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3 years ago

6

Eric wants to find out the most popular after-school activity among high school seniors at his school. What is the population in

Eric’s survey?

2 answers:

kolbaska11 [484]3 years ago

8 0

Answer:

all high school seniors at Eric’s school

Step-by-step explanation:

damaskus [11]3 years ago

3 0

Answer: The population in Eric's survey is the high school seniors.

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There are 158 students registered for American History classes. There are twice as many students registered in second period as

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There are 28 students in first period and 56 students in sceond period and 74 students in third period

<em><u>Solution:</u></em>

Let the number of students in first period be "x"

Let the number of students in second period be "y"

Let the number of students in third period be "z"

<em><u>There are 158 students registered for American History classes.</u></em>

Therefore,

x + y + z = 158 ---------- eqn 1

<em><u>There are twice as many students registered in second period as first period</u></em>

number of students in second period = twice of number of students in first period

y = 2x ------- eqn 2

<em><u>There are 10 less than three times as many students in third period as in first period</u></em>

number of students in third period = 3 times number of students in first period - 10

z = 3x - 10 ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

x + 2x + 3x - 10 = 158

6x = 168

<h3>x = 28</h3>

<em><u>Substitute x = 28 in eqn 2</u></em>

y = 2(28)

<h3>y = 56</h3>

<em><u>Substitute x = 28 in eqn 3</u></em>

z = 3(28) - 10

z = 84 - 10

<h3>z = 74</h3>

Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period

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2 years ago

the probability tree diagram shows the probabilities that a football team will win their first two games find the probability of

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Answer:1/2

Step-by-step explanation:

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2 years ago

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A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor

Leya [2.2K]

Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<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:

$\mu = 700, \sigma = 50$ .

The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:

X = 790:

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

$Z = \frac{790 - 700}{50}$

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

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

$Z = \frac{575 - 700}{50}$

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

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Help me! <br>This is Algebra problem

aleksklad [387]

Answer:

They paid 16 3/4% tax

Step-by-step explanation:

9%+1 3/4%+6%=16 3/4%

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