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jasenka [17]
3 years ago
6

Two random variables x and y are independent if the value of x does not affect the value of y. If the variables are not​ indepen

dent, they are dependent. A new random variable can be formed by finding the sum or difference of random variables. If a random variable x has mean μx and a random variable y has mean μy then the means of the sum and difference of the variables are given by the following equations. μx+y=μx+μy μx−y=μx−μy The distribution of a​ test's scores for​ college-bound male seniors has a mean of 1528 and a standard deviation of 319. The distribution of a​ test's scores for​ college-bound female seniors has a mean of 1498 and a standard deviation of 304. One male and one female are randomly selected. Assume their scores are independent.What is the average sum of their scores? What is the average difference of their scores?
Mathematics
1 answer:
jeyben [28]3 years ago
6 0

Answer: \mu_{x+y} = 3026

              \mu_{x-y}= 30

Step-by-step explanation: Average sum of the female and male's test score is the sum of expected value of each gender:

\mu_{x+y}=\mu_{x}+\mu_{y}

Assuming x represents the random male selected and y represents the random female selected:

\mu_{x+y}=1528+1498

\mu_{x+y}= 3026

The average sum of their scores is 3026.

Average difference is the difference between the expected value (mean) of each gender:

\mu_{x-y}=\mu_{x}-\mu_{y}

\mu_{x-y}= 1528 - 1498

\mu_{x-y}= 30

The average difference of their scores is 30.

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The x-intercept is 2/3 and the y-intercept is -1/3

Step-by-step explanation:

In order to find this, we need to note that the x-intercept is the value of x, when y = 0. So we plug in y = 0 and see what the x value is.

3x - 6y = 2

3x - 6(0) = 2

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The y-intercept is the opposite. It is the y value when x = 0. So we do the same with the other term.

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y = -1/3

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162.4

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2 years ago
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y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945

\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\  ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
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(b)
Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
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2 years ago
C=2(3.14)r; c=14(3.14) solve for r
tekilochka [14]

Answer:

c=2(3.14)r

given c=14(3.14), then r

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r=c/2(3.14)

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