Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent. What is the 95 percent confidence interval for this single-sample t test?

a. [6.94, 11.66]
b. [11.64, 16.36]
c. [10.47, 15.19]
d. [12.99, 15.01]

b)  95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data  a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14  

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are  the 95 percent confidence interval for the population mean  'μ'

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

Conclusion:-

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl