Suppose 10,000 students have taken a certain course and all of them were invited to evaluate the course. Out of the 10,000, we see that 496 responded to the course evaluation. And out of those 496, there were 130 that gave the course the highest rating (that is, 5 on a scale from 1 to 5). We want an approximate confidence interval for p, the proportion of all 10,000 students that would give this course the highest rating. We estimate p to be 130/496 or about 26.2%. Is a z-interval for a proportion applicable here

walter needs to make at least 5 more cookies but no more than 40. 5 ≥ x ≥ 40;

step-by-step explanation:

walter has to make a minimum of 20 cookies. he already has 15 cookies made, so he must make 5 more to meet the minimum.

the maximum amount of cookies he can make is 55. he already has 15, so he has to make no more than 40 more cookies to meet the maximum.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt