Simplify 6^3/6^2 A. 6 B. 1 1/2 C. 18 D. 256

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

4 years ago

This is for leadership

Nikolay [14]

Answer:

I don't think many schools do Powderpuff which is where boys cheer and girls play football. You can do a live Insta and congratulate the students who made the team and mail them gift cards and/or gift baskets. You can also buy basket balls and footballs for each member and decorate it with their names and numbers. I'm also in Student Council so, if you have any other questions feel free to post them!!

6 0

3 years ago

Can someone please help me with these questions!!!!

IgorLugansk [536]

Answer: B

Step-by-step explanation:

The only labeled angle is 111

We can conclude that angle 3 is also 111

1 and 2 are both less than 90

Therefore, 1 is 69, 2 is 69, and 3 is 111

5 0

3 years ago

A box is 12 centimeters wide ,12 centimeters long, and 15 centimeters tall. What is the total surface area of 4 such boxes?

Grace [21]

First you want to find the surface area of one box
SA=2(l*w)+2(h*l)+2(h*w)
SA=2(12*12)+2(15*12)+2(15*12)
SA=2(144)+2(180)+2(180)
SA=288+360+360
SA=1008 cm² is for one box
Multiply by 4 to get for 4 boxes
4*1008=4032 cm²

4 0

3 years ago

Find mean, median, mode for the following. The following table of grouped data represents the weight (in pounds) of 100 Computer

ExtremeBDS [4]

Answer:

A.(for weight in pounds)

3, 4, 5, 6, 7

Median: 5

Mode: There are no mode in the following data set.

Mean: 3 + 4 + 5 + 6 + 7 = 25/5 = 25

B.(for number of computers)

4, 8, 18, 25, 45

Median: 18

Mode: There are no mode in the following data set.

Mean: 4 + 8 + 18 + 25 + 45 = 100/5 = 20

~hope this helps~

I would highly appreciate it if you b=would give me the brainliest~ :)

5 0

3 years ago