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
This question is incomplete.
The complete question says;
The two-way table shows the number of hours students studied and whether they studied independently or with a study group.
What is the relative frequency of students that studied independently for more than 2 hours to the total number of students that studied independently?
a) 0.4 c) 0.25
b) 0.33 d) 0.11
Table is attached as image
Answer: C (0.25)
The number of students that studied for more than 2 hours as given in the table are 4.
The total number of people that studied independently include those that studied less than 2 hours and those that studied for more than 2 hours.
Those that studied less than 2 hours independently are 12.
Those that studied more than 2 hours independently are 4.
Hence the total number of people that studied independently is 16.
Therefore the relative frequency of students that studied independently for more than 2 hours to the total number of students that studied independently would be = 4/16 = 1/4 = 0.25.
Answer:
Step-by-step explanation:
Multiply the two polynomials by multiplying each term
Answer:
A
Step-by-step explanation: I believe it's A because It's the opposite but I'm not 100% certain, because I don't remember inverse statements.
Answer:
a. P(male) = 0.4
b. P(no sport and male) = 0.1
c. Unclear question (isn't it the same as b.?)
Step-by-step explanation:
The data below is what I've worked according to, which isn't very clear from the question so the answers are only correct if this is the correct table of data;
![\left[\begin{array}{ccc}&No \ Sports&Sports\\Female&10&32\\Male&7&21\end {array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26No%20%5C%20Sports%26Sports%5C%5CFemale%2610%2632%5C%5CMale%267%2621%5Cend%20%7Barray%7D%5Cright%5D)
a.
b.
Using the tree diagram in the picture;
