A plant manager has collected productivity data for a sample of workers, intending to see whether there is a difference in the number of units they produce on Monday versus Thursday. The results (units produced) are in given data . Using the 0.01 level of significance, evaluate the null hypothesis that there is no difference in worker productivity between the two days. Identify and interpret the p-value for the test, then construct and interpret the 99% confidence interval for the mean difference in productivity between the days.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

the answers c

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

answer4 to the second power is 16

and 4 to the third power is 64,