(HELP PLEASE) A field 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 2% of 76%. 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

Step-by-step explanation:Confidence Level:

The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%.

Margin of Error:

The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .

Confidence Interval:

Lower Limit = 70% - 4% = 66%

Upper Limit = 70% + 4% = 74%

Interpretation:

The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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