Alarge disaster cleaning company estimates that 30 percent of the jobs it bids on are finished within the bid time. looking at a random sample of 8 jobs that it has contracted, find the probability that x (number of jobs finished on time) is within one standard deviation of the mean.

68.26% probability that the number of jobs finished on time is within 1 standard deviation of the mean.

Step-by-step explanation:

Problems of normally distributed samples can be solved using 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.

Looking at a random sample of 8 jobs that it has contracted, find the probability that x (number of jobs finished on time) is within one standard deviation of the mean.

Within 1 standard deviation of the mean is from Z = -1 to Z = 1. So this probability is the pvalue of Z = 1 subtracted by the pvalue of Z = -1.

Z = 1 has a pvalue of 0.8413

Z = -1 has a pvalue of 0.1587

So there is a 0.8413 - 0.1587 = 0.6826 = 68.26% probability that the number of jobs finished on time is within 1 standard deviation of the mean.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.