A new cola company is testing to see what proportion of their cans contain at least 12 oz. If they want to be within 3% of the actual percentage, how many cans should they measure to be 90% confident

752

Step-by-step explanation:

Given that,

Margin of error = 3% = 0.03

confidence level = 90% = 0.90

therefore from the z-table

z = 1.645

Now since no prior estimate of p is given, so we say p = 0.5

Sample size required will be

n = 1.645² × 0.5 ×(1-0.5) / 0.03² = 751.67

n = 751.67 ≈ 752

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100