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Crazy boy [7]
2 years ago
11

FIND THE INDICATED PROBABILITY FOR THE FOLLOWING:

Mathematics
1 answer:
luda_lava [24]2 years ago
3 0

The value of the probability P(A) is 0.40

<h3>How to determine the probability?</h3>

The given parameters about the probability are

P(A or B) = 0.6

P(B) = 0.3

P(A and B) = 0.1

To calculate the probability P(A), we use the following formula

P(A and B) = P(A) + P(B) - P(A or B)

Substitute the known values in the above equation

0.1 = 0.3 + P(A) - 0.6

Collect the like terms

P(A)= 0.1 - 0.3 + 0.6

Evaluate the expression

P(A)= 0.4

Hence, the value of the probability P(A) is 0.40

Read more about probability at

brainly.com/question/25870256

#SPJ1

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Please help, will mark brainliest .
yan [13]

18/3=6

24/3=8

21/3=7

The missing side is 7

4 0
3 years ago
Rectangle ABCD is dilated by a scale factor of 1/2 with a center of dilation at the origin. What are the coordinates of the imag
Alex Ar [27]
A. (-2,3) because Rectangle ABCD is dilated by a scale factor of 1/2.
6 0
3 years ago
Read 2 more answers
A square lawn has area 98 ft². A sprinkler placed at the center of the lawn sprays water in a circular pattern as
torisob [31]

The radius of the circle exists approximately 4.95 feet.

<h3>How to estimate the radius of the circle?</h3>

Given the area of the square as A = 98 ft²

The side length of the square choice is equivalent to the diameter of the circle.

First, we must obtain the length of the square.

Area of a square = L^2

98 = L²

L = √98

L = 9.89 ft

Thus the diameter of the circle exists at 9.89 ft

radius = diameter/2

radius = 9.89/2

radius ≈ 4.95 feet

Therefore the radius of the circle exists approximately 4.95 feet.

To learn more about the radius of circle refer to:

brainly.com/question/19568457

#SPJ9

3 0
1 year ago
Jesse has two packages of 36TACKS and 16 TACKS. How many garage sale posters can she put up if she uses 4TACKS for each poster?
hodyreva [135]

Answer: 21 posters or 13 posters, please read the explanation since your wording was confusing.


Step-by-step explanation: Okay if you have two packs of 36 and 16 which is 104, you divide by 4 tacks. You end up getting 21 posters.


If you mean there are only 36 and 16 you get 52 and divide by 4. You end up having 13 posters.


7 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
3 years ago
Read 2 more answers
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