A circle is dialated with a scale factor of 2/3. what is the area of the new circle if the area of the original circle is 45 squ
are inches? Show your work. :)
1 answer:
Original Circle Area = 45
A = pi * r^2
Plug in what we know:
45 = 3.14 * r^2
Divide 3.14 to both sides:
14.3312102 = r^2
Find the square root of both sides:
r = 3.78565849
Find 2/3 of this:
2/3 * 3.78565849 = 2.52377233
So this is the radius of the newly dilated circle.
A = pi * r^2
A = 3.14 * 2.52377233^2
A = 3.14 * 6.36942677
A = 20
So the area of the newly dilated circle is 20 square inches.
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The answer is 18.
Work:
[i]The range is the gap between the smallest value and the largest value. [/i]
The smallest is 68.
The largest is 86.
The gap between them is 86-68=18.
We're told that
![P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995](https://tex.z-dn.net/?f=P%28A%29%3D%5Cdfrac1%7B200%7D%3D0.005%5Cimplies%20P%28A%5EC%29%3D0.995)
![P(B\mid A)=0.7](https://tex.z-dn.net/?f=P%28B%5Cmid%20A%29%3D0.7)
![P(B\mid A^C)=0.05](https://tex.z-dn.net/?f=P%28B%5Cmid%20A%5EC%29%3D0.05)
a. We want to find
. By definition of conditional probability,
![P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}](https://tex.z-dn.net/?f=P%28A%5Cmid%20B%29%3D%5Cdfrac%7BP%28A%5Ccap%20B%29%7D%7BP%28B%29%7D)
By the law of total probability,
![P(B)=P(B\cap A)+P(B\cap A^C)=P(B\mid A)P(A)+P(B\mid A^C)P(A^C)](https://tex.z-dn.net/?f=P%28B%29%3DP%28B%5Ccap%20A%29%2BP%28B%5Ccap%20A%5EC%29%3DP%28B%5Cmid%20A%29P%28A%29%2BP%28B%5Cmid%20A%5EC%29P%28A%5EC%29)
Then
![P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657](https://tex.z-dn.net/?f=P%28A%5Cmid%20B%29%3D%5Cdfrac%7BP%28B%5Cmid%20A%29P%28A%29%7D%7BP%28B%5Cmid%20A%29P%28A%29%2BP%28B%5Cmid%20A%5EC%29P%28A%5EC%29%7D%5Capprox0.0657)
(the first equality is Bayes' theorem)
b. We want to find
.
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since
.
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area of a triangle is 1/2 x base x height
base = 19
height = 3
19 *3 = 57
1/2 x 57 = 28.5 square ft.
The answer is B.
When you set up a ratio of all the corresponding sides, the ratios all equal 2:1