Arandom sample of 120 sat scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. if we want to provide a 95% confidence interval for the sat scores, the degrees of freedom for reading the critical values of the "t" value is:

1.980

Step-by-step explanation:

Given that a random sample of 120 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240

Here nothing mentioned about the distribution but sample size is very large.

Since population std deviation is not known, only sample std deviation is used.

t test will be more appropriate.

df = degrees of freedom = n-1=119

For two tailed 95% level, t critical value for 119 df would be

1.980

Using this critical value confidence interval would be formed.

you need to show us the graph.

step-by-step explanation:

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.