Slope=rise/run
this one
slope=increase in money/increase in time

from 412 to 549 that is 137, increase 137
from 1990 to 1999, that is 9 years, increase 9 years
slope=$/t
slope=137/9
slope=15 and 2/9

5 0

3 years ago

PLESSSE HELP ME WITH THIS QUESTION TY

Charra [1.4K]

TV is the same length as VR, so VR=7

4 0

3 years ago

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation

melamori03 [73]

Answer:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The point estimate of the population mean is $\hat \mu = \bar X In order to calculate the critical value [tex]t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=64-1=63$

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".

$\bar X=1400$ represent the sample mean

$\mu$ population mean (variable of interest)

s=240 represent the sample standard deviation

n=64 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The point estimate of the population mean is $\hat \mu = \bar X In order to calculate the critical value [tex]t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=64-1=63$

5 0

4 years ago