One of the questions in the Pew Internet & American Life Project asked adults if they used the Internet at least occasionally. The results showed that 454 out of 478 adults aged 18–29 answered Yes; 741 out of 833 adults aged 30–49 answered Yes; and 1,058 out of 1,644 adults aged 50 and over answered Yes. If required, round your answers to four decimal places. (a) Develop a point estimate of the proportion of adults aged 18–29 who use the Internet. (b) Develop a point estimate of the proportion of adults aged 30–49 who use the Internet. (c) Develop a point estimate of the proportion of adults aged 50 and over who use the Internet. (d) Comment on any apparent relationship between age and Internet use. The input in the box below will not be graded, but may be reviewed and considered by your instructor.

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

765238725478629038640962136509762139056074

step-by-step explanation:

64087235609746097215074621039754021037

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.