The standard deviation of a simple random sample of 15 customer waiting times is found to be 4.8 minutes. Find the test statistic to test the claim that the standard deviation of all customer
waiting times is greater than 3.5 minutes. Use a 0.01 significance level.

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= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5

step-by-step explanation:

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6