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Basile [38]
3 years ago
14

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!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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Which of the following<br> represents 9!<br> A. 9<br> B. 9*8*7*6*5*4*3*2*1<br> C. 9+8+7+6+5+4+3+2+1
Zielflug [23.3K]
The answer is B. Btw don’t click on the link provided by the other answer provider it’s a virus.
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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.

4 0
3 years ago
Will give brainlst <br> Have a nice day
Zarrin [17]

Answer:

ok where are you from which country

5 0
3 years ago
Please help math thank you
Alina [70]

a: 30

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5 0
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
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