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Furkat [3]
2 years ago
12

There's supposed to be 2 ansers for x, by the way​

Mathematics
2 answers:
ololo11 [35]2 years ago
4 0
  • 3tanx=-4
  • tanx=-4/3

Now

we know tan53=4/3

But it's negative

So tan is either in 2nd or 4th quadrant

For 2nd Quarter

  • x=π/2+53=90+53=143°

For 4th quadrant

  • x=3π/2+53=270+53=323°
lorasvet [3.4K]2 years ago
3 0

Answer:

  • x = 143° and x = 323°

Step-by-step explanation:

<u>Solving</u>

  • 3tanx = -4
  • tanx = -4/3
  • tan values are negative in the 2nd and 4th quadrant
  • The base value is x = 53°

In <u>Quadrant II</u>,

  • 90° + 53°
  • ⇒ <u>x = 143°</u>

In <u>Quadrant IV</u>,

  • 270° + 53°
  • ⇒ <u>x = 323°</u>
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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%

8 0
3 years ago
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Answer:

Step-by-step explanation:

A/4=72/12

A=4(72)/12

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