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Jared writes a multiplication expression with eight rational factors. Half of the factors are positive and half are negative. Is

soldi70 [24.7K]

Answer:

negative they are 4 negative factors

3 0

3 years ago

Por favor denme una ecuación a esta pregunta:

Katena32 [7]

Answer:

x + 13 = 25

Step-by-step explanation:

Se puede representar "un número" con x.

x

"Adiciona" significa +.

x + 13

"El resultado" significa =.

x + 13 = 25

7 0

3 years ago

Ms. Abrams receives a 4% commission plus a salary of $275.00 per week. If her total sales for one week are $2,880.00, how much d

Marta_Voda [28]

390.20 is the answer because 4%of 2880 is 115.20 plus 275.00 is 390.20

8 0

2 years ago

Which graph represents the function f(x) = 1/3 |x|

lisabon 2012 [21]

Answer:

Step-by-step explanation:

your answer is the 2 option

3 0

3 years ago

Read 2 more answers

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

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

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

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

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=\pm 60$ $ is n=28.

5 0

2 years ago