Step-by-step explanation:
All ik is that its gonna take 10 minutes to drain.
The answer for this question is "y = -5/3 and x = 8/3"
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
Answer:
(A) Yes, since the test statistic is in the rejection region defined by the critical value, reject the null. The claim is the alternative, so the claim is supported.
Step-by-step explanation:
Null hypothesis: The wait time before a call is answered by a service representative is 3.3 minutes.
Alternate hypothesis: The wait time before a call is answered by a service representative is less than 3.3 minutes.
Test statistic (t) = (sample mean - population mean) ÷ sd/√n
sample mean = 3.24 minutes
population mean = 3.3 minutes
sd = 0.4 minutes
n = 62
degree of freedom = n - 1 = 62 - 1 = 71
significance level = 0.08
t = (3.24 - 3.3) ÷ 0.4/√62 = -0.06 ÷ 005 = -1.2
The test is a one-tailed test. The critical value corresponding to 61 degrees of freedom and 0.08 significance level is 1.654
Conclusion:
Reject the null hypothesis because the test statistic -1.2 is in the rejection region of the critical value 1.654. The claim is contained in the alternative hypothesis, so it is supported.
