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What is the sum of the areas of circle C and circle D?

Morgarella [4.7K]

Assuming that the center of each circle is at point C and D, then both have a radius of 7cm. We know that the formula for solving area of the circle is A=pi*r².
Then we solve area of once circle multiplied by two to get the total area of the given circles. The solution is shown below:
C1 area=3.1416*7²
C1 area=153.9384 cm²
Summation of the area of circle C and D is shown below:
Sum area=2*153.9384
Sum area=307.8768cm²

The answer is 98pi.

4 0

3 years ago

Read 2 more answers

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd =

swat32

Answer:

Test statistic, $t_{s} = -0.603$ (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

$\bar{d} = \frac{\sum x-y}{n}$

$\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625$

Standard deviation: $SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1} }$

$\sum d^{2} = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2} = 63$

$SD = \sqrt{\frac{63 - 8 * (-0.625)^2}{8-1} }$

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

$t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n} } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8} }$

$t_{s} = -0.6033$

6 0

2 years ago

Write an expression that equals 56. Include an exponent.

Troyanec [42]

There are many possibilities. An example is $1^{1} + 55 = 56$ .

6 0

3 years ago

pascal wanted the area of the door to be 54 square feet and the width still to be 2/3 the length. what would the dimensions of t

dlinn [17]

The floor wood be 81 dimensions.

8 0

2 years ago

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If you have one quarter, nine pennies, six dimes, and one nickle, how much money would you have?

Nady [450]

Answer:

99 cents

Step-by-step explanation:

one quarter=25 cents

nine pennies= 9 cents

six dimes= 60 cents

one nickle= 5 cents

25+9+60+5=85+14=99

3 0

2 years ago

Read 2 more answers