The cost of 5 gallons of ice cream has a variance of 49 with a mean of 30 dollars during the summer. What is the probability that the sample mean would differ from the true mean by less than 0.6 dollars if a sample of 145 5-gallon pails is randomly selected?

0.303 or 30.3%

Step-by-step explanation:

We are given;

Variance; Var = 49

Sample size; n = 145

Standard error from mean; SEM = 0.6

Now, standard deviation is given by the formula;

SD = √Var

SD = √49

SD = 7

Now, the standard error from standard deviation is;

SE = SD/√n

SE = 7/√145

SE = 0.5813

z = Number of standard errors away from the mean = SEM/SE = 0.6/0.5813 = 1.0322

Using online P-value from z-score calculator with the z-score above and an assumed significance level of 0.05, we have;

P-value ≈ 0.303

Probability = 0.303 or 30.3%

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47