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Tomtit [17]
3 years ago
8

A point in the figure is selected at random. Find the probability that the point will be in the part that is NOT shaded.

Mathematics
2 answers:
Dovator [93]3 years ago
7 0

Answer:

B

Step-by-step explanation:

So the question really is what is the probability that the point is in the square? You cannot pick something surrounding the circles because you do not know anything about that region.

This is solved by comparing areas. There are 2 full circles there. 1 circle has an area of pi*r^2

Total shaded area = 2 * pi * r^2.

Let r = 3   This is a completely random choice. No matter what number you choose for r, the answer will come out the same. When you get along a little further in math, you will find that you can just use letters.

Total shaded area = 2 * 3.14 * 3^2

Total shaded area = 2 * 3.14 * 9

Total shaded area = 56.55

Total area of the square.

s = 2*r

s = 2*3

s = 6

Area of the square = 6^2 = 36

Total area of both regions = 56.55 + 36 = 92.55

Answer

% = (area of square) * 100% / Total area of both regions

% = 3600 / 92

% = 39%

Obviously the answer I get is not offered. The closest answer is B so I will choose that.

==========================

This is how this question would be done without using any value for r.

Area of shaded region = 2 pi r^2 = 6.28 r^2

Area of square  = d^2 where d = 2r

Area of square = 4 r^2

Total area = 6.28 r^2 + 4r^2 = 10.28 r^2

Area of square to total = (4r^2/10.28 r^2 ) * 100%

Area of square to total (as a %) = 4/10.28  * 100 = 39% which gives the same answer.

alex41 [277]3 years ago
5 0
B cause that’s what I got
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3 years ago
The senate in a certain state is comprised of 51 republicans, 48 democrats, and 1 independent. How many committees can be formed
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Calcula el RIQ de las temperaturas en las Ciudades de Kansas, Misouri las temperaturas: 23,25,28,28,32,33,35 y en Paradise, Mich
sasho [114]

Respuesta:

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Paradise, Michigan = 3

Explicación paso a paso:

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3 years ago
I need help!!
netineya [11]

Answer:

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Step-by-step explanation:

log_{5} {5}^{(2x - 9)}   +  log_{3} {729}^{5}   \\  \\  =  log_{5} {5}^{(2x - 9)}   +  log_{3} {( {3}^{6} )}^{5}   \\  \\  =  log_{5} {5}^{(2x - 9)}   +  log_{3} { {3}}^{6 \times 5}   \\  \\  =  log_{5} {5}^{(2x - 9)}   +  log_{3} { {3}}^{30}    \\  \\ =(2x - 9)  log_{5} {5} + 30 log_{3} { {3}}   \\  \\  = (2x - 9).1 + 30(1 )\\ ( \because \: log_{a} { {a}}    = 1) \\  \\  = 2x - 9 + 30 \\  \\  = 2x +21

7 0
3 years ago
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