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3 years ago

Solve and check the following equation. 8d + 21 = 12d + 5 d =

Mathematics

1 answer:

olganol [36]3 years ago

3 0

Subtract 8d from both sides

21 = 12d + 5 - 8d

Simplify 12d + 5 - 8d to 4d + 5

21 = 4d + 5

Subtract 5 from both sides

21 - 5 = 4d

Simplify 21 - 5 to 16

16 = 4d

Divide both side by 4

16/4 = d

Simplify 16/4 to 4

4 = d

Switch sides

d = 4

<u>Check answer</u>

8d + 21 =12d + 5

Let d = 4

8 × 4 + 21 = 12 × 4 + 5

Simplify 8 × 4 to 32

32 + 21 = 12 × 4 + 5

Simplify 12 × 4 to 48

32 + 21 = 48 + 5

Simplify 32 + 21 to 53

53 = 48 + 5

Simplify 48 + 5 to 53

53 = 53

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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

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