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marusya05 [52]
3 years ago
8

I need to find the area, I just need to show my work any answers can help.​

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
4 0
Area of rectangle: 15 x 7 = 105
area of semicircle: 1/2 (3.14r^2)
radius is half of diameter so it would be 7.5
1/2 (3.14(7.5)^2)
=1/2 (176.71)
=88.355
rounded is 88.4
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The probability of event a is x, and the probability of event b is y. if the two events are independent, which condition must be
nalin [4]

Condition (A) P(B/A) = y is true.

<h3>What is probability?</h3>
  • Probability is an area of mathematics that deals with numerical descriptions of how probable an event is to occur or how likely a statement is to be true.

To find the true condition:


If two events are independent, then:

  • Pr(A∩B) = Pr(A) . Pr(B)

Use formulas for conditional probabilities:

  • Pr(A/B) = Pr(A∩B) / Pr(B)
  • Pr(B/A) = Pr(B∩A) / Pr(A)

For independent events these formulas will be:

  • Pr(A/B) = Pr(A∩B) / Pr(B) = Pr(A) . Pr(B) / Pr(B) = Pr(A)
  • Pr(B/A) = Pr(B∩A) / Pr(A) = Pr(B) . Pr(A) / Pr(A) = Pr(B)

Now in your case, Pr(A) = x and Pr(B) = y.

  • Pr(A/B) = x, Pr(B/A) = y, Pr(A∩B) = x.y

Therefore, condition (A) P(B/A) = y is true.

Know more about probability here:

brainly.com/question/25870256

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The complete question is given below:

The probability of event A is x, and the probability of event B is y. If the two events are independent, which of these conditions must be true?

a. P(B|A) = y

b. P(A|B) = y

c. P(B|A) = x

d. P(A and B) = x + y

e. P(A and B) = x/y

5 0
1 year ago
Please please answer What is 4328 divided by 4
Lerok [7]
It is 1082. 

I hope it helps, you can do more on a calculator. :)))
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In 2006, the population of Tewksbury, Rhode Island was 25,000, and it was growing at an annual rate of 2.2%. What is the growth
Shtirlitz [24]
Tewksbury Population In 2011

=
25000 {( {1.022)}^{2011 - 2006} }
= 27,874
7 0
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1/8 plus 3/4: plz tell me I need help
xenn [34]
The answer should be 7/8
6 0
3 years ago
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F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))
Marrrta [24]

Given:

f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6

Required:

We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.

Explanation:

Given equation is

F(x)=\sqrt[4]{f(x)g(x)}.F(x)=(f(x)g(x))^{\frac{1}{4}}F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}

Differentiate the given equation for x.

Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{  Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.

F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)

Replace x=3 in the equation.

F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}F^{\prime}(3)=\frac{-6-1}{4}F^{\prime}(3)=\frac{-7}{4}

Final answer:

F^{\prime}(3)=\frac{-7}{4}

8 0
1 year ago
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