Answer:
And the best option would be:
c. 1450 +/- 12
Step-by-step explanation:
Information provided
represent the sample mean for the SAT scores
population mean (variable of interest)
represent the sample variance given
n=25 represent the sample size
Solution
The confidence interval for the true mean is given by :
(1)
The sample deviation would be
The degrees of freedom are given by:
The Confidence is 0.954 or 95.4%, the value of
and
, assuming that we can use the normal distribution in order to find the quantile the critical value would be
The confidence interval would be
And the best option would be:
c. 1450 +/- 12
Answer:
![-8\left[\begin{array}{ccc}-10&-2&-5\\9&-7&-1\\9&-4&2\end{array}\right]](https://tex.z-dn.net/?f=-8%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-10%26-2%26-5%5C%5C9%26-7%26-1%5C%5C9%26-4%262%5Cend%7Barray%7D%5Cright%5D)
![-\left[\begin{array}{ccc}-80&-56&-40\\72&-56&-8\\72&-32&16\end{array}\right]](https://tex.z-dn.net/?f=-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-80%26-56%26-40%5C%5C72%26-56%26-8%5C%5C72%26-32%2616%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{ccc}56&50&-14\\-120&92&62\\-78&26&2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D56%2650%26-14%5C%5C-120%2692%2662%5C%5C-78%2626%262%5Cend%7Barray%7D%5Cright%5D)

<u>---------------------------</u>
<u>hope it helps..</u>
<u>have a great day!!</u>
Answer:
95
Step-by-step explanation:
The smallest number that divided by 6, 15, and 18 leavses 5 in each case is 95