carnival.

Can you calculate the exact number of visitors attending the carnival? It is practically impossible.

Let us consider another situation.

The finance minister of a certain country presents an annual budget.

The minister allocates a certain amount under the head "Education."

Can the amount allocated be absolutely accurate?

It can only be a reasonably good estimate of the expenditure the country needs for education during that year.

In the above-mentioned scenarios, a rough estimate will help us plan things properly.

Here we do not need exact figures.

In all these kinds of situations where the round-off value is as good as the exact value, we use estimates. 

Why Do We Estimate in Math?

Often in math, we focus on getting an accurate answer.

But in some places, the estimation gives a quick and almost realistic answer.

Estimation saves time and we can achieve a better picture of things in a simpler way.

Hence, to get a quick and reliable answer, we estimate in math.

How to Estimate a Number?

Here is the summary on how to estimate a number in general:

• Estimation is done when an almost realistic value is enough to solve instead of the exact answer.

Rules to be followed while rounding off :

If the digit to the right of the place value you are rounding off is 0-4 i.e., 0,1,2,3,4: we leave the digit alone (round down).

If the digit to the right of the place value you are rounding off is 5-9 i.e., 5,6,7,8,9: we increase the digit by 1 (round up).

To estimate a sum, round off all the numbers to the same place value, finally add them.

The crude rule to round off a product is to estimate all factors to the nearest round number.

Interactive Questions

Let's summarize.

This mini-lesson targeted the fascinating concept of estimation. The math journey around estimation starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

About Cuemath

At  Cuemath , our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

Frequently Asked Questions (FAQs)

1. what is the difference between exact and estimate.

The estimate is the nearest rounded off value to the exact value.

2. Why do we estimate in math?

3. what is the advantage of estimation.

Estimation helps save time and money before investing in real.

By estimating, we can benefit as we can save money and time on huge projects.

Estimation helps in doing a reality check of any project.

4. How do you estimate sums and differences?

To get an estimate of the sum, round off all the numbers to the same place value, then add them.

To estimate the difference, round off all the numbers to the same place value, then subtract them accordingly.

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6.5: Estimating Population Proportion

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SECTION OBJECTIVES

By the end of this chapter, the student should be able to:

• Calculate and interpret confidence intervals for estimating a population proportion.
• Calculate the sample size required to estimate a population proportion given a desired confidence level and margin of error.

During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).

Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.

The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? There is no mention of a mean or average!

To form a proportion, take \(X\), the random variable for the number of successes and divide it by \(n\), the number of trials (or the sample size). The random variable \(\hat{P} \) (read "P hat") is that proportion,

\[\hat{P} = \dfrac{X}{n}\nonumber \]

When \(n\) is large and \(p\) is not close to zero or one, we can use the normal distribution to approximate the number of successes.

\[X \sim N(np, \sqrt{npq})\nonumber \]

If we divide the random variable, the mean, and the standard deviation by \(n\), we get a normal distribution of proportions with \(P′ \), called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by \(n\).)

\[\dfrac{X}{n} = \hat{P} \sim N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber \]

Using algebra to simplify:

\[\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber \]

\(\hat{P}\) follows a normal distribution for proportions:

Calculating the Error Bound

The error bound (EBP) for a proportion is

\[EBP = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{\hat{p}\hat{q}}{n}}\right)\nonumber \]

where \(\hat{q}\ = 1 - \hat{p}\).

This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is \(\dfrac{\sigma}{\sqrt{n}}\). For a proportion, the appropriate standard deviation is

\[\sqrt{\dfrac{pq}{n}}.\nonumber \]

However, in the error bound formula, we use

\[\sqrt{\dfrac{\hat{p}\hat{q}}{n}}\nonumber \]

as the standard deviation, instead of

In the error bound formula, the sample proportions \(\hat{p}\) and \(\hat{q}\) are estimates of the unknown population proportions p and q . The estimated proportions \(\hat{p}\) and \(\hat{q}\) are used because \(p\) and \(q\) are not known. The sample proportions \(\hat{p}\) and \(\hat{q}\) are calculated from the data: \(\hat{p}\) is the estimated proportion of successes, and \(\hat{q}\) is the estimated proportion of failures.

Which Calculator to Use?

There are many online calculators that can be used to compute the Margin of Error. For example, you can use this one :

Also, you are encouraged to ask your instructor about which calculator is allowed/recommended for this course.

Example \(\PageIndex{1}\)

Use the calculator  provided above to verify the following statements:

When \(\alpha=0.1, n=200, \hat{p}=0.43\) the EBP is \(0.0577\)

When \(\alpha=0.05, n=100, \hat{p}=0.81\) the EBP is \(0.0768\)

When \(\alpha=0.01, n=250, \hat{p}=0.57\) the EBP is \(0.0806\)

Exercise \(\PageIndex{1}\)

Find EBP when \(\alpha=0.07, n=168, \hat{p}=0.82\).

Interactive Exercise \(\PageIndex{1}\)

Constructing the confidence interval.

The confidence interval has the form

\[(\hat{p} – EBP, \hat{p} + EBP).\nonumber \]

• \(EBP\) is error bound for the proportion.
• \(\hat{p} = \dfrac{x}{n}\)
• \(\hat{p} =\) the estimated proportion of successes (\hat{p} is a point estimate for p , the true proportion.)
• \(x =\) the number of successes
• \(n =\) the size of the sample

The confidence interval can be used only if the number of successes \(n\hat{p}\) and the number of failures \(n\hat{q}\) are both greater than five.

The graph gives a picture of the entire situation.

\[CL + \dfrac{\alpha}{2} + \dfrac{\alpha}{2} = CL + \alpha = 1.\nonumber \]

There are many online calculators that can be used to compute the confidence intervals. For example, you can use this one :

Example \(\PageIndex{2}\)

Use the calculator  provided above to verify the following:

Confidence Level (%): \(95\)

Sample Size: \(197\)

Number of Successes: \(61\)

95% Confidence Interval: \((0.2450, 0.3742)\)

Exercise \(\PageIndex{2}\)

Find a 90% confidence interval when the sample size is 250 and the number of successes is 85.

(0.2907, 0.3893)

Interactive Exercise \(\PageIndex{2}\)

Writing the interpretation.

The interpretation should clearly state the confidence level (\(CL\)), explain what population parameter is being estimated (here, a population proportion ), and state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population proportoin (include the context of the problem) is between ___ and ___ ."

Example \(\PageIndex{3}\)

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.

To calculate the confidence interval, you must find \(\hat{p}\), \(\hat{q}\), and \(EBP\).

• \(n = 500\)
• \(x =\) the number of successes \(= 421\)

\[\hat{p} = \dfrac{x}{n} = \dfrac{421}{500} = 0.842\nonumber \]

• \(\hat{p} = 0.842\) is the sample proportion; this is the point estimate of the population proportion.

\[\hat{q} = 1 – \hat{p} = 1 – 0.842 = 0.158\nonumber \]

Since \(CL = 0.95\), then

\[\alpha = 1 – CL = 1 – 0.95 = 0.05\left(\dfrac{\alpha}{2}\right) = 0.025.\nonumber \]

\[z_{\dfrac{\alpha}{2}} = z_{0.025 = 1.96}\nonumber \]

Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find \(z_{0.025}\). Remember that the area to the right of \(z_{0.025}\) is \(0.025\) and the area to the left of \(z_{0.025}\) is \(0.975\). This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.

\[EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{\hat{p}\hat{q}}{n}} = (1.96)\sqrt{\dfrac{(0.842)(0.158)}{500}} = 0.032\nonumber \]

\[\hat{p} – EBP = 0.842 – 0.032 = 0.81\nonumber \]

\[\hat{p} + EBP = 0.842 + 0.032 = 0.874\nonumber \]

The confidence interval for the true binomial population proportion is \((\hat{p} – EBP, \hat{p} +EBP) = (0.810, 0.874)\).

Interpretation:  We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.

Explanation of 95% Confidence Level:  Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.

Exercise \(\PageIndex{3}\)

Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.

(0.3315, 0.4525)

Interactive Exercise \(\PageIndex{3.1}\)

Interactive exercise \(\pageindex{3.2}\), example \(\pageindex{4}\).

For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.

\(x = 300\) and \(n = 500\)

\[\hat{p} = \dfrac{x}{n} = \dfrac{300}{500} = 0.600\nonumber \]

\[\hat{q} = 1 − \hat{p} = 1 − 0.600 = 0.400\nonumber \]

Since \(CL = 0.90\), then

\[\alpha = 1 – CL = 1 – 0.90 = 0.10\left(\dfrac{\alpha}{2}\right) = 0.05\]

\[z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber \]

Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find \(z_{0.05}\). Remember that the area to the right of \(z_{0.05}\) is 0.05 and the area to the left of \(z_{0.05}\) is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.

\[EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{\hat{p}\hat{q}}{n}} = (1.645)\sqrt{\dfrac{(0.60)(0.40)}{500}} = 0.036\nonumber \]

\[\hat{p} – EBP = 0.60 − 0.036 = 0.564\nonumber \]

\[\hat{p} + EBP = 0.60 + 0.036 = 0.636\nonumber \]

The confidence interval for the true binomial population proportion is \((\hat{p} – EBP, \hat{p} +EBP) = (0.564,0.636)\).

Interpretation

• We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
• Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.

Explanation of 90% Confidence Level:  Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.

Exercise \(\PageIndex{4}\)

A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.

• Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
• In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.

(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%

Interactive Exercise \(\PageIndex{4.1}\)

Interactive exercise \(\pageindex{4.2}\), example \(\pageindex{5}\).

A random sample of 29 statistics students was asked: “Have you smoked a cigarette in the past week?” Eight students reported smoking within the past week. Find a 95% confidence interval for the true proportion of statistics students who smoke.

Eight students out of 29 reported smoking within the past week, so \(x = 8\) and \(n = 29\).

\[\hat{p} = \dfrac{x}{n} = \dfrac{8}{29} \approx 0.276\nonumber \]

\[\hat{q} = 1 – \hat{p} = 1 – 0.276 = 0.724\nonumber \]

Since \(CL = 0.95\), we know \(\alpha = 1 – 0.95 = 0.05\) and \(\dfrac{\alpha}{2} = 0.025\).

\[z_{0.025} = 1.96\nonumber \]

\(EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{\hat{p}\hat{q}}{n}} = (1.96)\sqrt{\dfrac{0.276(0.724)}{29}} \approx 0.163\)

\[\hat{p} – EPB = 0.276 – 0.163 = 0.113\nonumber \]

\[\hat{p} + EPB = 0.276 + 0.163 = 0.439\nonumber \]

We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.

Exercise \(\PageIndex{5}\)

Out of a random sample of 69 freshmen at State University, 33 students have declared a major. Find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.

We have \(x = 33\) and \(n = 69\).

\[\hat{p} = 33/69 \approx 0.478\nonumber \]

\[\hat{q} = 1 – \hat{p} = 1 – 0.478 = 0.522\nonumber \]

Since \(CL = 0.96\), we know \(\alpha = 1 – 0.96 = 0.04\) and \(\dfrac{\alpha}{2} = 0.02\).

\[z_{0.02} = 2.054\nonumber \]

\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (2.054)\left(\sqrt{\dfrac{(0.478)(0.522)}{69}}\right) - 0.124\nonumber \]

\[\hat{p} – EPB = 0.478 – 0.124 = 0.354\nonumber \]

\[\hat{p} + EPB = 0.478 + 0.124 = 0.602\nonumber \]

We are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.

Interactive Exercise \(\PageIndex{5.1}\)

Interactive exercise \(\pageindex{5.2}\), example \(\pageindex{6}\).

The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.

Using “plus-four,” we have \(x = 13 + 2 = 15\) and \(n = 50 + 4 = 54\).

\[p′ = 1554 \approx 0.278\nonumber \]

\[q′ = 1 – p′ = 1 − 0.241 = 0.722\nonumber \]

Since \(CL = 0.90\), we know \(\alpha = 1 – 0.90 = 0.10\) and \(\dfrac{\alpha}{2} = 0.05\).

\[z_{0.05} = 1.645\nonumber \]

\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.278)(0.722)}{54}}\right) \approx 0.100\nonumber \]

\[p′ – EPB = 0.278 – 0.100 = 0.178\nonumber \]

\[p′ + EPB = 0.278 + 0.100 = 0.378\nonumber \]

We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER. Arrow down to \(x\) and enter 15. Arrow down to \(n\) and enter 54. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER.

The confidence interval is (0.178, 0.378).

Calculating the Sample Size \(n\)

If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is

Solving for \(n\) gives you an equation for the sample size.

\[n = \dfrac{\left(z_{\frac{\alpha}{2}}\right)^{2}(\hat{p}\hat{q})}{EBP^{2}}\nonumber \]

Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.

From the problem, we know that \(\bf{EBP = 0.03}\) (3%=0.03) and \(z_{\dfrac{\alpha}{2}} z_{0.05} = 1.645\) because the confidence level is 90%.

However, in order to find \(n\), we need to know the estimated (sample) proportion \(\hat{p}\). Remember that \(\hat{q} = 1 – \hat{p}\). But, we do not know \(\hat{p}\) yet. Since we multiply \(\hat{p}\) and \(\hat{q}\) together, we make them both equal to 0.5 because \(\hat{p}\hat{q} = (0.5)(0.5) = 0.25\) results in the largest possible product. (Try other products: \((0.6)(0.4) = 0.24\); \((0.3)(0.7) = 0.21\); \((0.2)(0.8) = 0.16\) and so on). The largest possible product gives us the largest \(n\). This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size \(n\), use the formula and make the substitutions.

\[n = \dfrac{z^{2}\hat{p}\hat{q}}{EBP^{2}}\nonumber \]

\[n = \dfrac{1.645^{2}(0.5)(0.5)}{0.03^{2}} = 751.7\nonumber \]

Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.

Exercise \(\PageIndex{6}\)

Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?

271 customers should be surveyed. Check the Real Estate section in your local

Interactive Exercise \(\PageIndex{6}\)

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Watch CBS News

Teens come up with trigonometry proof for Pythagorean Theorem, a problem that stumped math world for centuries

By Bill Whitaker

May 5, 2024 / 7:00 PM EDT / CBS News

As the school year ends, many students will be only too happy to see math classes in their rearview mirrors. It may seem to some of us non-mathematicians that geometry and trigonometry were created by the Greeks as a form of torture, so imagine our amazement when we heard two high school seniors had proved a mathematical puzzle that was thought to be impossible for 2,000 years. 

We met Calcea Johnson and Ne'Kiya Jackson at their all-girls Catholic high school in New Orleans. We expected to find two mathematical prodigies.

Instead, we found at St. Mary's Academy , all students are told their possibilities are boundless.

Come Mardi Gras season, New Orleans is alive with colorful parades, replete with floats, and beads, and high school marching bands.

In a city where uniqueness is celebrated, St. Mary's stands out – with young African American women playing trombones and tubas, twirling batons and dancing - doing it all, which defines St. Mary's, students told us.

Junior Christina Blazio says the school instills in them they have the ability to accomplish anything. 

Christina Blazio: That is kinda a standard here. So we aim very high - like, our aim is excellence for all students. 

The private Catholic elementary and high school sits behind the Sisters of the Holy Family Convent in New Orleans East. The academy was started by an African American nun for young Black women just after the Civil War. The church still supports the school with the help of alumni.

In December 2022, seniors Ne'Kiya Jackson and Calcea Johnson were working on a school-wide math contest that came with a cash prize.

Ne'Kiya Jackson: I was motivated because there was a monetary incentive.

Calcea Johnson: 'Cause I was like, "$500 is a lot of money. So I-- I would like to at least try."

Both were staring down the thorny bonus question.

Bill Whitaker: So tell me, what was this bonus question?

Calcea Johnson: It was to create a new proof of the Pythagorean Theorem. And it kind of gave you a few guidelines on how would you start a proof.

The seniors were familiar with the Pythagorean Theorem, a fundamental principle of geometry. You may remember it from high school: a² + b² = c². In plain English, when you know the length of two sides of a right triangle, you can figure out the length of the third.

Both had studied geometry and some trigonometry, and both told us math was not easy. What no one told  them  was there had been more than 300 documented proofs of the Pythagorean Theorem using algebra and geometry, but for 2,000 years a proof using trigonometry was thought to be impossible, … and that was the bonus question facing them.

Bill Whitaker: When you looked at the question did you think, "Boy, this is hard"?

Ne'Kiya Jackson: Yeah. 

Bill Whitaker: What motivated you to say, "Well, I'm going to try this"?

Calcea Johnson: I think I was like, "I started something. I need to finish it." 

Bill Whitaker: So you just kept on going.

Calcea Johnson: Yeah.

For two months that winter, they spent almost all their free time working on the proof.

CeCe Johnson: She was like, "Mom, this is a little bit too much."

CeCe and Cal Johnson are Calcea's parents.

CeCe Johnson:   So then I started looking at what she really was doing. And it was pages and pages and pages of, like, over 20 or 30 pages for this one problem.

Cal Johnson: Yeah, the garbage can was full of papers, which she would, you know, work out the problems and-- if that didn't work she would ball it up, throw it in the trash. 

Bill Whitaker: Did you look at the problem? 

Neliska Jackson is Ne'Kiya's mother.

Neliska Jackson: Personally I did not. 'Cause most of the time I don't understand what she's doing (laughter).

Michelle Blouin Williams: What if we did this, what if I write this? Does this help? ax² plus ….

Their math teacher, Michelle Blouin Williams, initiated the math contest.

Bill Whitaker: And did you think anyone would solve it?

Michelle Blouin Williams: Well, I wasn't necessarily looking for a solve. So, no, I didn't—

Bill Whitaker: What were you looking for?

Michelle Blouin Williams: I was just looking for some ingenuity, you know—

Calcea and Ne'Kiya delivered on that! They tried to explain their groundbreaking work to 60 Minutes. Calcea's proof is appropriately titled the Waffle Cone.

Calcea Johnson: So to start the proof, we start with just a regular right triangle where the angle in the corner is 90°. And the two angles are alpha and beta.

Bill Whitaker: Uh-huh

Calcea Johnson: So then what we do next is we draw a second congruent, which means they're equal in size. But then we start creating similar but smaller right triangles going in a pattern like this. And then it continues for infinity. And eventually it creates this larger waffle cone shape.

Calcea Johnson: Am I going a little too—

Bill Whitaker: You've been beyond me since the beginning. (laughter) 

Bill Whitaker: So how did you figure out the proof?

Ne'Kiya Jackson: Okay. So you have a right triangle, 90° angle, alpha and beta.

Bill Whitaker: Then what did you do?

Ne'Kiya Jackson: Okay, I have a right triangle inside of the circle. And I have a perpendicular bisector at OP to divide the triangle to make that small right triangle. And that's basically what I used for the proof. That's the proof.

Bill Whitaker: That's what I call amazing.

Ne'Kiya Jackson: Well, thank you.

There had been one other documented proof of the theorem using trigonometry by mathematician Jason Zimba in 2009 – one in 2,000 years. Now it seems Ne'Kiya and Calcea have joined perhaps the most exclusive club in mathematics. 

Bill Whitaker: So you both independently came up with proof that only used trigonometry.

Ne'Kiya Jackson: Yes.

Bill Whitaker: So are you math geniuses?

Calcea Johnson: I think that's a stretch. 

Bill Whitaker: If not genius, you're really smart at math.

Ne'Kiya Jackson: Not at all. (laugh) 

To document Calcea and Ne'Kiya's work, math teachers at St. Mary's submitted their proofs to an American Mathematical Society conference in Atlanta in March 2023.

Ne'Kiya Jackson: Well, our teacher approached us and was like, "Hey, you might be able to actually present this," I was like, "Are you joking?" But she wasn't. So we went. I got up there. We presented and it went well, and it blew up.

Bill Whitaker: It blew up.

Calcea Johnson: Yeah. 

Ne'Kiya Jackson: It blew up.

Bill Whitaker: Yeah. What was the blowup like?

Calcea Johnson: Insane, unexpected, crazy, honestly.

It took millenia to prove, but just a minute for word of their accomplishment to go around the world. They got a write-up in South Korea and a shout-out from former first lady Michelle Obama, a commendation from the governor and keys to the city of New Orleans. 

Bill Whitaker: Why do you think so many people found what you did to be so impressive?

Ne'Kiya Jackson: Probably because we're African American, one. And we're also women. So I think-- oh, and our age. Of course our ages probably played a big part.

Bill Whitaker: So you think people were surprised that young African American women, could do such a thing?

Calcea Johnson: Yeah, definitely.

Ne'Kiya Jackson: I'd like to actually be celebrated for what it is. Like, it's a great mathematical achievement.

Achievement, that's a word you hear often around St. Mary's academy. Calcea and Ne'Kiya follow a long line of barrier-breaking graduates. 

The late queen of Creole cooking, Leah Chase , was an alum. so was the first African-American female New Orleans police chief, Michelle Woodfork …

And judge for the Fifth Circuit Court of Appeals, Dana Douglas. Math teacher Michelle Blouin Williams told us Calcea and Ne'Kiya are typical St. Mary's students.  

Bill Whitaker: They're not unicorns.

Michelle Blouin Williams: Oh, no no. If they are unicorns, then every single lady that has matriculated through this school is a beautiful, Black unicorn.

Pamela Rogers: You're good?

Pamela Rogers, St. Mary's president and interim principal, told us the students hear that message from the moment they walk in the door.

Pamela Rogers: We believe all students can succeed, all students can learn. It does not matter the environment that you live in. 

Bill Whitaker: So when word went out that two of your students had solved this almost impossible math problem, were they universally applauded?

Pamela Rogers: In this community, they were greatly applauded. Across the country, there were many naysayers.

Bill Whitaker: What were they saying?

Pamela Rogers: They were saying, "Oh, they could not have done it. African Americans don't have the brains to do it." Of course, we sheltered our girls from that. But we absolutely did not expect it to come in the volume that it came.  

Bill Whitaker: And after such a wonderful achievement.

Pamela Rogers: People-- have a vision of who can be successful. And-- to some people, it is not always an African American female. And to us, it's always an African American female.

Gloria Ladson-Billings: What we know is when teachers lay out some expectations that say, "You can do this," kids will work as hard as they can to do it.

Gloria Ladson-Billings, professor emeritus at the University of Wisconsin, has studied how best to teach African American students. She told us an encouraging teacher can change a life.

Bill Whitaker: And what's the difference, say, between having a teacher like that and a whole school dedicated to the excellence of these students?

Gloria Ladson-Billings: So a whole school is almost like being in Heaven. 

Bill Whitaker: What do you mean by that?

Gloria Ladson-Billings: Many of our young people have their ceilings lowered, that somewhere around fourth or fifth grade, their thoughts are, "I'm not going to be anything special." What I think is probably happening at St. Mary's is young women come in as, perhaps, ninth graders and are told, "Here's what we expect to happen. And here's how we're going to help you get there."

At St. Mary's, half the students get scholarships, subsidized by fundraising to defray the $8,000 a year tuition. Here, there's no test to get in, but expectations are high and rules are strict: no cellphones, modest skirts, hair must be its natural color.

Students Rayah Siddiq, Summer Forde, Carissa Washington, Tatum Williams and Christina Blazio told us they appreciate the rules and rigor.

Rayah Siddiq: Especially the standards that they set for us. They're very high. And I don't think that's ever going to change.

Bill Whitaker: So is there a heart, a philosophy, an essence to St. Mary's?

Summer Forde: The sisterhood—

Carissa Washington: Sisterhood.

Tatum Williams: Sisterhood.

Bill Whitaker: The sisterhood?

Voices: Yes.

Bill Whitaker: And you don't mean the nuns. You mean-- (laughter)

Christina Blazio: I mean, yeah. The community—

Bill Whitaker: So when you're here, there's just no question that you're going to go on to college.

Rayah Siddiq: College is all they talk about. (laughter) 

Pamela Rogers: … and Arizona State University (Cheering)

Principal Rogers announces to her 615 students the colleges where every senior has been accepted.

Bill Whitaker: So for 17 years, you've had a 100% graduation rate—

Pamela Rogers: Yes.

Bill Whitaker: --and a 100% college acceptance rate?

Pamela Rogers: That's correct.

Last year when Ne'Kiya and Calcea graduated, all their classmates went to college and got scholarships. Ne'Kiya got a full ride to the pharmacy school at Xavier University in New Orleans. Calcea, the class valedictorian, is studying environmental engineering at Louisiana State University.

Bill Whitaker: So wait a minute. Neither one of you is going to pursue a career in math?

Both: No. (laugh)

Calcea Johnson: I may take up a minor in math. But I don't want that to be my job job.

Ne'Kiya Jackson: Yeah. People might expect too much out of me if (laugh) I become a mathematician. (laugh)

But math is not completely in their rear-view mirrors. This spring they submitted their high school proofs for final peer review and publication … and are still working on further proofs of the Pythagorean Theorem. Since their first two …

Calcea Johnson: We found five. And then we found a general format that could potentially produce at least five additional proofs.

Bill Whitaker: And you're not math geniuses?

Bill Whitaker: I'm not buying it. (laughs)

Produced by Sara Kuzmarov. Associate producer, Mariah B. Campbell. Edited by Daniel J. Glucksman.

Bill Whitaker is an award-winning journalist and 60 Minutes correspondent who has covered major news stories, domestically and across the globe, for more than four decades with CBS News.

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A rafah evacuation would solve the problem biden pretends to care about.

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There’s a clear and obvious answer to the humanitarian worries President Biden is trying to blackmail Israel with as a full-scale Rafah invasion looms: Simply let Israel fully evacuate civilians from the Palestinian city and leave behind Hamas and its battalions. 

So why on earth isn’t the president demanding that of Hamas leadership, instead of threatening to cut off arms to Israel?

Yes, there are logistical problems. 

If Israel were to oversee such an evacuation, Hamas might well engage in one of its ugly trademarks: firing on and attempting to kill or otherwise interfere with those trying to leave. 

We saw the monstrous philosophy in action this week , with Hamas firing on the Kerem Shalom crossing once it was opened to aid flows. 

Then there’s Egypt’s refusal to take any Palestinian refugees . 

But if Biden were not busily courting the pro-Hamas vote , he could well crack the whip over Cairo on that issue. 

Don’t forget that the line between Hamas and the Gaza civilian population is blurry; like most terrorist groups, it hides among civilians and counts some civilians as semi-members and supporters. 

As for the canard that a mass evacuation of Rafah would lead to permanent displacement of its citizens at the hands of Israel, it’s transparent nonsense. 

It is in Israel’s best and immediate interest to have a settled population of Palestinian civilians free from Hamas; why would its government stand in the way of that happening?

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At least 110,000 civilians have already fled since Monday, per UN estimates, which is around 10% of the estimated population of refugees. 

It’s time to let the rest go as well. 

The usual global suspects are, of course, screeching that even the merest thought of evacuating Rafah amounts to a war crime on Israel’s part. 

And this is, of course, exactly backward. 

The danger to Palestinian civilians here comes from Hamas’ tactical playbook, which sees them as no more than bullet shields. 

The situation in Rafah points out, yet again, that the world’s only Jewish state is held to a different and much higher standard when it comes to warfighting. 

One that, in this case, will cause unthinkable suffering among the very people Weepy Joe claims to be so concerned about. 

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