450 high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 2.96, and the standard deviation was 0.21. What is the margin of error, assuming a 70% confidence level, to the nearest hundredth

The margin of error is of 0.01.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 1.037.

The margin of error is of:

In which is the standard deviation of the population and n is the size of the sample.

Standard deviation was 0.21.

This means that

Sample of 450:

This means that

What is the margin of error, assuming a 70% confidence level, to the nearest hundredth?

The margin of error is of 0.01.

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950