Alarge city newspaper periodically reports the mean cost of dinner for two people at restaurants in the city. the newspaper staff will collect data from a random sample of restaurants in the city and estimate the mean price using a 90 percent confidence interval. in past years, the standard deviation has always been very close to $35. assuming that the population standard deviation is $35, which of the following is the minimum sample size needed to obtain a margin of error of no more than $5 ? a) 90
b) 112
c) 133
d)147

C) 133

Step-by-step explanation:

The formula to find the sample size is given by :-

, where z* = Critical z-value

= Population standard deviation for prior study.

E= Margin of error.

As per given , we have

E= 5

The critical z-value for 90% confidence level is 1.645.

Substitute al;l the value sin the above formula  , we get

Hence, the minimum sample size needed is 133.

Thus , the correct answer is : C) 133

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

24

step-by-step explanation: