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3 years ago

Help please , i need the perimeter and area

Mathematics

1 answer:

Ludmilka [50]3 years ago

7 0

Answer:

Area = 3.5325 ft² ≈ 3.53 ft²

Perimeter = 7.71 ft

Step-by-step explanation:

This is a semi-circle so we just divide the circumference of the original circle by 2

C = 2πr

C = 2 * 3.14 * 1.5

C = 3.14 * 3

C = 9.42

9.42 / 2 = 4.5 + 0.21 = 4.71 ft

4.71 ft + 3 = 7.71 ft

Area:

A = πr²

A = 3.14 * 2.25

A = 7.065 ft²

7.065 ft² = 3.5 + 0.0325 = 3.5325 ft²

If my answer is incorrect, pls correct me!

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-Chetan K

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3 years ago

Someone be as kind as to offer some help?

RideAnS [48]

Answer:

x = 4 and y = 0

Step-by-step explanation:

Using the elimination method:

4(-x + 5y=-4) --> -4x + 20y = -16.

Now we add -4x + 20y = 016 and 4x + 3y = 16. This gives us 23y = 0 or just y =0.

Now we can find x by plugging in y.

-4x + 0 = -16.

Divide by -4 on both sides -> x = 4.

If you want to double check you can plug your values into -4x + 20y = -16.

-4(4) +20(0) = -16.

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3 years ago

Will give brainlst Have a nice day

Zarrin [17]

Answer:

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5 0

3 years ago

Please help math thank you

Alina [70]

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4 years ago

Read 2 more answers

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

3 years ago