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Talja [164]
2 years ago
11

Plese HELP !!!

Mathematics
1 answer:
beks73 [17]2 years ago
6 0

Answer:

-9/8

Step-by-step explanation:

\frac{-2a^{-3}b^2}{a} \\ \\ =-\frac{2b^2}{a^4} \\ \\ =-\frac{2(3)^2}{(-2)^4} \\ \\ =-\frac{18}{16} \\ \\ =-\frac{9}{8}

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Answer:

y = (x - 1) / 2            (First answer choice)

Step-by-step explanation:

1) Start with f(x) = 2x + 1

2) Replace f(x) with y:  y = 2x + 1

3) Interchange x and y:  x = 2y + 1

4) Solve this result for y:  2y = x - 1, or y = (x - 1) / 2       (This matches the first                      

                                                                                          answer choice.)

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Candice bought 3 shirts.Each shirt Cost the same and was discounted by 3.66.Candice paid a total of 62.31 before tax. How much d
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$21.99 each

1. 3.66+62.31 (since it was discounted)

2. 65.97/3 (for each shirt)

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Solve (4 - x ≤ -1) ∩ (2 + 3x ≥ 17). Click on the graph until the solution set appears.
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15 points!!! Help please
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Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks
Gnesinka [82]

Answer:

a) 3.75

b) 23.61% probability that fewer than 3 tanks will be found to be leaking

c) 0% the probability that at least 600 of these tanks are leaking

Step-by-step explanation:

For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

To solve question c), i am going to approximate the binomial distribution to the normal.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

It is estimated that 25% of these tanks leak.

This means that p = 0.25

15 tanks chosen at random

This means that n = 15

a.What is the expected number of leaking tanks in such samples of 15?

E(X) = np = 15*0.25 = 3.75

b.What is the probability that fewer than 3 tanks will be found to be leaking?

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361

23.61% probability that fewer than 3 tanks will be found to be leaking

c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?

Now we have n = 2000. So

\mu = E(X) = np = 2000*0.25 = 500

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36

This probability is 1 subtracted by the pvalue of Z when X = 600. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{600 - 500}{19.36}

Z = 5.16

Z = 5.16 has a pvalue of 0.

0% the probability that at least 600 of these tanks are leaking

4 0
3 years ago
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