Asample of 100 receipts from a restaurant showed that 12 had errors. what is the 95% confidence interval for the proportion of all receipts that have error? note that z 0.025= 1.96 and t 0.025= 1.984 for 99 degrees of freedom?

0.03919

Step-by-step explanation:

You take 1.984+1.96=3.944

3.944-0.025=3.919

3.919 into the percentage =0.03919

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer