Atelevision news program conducts a call in poll about a proposed city ban on handgun ownership. of the 2372 calls,1921 oppose the ban. the station, following recommended practice, makes a confidence statement: "81% of the channel 13 pulse poll sample opposed the ban. we can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result." is this conclusion justified?

We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result.

is true.

Step-by-step explanation:

Given that a television news program conducts a call in poll about a proposed city ban on handgun ownership. Of the 2372 calls,1921 oppose the ban. The station, following recommended practice, makes a confidence statement:

Sample proportion p =

q=1-p =

Std error of p =

Margin of error for 95%=1.96*std error = 0.0158

Since margin of error is 0.0158 i.e. nearly 15.8% and proportion is 0.8099 = 81%

the statement is correct.

We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result.

is true.

(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

b

step-by-step explanation: