Spencer randomly samples the out-of-state tuition and fees for 25 public universities and found a mean of $21,032 with a standard deviation of $2,494. Using the alternative hypothesis that µ < $21,706, Spencer found a z test statistic of -1.35. Given the data provided, what is the p-value of the z-test statistic?

0.0885

Step-by-step explanation:

To find :  p-value of the z-test statistic.

p-value of the z-statistic depends on the alternative hypothesis.

Here , Alternative hypothesis :  µ < $21,706 , which is a left tailed hypothesis.

So we will check the area to the left side of the z-statistic.

Since z= -1.35

P(Z<z) =P(Z<-1.35)= 1- P(Z<1.35)

= 1- 0.9115

= 0.0885

Hence, the required p-value = 0.0885

(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