PLEASE HELP!! The AHS Football team did a weigh in at the start of training camp. The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg. Which of the following curves represents the data? A, B, Or C??

Choice C

Explanation:

The mean is 98, so that's the center of the distribution. This is where the highest point occurs because the most people are clustered around here. This rules out choice A and the answer is between B or C.

One standard deviation above the mean is 98+6 = 104. All we do is add on the standard deviation (6) to the mean (98) to get 104.

One standard deviation below the mean is 98-6 = 92

The graph that has tickmarks at 92 and 104 indicate we are exactly 1 standard deviation from the mean (98)

Similarly, 2 standard deviations above the mean is at 98+2*6 = 98+12 = 110. Or you could say 104+6 = 110.

2 standard deviations below the mean would get us to 98-2*6 = 98-12 = 86

The same idea applies to 3 standard deviations as well.

All of this matches with graph C. We can eliminate graph B by noticing that something like 95 is not a result of 98-6.

56

step-by-step explanation:

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.