I haven’t got there yet sorry
Answer:
18.67% probability that the sample proportion does not exceed 0.1
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For the sampling distribution of a sample proportion, we have that 
In this problem, we have that:

What is the probability that the sample proportion does not exceed 0.1
This is the pvalue of Z when X = 0.1. So



has a pvalue of 0.1867
18.67% probability that the sample proportion does not exceed 0.1
Answer:
7 and 8 im pretty sure...
Step-by-step explanation:
sorry if its wrong..
f(x) = x² + 7
g(x) = -3x + 1
h(x) = 12/x
j(x) = 2x + 9
a. g(10) = -3(10) + 1 = -20
b. f(3) = 3² + 7 = 9 + 7 = 16
c. h(-2) = 12/(-2) = -6
d. j(7) = 2(7) + 9 = 23
e. h(a) = 12/a
f. g(b+c) = -3(b+c) + 1 = -3b - 3c + 1
g. f(h(x)) = f(12/x) = (12/x)² + 7 = 7 + 144/x²
h. 16 = g(x) = -3x + 1, -3x = 15, x = -5
i. -2 = h(x) = 12/x, x = 12/(-2), x = -6
j. 23 = f(x) = x²+7, 16=x², x=±4