A random sample of 64 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 "t" statistic is:.

Degree of freedom = 63

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 64

Sample mean = 1400

Sample standard deviation = 240

We need to make a 95% confidence interval.

We use the t-statistic for the same.

Degree of freedom:

Degree of freedom  for reading the critical values of t-statistic is

Thus, the degree of freedom is 63.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

96 39 cause 135 -96=39 im smart (i used a calculater)