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3 years ago

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g The probability 0.15. The probability that is 0.085. What is the probability that a undergraduate is a Junior and a Nursing ma

jor

1 answer:

Crazy boy [7]3 years ago

3 0

<em>Question:</em>

<em>The probability that a UCR undergraduate majors in Nursing is 0.15. The probability that a UCR undergraduate is a Junior is 0.20. The probability that a UCR undergraduate is a Junior or a Nursing major is 0.085. What is the probability that a UCR undergraduate is a Junior and a Nursing major?</em>

Answer:

$P(J\ and\ N) = 0.265$

Step-by-step explanation:

Given

Represent Junior with J and Nursing Major with N.

So, we have:

$P(N) = 0.15$

$P(J) = 0.20$

$P(J\ or\ N) = 0.085$

Required

Determine $P(J\ and\ N)$

In probability, for events A and B

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

In this case:

$P(J\ and\ N) = P(J) + P(N) - P(J\ or\ N)$

Substitute values for P(J), P(N) and P(J or N)

$P(J\ and\ N) = 0.20 + 0.15 - 0.085$

$P(J\ and\ N) = 0.265$

<em>Hence, the probability that an undergraduate is a Junior and major in Nursing is 0.265</em>

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Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
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Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
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Make the fractions have common denominators.

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Simplify this equation.

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Multiply the right side by the reciprocal of 5/4.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}$

The 4's cancel out and we are left with:

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Factor $e^-^x$ out of the numerator.

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