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Scorpion4ik [409]
3 years ago
6

Please answer !!!!!!!!!!! Will mark brainliest !!!!!!!!!!!

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

f ⁻¹ (-2)=3

f ⁻¹ (1)= 0

Step-by-step explanation:

inverse functions are switching x and f(x) coordinates so f ⁻¹ (-2)=3 and f ⁻¹ (1)= 0

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ninety-five trillion, forty seven billion, eight hundred sixteen million, seven thousand, twenty six written in standard form
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I think it would be 95,047,816,007,026
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A film distribution manager calculates that 7% of the films released are flops. If the manager is correct, what is the probabili
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Answer:

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

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)}

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)}.

In this problem, we have that:

p = 0.07, n = 404. So

\mu = E(X) = np = 404*0.07 = 28.28

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{404*0.07*0.93} = 5.13

If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%?

This is the pvlaue of Z when X = 0.04*404 = 16.16. So

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

Z = \frac{16.16 - 28.28}{5.13}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

1 - 0.0091 = 0.9909

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

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