An SRS of 350 high school seniors gained an average of ¯ x = 21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 52 . (a) Find a 99 % confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval:

Upper bound: 26.45

Lower bound: 15.55            

Step-by-step explanation:

We are given the following in the question:

Sample mean, = 21

Sample size, n = 350

Alpha, α = 0.05

Population standard deviation, σ = 52

95% Confidence interval:

Putting the values, we get,

Upper bound: 26.45

Lower bound: 15.55

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation: