Suppose gre quantitative scores are normally distributed with a mean of 588588 and a standard deviation of 154154. a university plans to send letters of recognition to students whose scores are in the top 12%12%. what is the minimum score required for a letter of recognition? round your answer to the nearest whole number, if necessary.

The minimum score required for a letter of recognition is 769

Step-by-step explanation:

If GRE Quantitative scores are normally distributed with a mean of 588 and a standard deviation of 154, then, for a value x, the associated z-score is computed as (x-588)/154. On the other hand, we are looking for a value "a" such that P(X > a) = 0.12 or equivalently P(X < a) = 0.88. But this last expression is equivalent to P((X-588)/154 < (a-588)/154) = 0.88, i.e., (a-588)/154 is the 88th quantile of the standard normal distribution. Therefore (a-588)/154 = 1.1750, and so, a = 588 + (154)(1.1750) = 768.95

correct answer:

b.(19,-2)

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm