Asample of 50 women is obtained, and their heights (in inches) and pulse rates (in beats per minute) are measured. the linear correlation coefficient is 0.291 and the equation of the regression line is modifyingabove y with caret equals 18.1 plus 0.930 x, where x represents height. the mean of the 50 heights is 63.1 in and the mean of the 50 pulse rates is 74.9 beats per minute. find the best predicted pulse rate of a woman who is 73 in tall. use a significance level of alpha equals 0.01.

The best predicted pulse rate of a woman who is 73 in tall is 85.99 beats per minute .

Step-by-step explanation:

Given :The equation of the regression line :

To Find : Find the best predicted pulse rate of a woman who is 73 in tall.

Solution:

y represents the pulse rate

x represents the height

Now we are supposed to find the pulse rate of a woman who is 73 in tall.

So, Substitute x = 73 in equation.

So,  the best predicted pulse rate of a woman who is 73 in tall is 85.99 beats per minute .

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation: