(1,-1) (-3,-3)
-3 - -3/ -1 -1
0/ -2
Slope is -2
Let

denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by

, each independently and identically distributed with distribution

.
You want to find

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

Recall that if

, then the sampling distribution

with

being the size of the sample.
Transforming to the standard normal distribution, you have

so that in this case,

and the probability is equivalent to

One way to approach this problem is to go ahead and find the actual equation in slope-intercept form.
The slope of the line thru (3,6) and (5,4) is m = (6-4) / (3-5), or -2/2, or -1.
Subbing the known values of m, x and y, we get:
4 = -(5) + b. Then b = 9.
The correct equation (representing the data values in the table) is y = -x + 9.
Answer:

Step-by-step explanation:

C is the right answer answer but I gotta is a great time to go home to the right one or the other one is the right one and you have to answer me by the heart and
I hope this helps you!