Desperate need of **correct** 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 5% of 75%. 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. (i know the confidence interval is clearly at 90% but i don't know what the marigin of error is would it be between 70% and 80% or 5% of 75% which is 3.75% giving you 78.75% and 71.25%)

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

the answer is 13/48

step-by-step explanation: