The scores of a certain population on the wechsler intelligence scale for children (wisc) are thought to be normally distributed with mean m and standard deviation s = 10. a simple random sample of 25 children from this population is taken and each is given the wisc. the mean of the 25 scores is x = 104.32. based on these data, a 95% confidence interval for m is a) 104.32 â± 0.78. b) 104.32 â± 3.29. c) 104.32 â± 3.92. d) 104.32 â± 19.60

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.

answer: c

substitute in the values.

for a:

f(0) = 0 + 9

f(0) = 9

a is a solution.

for b:

f(7) = 14 + 9

f(7) = 23

b is a solution.

for c:

f (3) = 6 + 9

f(3) = 15

c is not a solution. it's supposed to be 14.

for d:

f(2) = 4 + 9

f(2)= 13

d is a solution.