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3 years ago

Write a linear equation from the graph.

Mathematics

2 answers:

Leya [2.2K]3 years ago

8 0

This was super close to getting it right.

The slope formula is m=(y2-y1)/(x2-x1) with m being the slope and (x1, y1) and (x2, y2) being points.

The points you are working with are (0,4) and (5, 1). When I calculate the slope, it doesn’t matter which y coordinate I use first, as long as I use the x coordinate from the same point first in the denominator.

I’m now going to calculate the slope:
m = (1-4)/(5-0) = -3/5
(5,1) is a point, and because I used 1 first up top I used 5 first on the bottom.

The y-intercept is right, as when x = 0, y = 4.

The equation of the line is y = -3/5x + 4
You can double check this by creating your own graph.

USPshnik [31]3 years ago

7 0

Answer:

x+y=4...........eq1

5x-y=1............eq2...

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3 years ago

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5.5 miles

Step-by-step explanation:

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2 years ago

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a 1

b 6

c 15

Step-by-step explanation:

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3 years ago

6. Describe how we could conduct a simulation to decide whether a simple random sample from this population could produce a samp

BigorU [14]

Step-by-step explanation:

We have proportion p = 50/100 = 0.50

Probability of success = 1-0.50 = 0.50

To conduct simulation, let's assume experiment is to be done 10 times

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2. We get a random sample and get the size of people who support an against smoking with kids in the car.

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2 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:

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

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

3 0

3 years ago