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enot [183]
4 years ago
15

PLEASE ANSWER ASAP!!! I WILL GIVE YOU 5 STARS AND BRAINLIEST!!

Mathematics
1 answer:
ehidna [41]4 years ago
5 0

Answer:

j(x) = 2^(x - 5) - 2

Step-by-step explanation:

To the left of the y-axis we see that the blue graph is 2 units higher (vertically) than is the red graph.  This translation is shown in the formula h(x) = e^x + 2.  Beyond that, the graph of f(x) = 2^x has been translated 5 units to the right.

Thus, the "new" function is j(x) = 2^(x - 5) - 2

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Answer:

I think the temperature went down 86° but I'm not sure

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3 years ago
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Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46
JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

M = \frac{50463-38737}{2} = 5863

Relating the margin of error with the sample size:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.64.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

For this problem, we have that \sigma = 14300, M = 5863. So

M = z\frac{\sigma}{\sqrt{n}}

5863 = 1.645\frac{14300}{\sqrt{n}}

5863\sqrt{n} = 1.645*14300

\sqrt{n} = \frac{1.645*14300}{5863}

(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2

n = 16

The sample size was of 16.

7 0
3 years ago
Four more than a number is more than 13
Pepsi [2]
X > 9 is the answer. This is the correct answer because you would subtract 4 from each side of the equation, which would leave X on one side and nine on the other. The inequality remains the same since you are not dividing at all by any negative numbers
8 0
3 years ago
The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

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Therefore the we find the probability that there are 3 or less errors on the page as

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

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

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heres one

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