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3 years ago

What are the solutions to the system of equations? (___ , 2) and ( _ , ___ )

Mathematics

1 answer:

alexandr1967 [171]3 years ago

8 0

Answer:

(1, 2) and (-1, -2)

Step-by-step explanation:

Since y = 2x, the first blank tells you

2x = 2

x = 1

____

Since the variables are squared, a negative solution is as good as a positive solution, so x = -1 and y = -2 will also work.

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2 years ago

(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at col

Ne4ueva [31]

Answer:

a) $n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

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

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

Step-by-step explanation:

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

$\sigma=16$ represent the population standard deviation

n represent the sample size

Solution to the problem

Part a

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 =5 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: