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.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3