All categories

3 years ago

a circle with radius of 9cm sits inside a circle with radius of 11cm. what is the area of the shaded region?

Mathematics

1 answer:

tatyana61 [14]3 years ago

3 0

Answer:

125.6 cm²

Step-by-step explanation:

Area inside the bigger circle but outside the smaller circle:

3.14 × (11² - 9²)

3.14 × 40

125.6 cm²

You might be interested in

ONLY ANSWER IF YOU KNOW THE ANSWER OR IF YOU ARE NIHHA :)) thanks

pishuonlain [190]

Answer:

Step-by-step explanation:

Let's just call the number x for simplicity.

So, 7x is 8 less than x².

Putting this into an equation would look like this

x² - 8 = 7x

It looks like we'll have to factor this to solve. Before we do that we need to move the 7x to the left side so that everything is together.

x² - 7x -8 = 0

Now, we can proceed. To factor we first need to find the factors of -8.

The factors of -8 are

-2 ⋅ 4, -4 ⋅ 2, -1 ⋅ 8, 1 ⋅ -8.

We need to find the pair of factors that adds up to -7. The only ones that do are -1 and 8.

So now that we have these we can create a pair of binomials using them. This will give us the factored form of this equation.

( x + 1 ) ( x - 8 )

To find the solutions we will have to set them to 0 and solve each of these binomials individually.

x - 1 = 0

x = 1

So, one of the solutions is 1. It's not the one we want, since it's positive.

x - 8 = 0

x = 8

This is the one we want since it is positive.

8 0

3 years ago

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.5 parts

DochEvi [55]

Answer:

We accept H₀ with the information we have, we can say level of ozone is under the major limit

Step-by-step explanation:

Normal Distribution

population mean = μ₀ = 7.5 ppm

Sample size n = 16 df = n - 1 df = 15

Sample mean = μ = 7.8 ppm

Sample standard deviation = s = 0.8

We want to find out if ozono level, is above normal level that is bigger than 7.5

1.- Hypothesis Test

null hypothesis H₀ μ₀ = 7.5

alternative hypothesis Hₐ μ₀ > 7.5

2.-Significance level α = 0.01 we will develop one tail-test (right)

then for df = 15 and α = 0,01 from t -student table we get

t(c) = 2.624

3.-Compute t(s)

t(s) = ( μ - μ₀ ) / s /√n ⇒ t(s) = ( 7.8 - 7.5 )*4/0.8

t(s) = 0.3*4/0.8

t(s) = 1.5

4.-Compare t(s) and t(c)

t(s) < t(c) 1.5 < 2.64

Then t(s) is inside the acceptance region. We accept H₀

7 0

3 years ago

Match each slope and y-intercept with the appropriate linear equation.

Bumek [7]

Answer:

m= 4, b = 3 matches with y = 4x +3

m=5, b = -2 matches with y = 5x - 2

m = 3, b = 4 matches with y = 3x + 4

m = -2, b=5 matches with y = -2x + 5

Step-by-step explanation:

Slope-intercept form is <em>y = mx + b</em><em> </em>in which m = slope and b = y-intercept

7 0

3 years ago

Read 2 more answers

CLICK ME!!!!!!!!!!! I need 8 measurements of your height and your arm span in inches. Please help, I need this done soon.

Anastasy [175]

Answer:

my arm is 21.8 inches and my height is 96.3 inches

Step-by-step explanation:

6 0

2 years ago

Please help!

quester [9]

Answer:

The third one

Step-by-step explanation:

Median - Middle number in terms of value

Mode - outcome that appears the most

Mean - average ( add the numbers together and divide that by the amount of numbers )

Data without Kevins score

13, 21, 28, 9, 31

First let's find the mean

To find the mean we simply add the numbers up and divide that by the total amount of numbers ( there are 5 numbers )

13 + 21 + 28 + 9 + 31 = 102

102/5=20.4

So the mean without Kevins score is 20.4

Next let's find the mode

The mode is the number that appears the most

Looking at the data no number appears more than once so there is no mode

Finally let's find the median.

To make it easier to find let's list the numbers in order from least to greatest

9, 13 , 21 , 28, 31

We then go to the middle number ( which would be 21 )

The mode is 21

So for data without Kevins score the mean would be 20.4, the median would be 21 and the mode would be none

Now let's add in Kevin's score and repeat this process

Data with Kevin's score

13, 21, 28, 9, 31, 9

First let's find the mean

( Note that there is 6 numbers this time )

13 + 21 + 28 + 9 + 31 + 9 = 111

111/6 = 18.5

So mean with Kevin's score = 18.5

Next let's find the mode

The number 9 appears the most so the mode is 9

Finally let's find the mean

Once again let's list the numbers from least to greatest

9, 9, 13 , 21 , 28, 31

Because there is an even amount of numbers we will take the sum of the two middle numbers and divide that by 2

The two middle numbers would be 21 and 13

21 + 13 = 34

34/2=17

So median = 17

In the end we get the following

Without Kevins score

Mean = 20.4

Median = 21

Mode = none

With Kevin's score

Mean = 18.5

Median = 17

Mode = 9

This information corresponds with the third table

5 0

3 years ago