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Genrish500 [490]
3 years ago
14

What is the algebraic expression for the following word phrase the sum of 15x and 4​

Mathematics
1 answer:
Tanya [424]3 years ago
8 0
15x + 4 would be the algebraic expression
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A) 100 degrees (congruent angle)
B) 45 degrees (congruent angle)
C) 90 degrees (right angle)
D) 99 degrees (congruent angle)
E) 30 degrees (congruent angle)
F) 30 degrees (180-150=30, that is to find the other angle that is congruent to f)
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It snowed from 9:43 PM to 11:37 PM. How long was it snowing?
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Central Allen High School's girls golf team has won 76% of its games this year. The team has won 19 games. How many total games
lawyer [7]

Answer: 25 games

Step-by-step explanation:

Take the fractions 76/100 and 19/x

Cross multiply. 76 times x equally 76x and 100 times 19 equals 1900.

Divide 76x by 76 which equals just x.

Divide 1900 by 76 which equals 25.

x=25

8 0
2 years ago
Which of the following describes the algebraic expression 5(x+2)-3
Pavel [41]

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3 0
3 years ago
An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th
butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that \pi = 0.2

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}

0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}

\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}

(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}

n = 5615

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now \pi = 0.12, n = 5615.

We have to find M.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

M = 2.81\sqrt{\frac{0.12*0.88}{5615}}

M = 0.012

7 0
2 years ago
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