Question

12 fred recorded the number of minutes he read each day, from monday through friday. his results are shown in the table.

day	number of minutes read
2	16
3	19
4	27
5	29

what is the correlation coefficient, to the nearest thousandth, and strength of the linear model of these data?
(1) 0.984 and strong (3) 0.984 and weak
(2) 0.968 and strong (4) 0.968 and weak

13 given ( f(x) = x^2 ), which function will shift ( f(x) ) to the left 3 units?
(1) ( g(x) = x^2 + 3 ) (3) ( j(x) = (x - 3)^2 )
(2) ( h(x) = x^2 - 3 ) (4) ( k(x) = (x + 3)^2 )

Explanation:

Step1: List data pairs

Let \(x\) = Day (1-5), \(y\) = Minutes (12,16,19,27,29). Pairs: (1,12),(2,16),(3,19),(4,27),(5,29).

Step2: Calculate means

\(\bar{x} = \frac{1+2+3+4+5}{5} = 3\), \(\bar{y} = \frac{12+16+19+27+29}{5} = 20.6\).

Step3: Compute numerator of r

\(\sum(x_i-\bar{x})(y_i-\bar{y}) = (1-3)(12-20.6)+(2-3)(16-20.6)+(3-3)(19-20.6)+(4-3)(27-20.6)+(5-3)(29-20.6)\)
= \((-2)(-8.6)+(-1)(-4.6)+0(-1.6)+1(6.4)+2(8.4)\)
= \(17.2 + 4.6 + 0 + 6.4 + 16.8 = 45\).

Step4: Compute denominators of r

\(\sum(x_i-\bar{x})^2 = (-2)^2+(-1)^2+0^2+1^2+2^2 = 4+1+0+1+4=10\)
\(\sum(y_i-\bar{y})^2 = (-8.6)^2+(-4.6)^2+(-1.6)^2+(6.4)^2+(8.4)^2\)
= \(73.96 + 21.16 + 2.56 + 40.96 + 70.56 = 209.2\)

Step5: Calculate r

\(r = \frac{45}{\sqrt{10 \times 209.2}} = \frac{45}{\sqrt{2092}} \approx \frac{45}{45.74} \approx 0.984\). Correlation >0.7 is strong.

Step6: Function shift rule

Left shift 3 units: \(f(x+3) = (x+3)^2\).

Answer:

1. (1) 0.984 and strong
2. (4) \(k(x) = (x + 3)^2\)