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In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co

Vera_Pavlovna [14]

B. 0.75 is the answer for this equation

7 0

3 years ago

PLEASE HELP ME NOW!!!! DUE VERY SOON! PLEASE I BEG YOU!

insens350 [35]

Step-by-step explanation:

I'm trying to solve this problem but there's something wrong

What's the radius of one ball?

7 0

3 years ago

Find FG FGH= 49 GH=31

Luba_88 [7]

Answer:

Step-by-step explanation:

From the given diagram;

FGH = 49

GH = 31

Using the expression to get FG;

FG + GH = FGH

FG + 31 = 49

FG = 49-31

FG = 18

Hence the measure of FG is 18

4 0

3 years ago

The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

dybincka [34]

Answer:

Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) 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".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n-1=12-1=11$

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for $t_{\alpha/2}=-2.201$

"=T.INV(1-0.025,11)" for $t_{1-\alpha/2}=2.201$

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

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

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval : "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

So the 95% confidence interval would be given by (3.646;4.472)

8 0

3 years ago

Graph this equation y=7/2x-2

Ludmilka [50]

To find the slope and y intercept, use the y=mx+b formula where m is the slope and b is the y intercept.
y=mx+b

Pull the values of m and b using the y=mx+b formula.
m=7/2,b=−2 where m is the slope and b is the y-intercept

6 0

3 years ago