A university's administrator proposes to do an analysis of the proportion of graduates who have not found employment in their major field one year after graduation. In previous years, the percentage averaged 15%. He wants the margin of error to be within 5% at a 99% confidence level. What sample size will suffice? Round to the nearest integer.

339

Step-by-step explanation:

We have that the formula would come being:

n = p (1-p) * [z (α / 2) / B] ^ 2

let p, percentage averaged (0.15)

B = 5%, 0.05 (bottom margin)

α = 1 - A

A, top margin 99%

α = 1 - 0.99 = 0.01

α / 2 = 0.005

z (0.005) = 2.58 (attached normal distribution table), the value is negative in the table but is taken as positive.

Replacing we have:

n = 0.15 * (1 - 0.15) * [2.58 / 0.05] ^ 2

n = 0.1275 * (2.58 / 0.05) ^ 2

n = 339.4764

Therefore the sample size is 339.

answer: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.

step-by-step explanation: if you weigh 60 pounds on earth, on the moon you will weigh about 10 pounds. if you weigh 120 pounds on earth, you will weigh about 20 pounds on the moon. since weight is caused by the mass of two objects that are near each other, the bigger the objects, the more the force of gravity.

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5