Jamil finds that he reads 1/5 of his 115-page book in 1/4 hour

Mathematics

1 answer:

Nadya [2.5K]3 years ago

6 0

Answer:

He will read 92 pages in an hour going by the current rate at which he reads

Step-by-step explanation:

The complete question is as follows;

Jamil found that he reads 1/5 of his 115 - page book in 1/4 of an hour. At this rate how many pages does he read in an hour

We proceed as follows;

If he reads 1/5 of 115 in 1/4 hour

1/5 of 115 pages refer to 1/5 * 115 = 23 pages

This means that he reads 23 pages in quarter of an hour

So in an hours the number of pages he will read will be 23 * 4 = 92 pages

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il63 [147K]

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3 years ago

Find the equation of the line that is perpendicular to x - y = -5 and passes through (4, 7).

Ivanshal [37]

X - y = -5
-y = -x - 5
y = x + 5...slope here is 1. A perpendicular line will have a negative reciprocal slope. All that means is flip ur slope and change the sign. So our perpendicular line will have a slope of -1.

y = mx + b
slope(m) = -1
(4,7)...x = 4 and y = 7
now we sub and find b, the y int
7 = -1(4) + b
7 = -4 + b
7 + 4 = b
11 = b

so ur perpendicular equation is : y = -x + 11 <==

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3 years ago

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got: