Agrocery store's receipts show that sunday customer purchases have a skewed distribution with a mean of $3030 and a standard deviation of $2121. suppose the store had 304304 customers this sunday. a) estimate the probability that the store's revenues were at least $9 comma 6009,600. b) if, on a typical sunday, the store serves 304304 customers, how much does the store take in on the worst 11% of such days?

A. we use the z statistic to solve this problem

z = (x – u) / s

We calculate the value of the sample mean u and standard deviation s:

u = $30 * 304 = $9120

s = $21 * 304 = $6384

z = (9,600 – 9120) / 6384

z = 0.075

From the normal tables using right tailed test,

P = 0.47

B. At worst 11% means P = 0.11, so the z value at this is z = -1.23

-1.23 = (x – 9120) / 6384

x = 1267.68

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

c

c

c

c

c

c

c

cc

c

step-by-step explanation: