Scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 114. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect. If 1 of the men is randomly selected, find the probability that his score is at least 582.5.

25.14% probability that his score is at least 582.5.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

If 1 of the men is randomly selected, find the probability that his score is at least 582.5.

This is 1 subtracted by the pvalue of Z when X = 582.5. So

has a pvalue of 0.7486

1 - 0.7486 = 0.2514

25.14% probability that his score is at least 582.5.

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

77

step-by-step explanation: