To see what are the factors use, we write each of the numbers, as product of prime factors:

As we can see, the 4 factors used to produce the numbers in the list are {2, 3, 5, 7}
Answer: {2, 3, 5, 7}
Answer:
The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7531
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 23.29 - 0.34 = 22.95 mpg
The upper end of the interval is the sample mean added to M. So it is 23.29 + 0.34 = 23.63 mpg
The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.
Adding Integers
If the numbers that you are adding have the same sign, then add the numbers and keep the sign.
Example:
-5 + (-6) = -11
Adding Numbers with Different Signs
If the numbers that you are adding have different (opposite) signs, then SUBTRACT the numbers and take the sign of the number with the largest absolute value.
Examples:
-6 + 5= -1
12 + (-4) = 8
Subtracting Integers
When subtracting integers, I use one main rule and that is to rewrite the subtracting problem as an addition problem. Then use the addition rules.
When you subtract, you are really adding the opposite, so I use theKeep-Change-Change rule.
The Keep-Change-Change rule means:
Keep the first number the same.
Change the minus sign to a plus sign.
Change the sign of the second number to its opposite.
Example:
12 - (-5) =
12 + 5 = 17
Multiplying and Dividing Integers
The great thing about multiplying and dividing integers is that there is two rules and they apply to both multiplication and division!
Again, you must analyze the signs of the numbers that you are multiplying or dividing.
The rules are:
If the signs are the same, then the answer is positive.
If the signs are different, then then answer is negative.
Answer:
confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.
(0.79 , 0.89)
Step-by-step explanation:
step 1:-
Given sample survey former UF students n = 1532
84% said they were still in love with Tim Tebow
p = 0.84
The survey sampling error

Given standard error of proportion = 2% =0.02
<u>Step 2</u>:-
The 99% of z- interval is 2.57
The 99% of confidence intervals are
p ± zₐ S.E (since sampling error of proportion = 

on simplification , we get
(0.84 - 0.0514 , 0.84 + 0.0514)
(0.79 , 0.89)
<u>conclusion</u>:-
confidence interval for the proportion of all former UF students who are still in love with Tim Tebow.
(0.79 , 0.89)
well, keeping in mind that a year has 12 months, that means that 8 months is 8/12 of a year, when Mrs Rojas pull her money out.
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\to \frac{8}{12}\dotfill &\frac{2}{3} \end{cases} \\\\\\ A=6000[1+(0.04)(\frac{2}{3})]\implies A=6000\left( \frac{77}{75} \right)\implies A=6160](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B8%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%28%5Cfrac%7B2%7D%7B3%7D%29%5D%5Cimplies%20A%3D6000%5Cleft%28%20%5Cfrac%7B77%7D%7B75%7D%20%5Cright%29%5Cimplies%20A%3D6160)
well, she put in 6000 bucks, got back 160 extra, that's the interest earned in the 8 months.
what if she had left her money for 1 whole year, then
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &1 \end{cases} \\\\\\ A=6000[1+(0.04)(1)]\implies A=6240](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cdotfill%20%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%281%29%5D%5Cimplies%20A%3D6240)
so had she left it in for a year, she'd have gotten 6240, namely 240 in interest, well, what fraction of a year's interest was earned? or worded differently, what fraction is 160(8 months) of 240(1 year)?
