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Sliva [168]
3 years ago
6

What is the equation of the line of reflection that maps

Mathematics
2 answers:
Usimov [2.4K]3 years ago
8 0

Answer:

y- axis

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y )

Consider the corresponding vertices

A(- 1, - 1 ) → A'(1, - 1 )

B(- 2, - 1 ) → B'(2, - 1 )

C(- 2, - 4) → C'(2, - 4 )

These coordinates satisfy the condition for reflection in the y- axis

Butoxors [25]3 years ago
5 0

Answer:

Y-axis

Step-by-step explanation:

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Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa
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Answer:

a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

b) 0.4129 = 41.29% probability that the mean return will be less than 8%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 8.7% and standard deviation 20.2%.

This means that \mu = 8.7, \sigma = 20.2

40 years:

This means that n = 40, s = \frac{20.2}{\sqrt{40}}

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?

This is 1 subtracted by the pvalue of Z when X = 13. So

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

By the Central Limit Theorem

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

Z = \frac{13 - 8.7}{\frac{20.2}{\sqrt{40}}}

Z = 1.35

Z = 1.35 has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

(b) What is the probability that the mean return will be less than 8%?

This is the pvalue of Z when X = 8. So

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

Z = \frac{8 - 8.7}{\frac{20.2}{\sqrt{40}}}

Z = -0.22

Z = -0.22 has a pvalue of 0.4129

0.4129 = 41.29% probability that the mean return will be less than 8%

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3 years ago
What is the average rate of change for this function for the interval from x=3 to x=5?
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Answer:

A. 49

Step-by-step explanation:

The average rate of change for the interval ranging from x = 3 to x = 5 for the given function represented in the table above can be calculated using:

Average rate of change = \frac{f(x2) - f(x1)}{x2 - x1}

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f(x1) = f(3) = 27

Thus,

Average rate of change = \frac{125 - 27}{5 - 3}

= \frac{98}{2}

Average rate of change = 49

Average rate of change of the given table values representing an exponential function is A. 49.

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Answer:

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