The amount of time workers spend commuting to their jobs each day in a large metropolitan city has a mean of 70 minutes and a standard deviation of 20 minutes. assuming the distribution of commuting times is known to be mound-shaped and symmetric, what percentage of these commuting times are between 50 and 110 minutes?

81.85% of the workers spend between 50 and 110 commuting to work

Step-by-step explanation:

We can assume that the distribution is Normal (or approximately Normal) because we know that it is symmetric and mound-shaped.

We call X the time spend from one worker; X has distribution N(μ = 70, σ = 20). In order to make computations, we take W, the standarization of X, whose distribution is N(0,1)

The values of the cummulative distribution function of the standard normal, which we denote , are tabulated. You can find those values in the attached file.

Using the symmetry of the Normal density function, we have that . Hece,

The probability for a worker to spend that time commuting is 0.8185. We conclude that 81.85% of the workers spend between 50 and 110 commuting to work.

(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