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marishachu [46]
3 years ago
6

84. Use properties of exponents to rewrite each expression with only positive, rational exponents. Then find the numerical value

of each expression when x = 9, y= 8, and z= 16. In each case, the expression evaluates to a rational number.
C. X-3/2 Y4/3 Z-3/4
Mathematics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

a) 1/27

b) 16

c) 1/8

Step-by-step explanation:

a) x^{-3/2}

One of the properties of the exponents tells us that when we have a negative exponent we can express it in terms of its positive exponent by turning it into the denominator (and changing its sign), so we would have:

x^{-3/2}=\frac{1}{x^{3/2} }

And now, solving for x = 9 we have:

\frac{1}{x^{3/2}}=\frac{1}{9^{3/2} }  =\frac{1}{27}

b) y^{4/3}

This is already a positive rational exponent so we are just going to substitute the value of y = 8 into the expression

y^{4/3}=8^{4/3}=16

c) z^{-3/4}

Using the same property we used in a) we have:

z^{-3/4}=\frac{1}{z^{3/4} }

And now, solving for z = 16 we have:

\frac{1}{z^3/4} } =\frac{1}{16^{3/4} } =\frac{1}{8}

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Using the binomial distribution, we have that:

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For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution  

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The parameters are:

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The expected number of <u>trials before q successes</u> is given by:

E = \frac{q(1-p)}{p}

The variance is:

V = \frac{q(1-p)}{p^2}

In this problem, 0.2 probability of a finding a person who attended the last football game, thus p = 0.2.

Item a:

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Thus:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

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The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

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