If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that the difference between the sample estimate and the actual market share for its new product was no more than 9%, what sample size would be needed?

We need a sample size of least 119

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

Sample size needed

At least n, in which n is found when

We don't know the proportion, so we use , which is when we would need the largest sample size.

Rounding up

We need a sample size of least 119

susie has 7 apples

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116