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3 years ago

Andrew scored an 89, 92, 97, 81, and 96 on five tests. What is the mean of Andrew's test scores?

2 answers:

8 0

To find the mean, add all the numbers together and divide by the amount of numbers there is

89 + 92 + 97 + 81 + 96 = 455

455/5 = 91

91 is Andrew's mean score, and is your answer

hope this helps

lbvjy [14]3 years ago

3 0

First off, to find the mean, you need to add the numbers to each other, then divide the total by the amount of numbers you had.

In this case, your total after adding them together would be, 374, now divide that by the amount of different numbers you have, 5, and you will get your answer. The answer would be 74.8, or rounded up to 75.

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3 years ago

Orwell building supplies' last dividend was $1.75. Its dividend growth rate is expected to be constant at 38.00% for 2 years, af

WITCHER [35]

Answer:

Estimated current stock price is $46.84

Step-by-step explanation:

First we have to find the value of dividend payment $t_1, t_2 \& t_3$

Given $D_0=$1.75$

Yearly growth rate for next 2 years( $g_1$ )=38.00%

$D_1=\$1.75\times 1.38=\$2.42$

$D_2=\$2.42\times1.38=\$3.34$

Growth after two years will 6% indefinitely

$g_2=6\%=\dfrac{6}{100}=0.06$

$D_3=\$3.34\times1.06=\$3.54$

Estimate of current stock price is

$=\dfrac{D_1}{(1+r)}+\dfrac{D_2}{(1+r)^{2}}+\dfrac{\dfrac{D_3}{(r-g_2)}}{(1-r)^{3}}$

$=\dfrac{\$2.42}{(1.12)}+\dfrac{\$3.34}{(1.12)^{2}}+\dfrac{\dfrac{\$3.54}{(0.12-0.06)}}{(1.12)^{3}}$

=$46.84

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3 years ago

The histogram shows the distributions of essay scores for high school sophomores and juniors in a contest. Which comparison of t

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Answer:

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Step-by-step explanation:

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3 years ago

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Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 4,000,000 and a mean

In-s [12.5K]

Using the normal distribution, it is found that there is a 0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 19000, \sigma = \sqrt{4000000} = 2000$

The probability that the life span of the monitor will be more than 20,179 hours is <u>one subtracted by the p-value of Z when X = 20179</u>, hence:

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

$Z = \frac{20179 - 19000}{2000}$

Z = 0.59.

Z = 0.59 has a p-value of 0.7224.

1 - 0.7224 = 0.2776.

0.2776 = 27.76% probability that the life span of the monitor will be more than 20,179 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

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2 years ago