Afield test for a new exam was given to randomly selected seniors. the exams were graded, and the sample mean and sample standard deviation were calculated. based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%.
is the confidence interval at 90%, 95%, or 99%? what is the margin of error? calculate the confidence interval and explain what it means in terms of the situation.

Margin of error: 6%; Confidence Interval Percentage: 90%; Confidence Interval: 74% and 86%

Step-by-step explanation:

Margin of error: 6%  --  It's already given in the question.

Confidence interval is at 90%. 9/10=.90

The exam creator is saying that they are 90% confident that the seniors test scores will be between 74% and 86%.

0.8 - 0.06 = 0.74

0.8 + 0.06 = 0.86

The confidence interval is (0.74, 0.86)

right

step-by-step explanation:

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1