I cant see the image...but I take it that the midpoint is (9,8) and the endpoint S is (10,10) and ur looking for the other endpoint R.
midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
(10,10)....x1 = 10 and y1 = 10
(x,y)....x2 = x and y2 = y
so we sub
(10 + x) / 2, (10 + y) / 2 = 9/8
(10 + x) / 2 = 9
10 + x = 9 * 2
10 + x = 18
x = 18 - 10
x = 8
(10 + y) / 2 = 8
10 + y = 8 * 2
10 + y = 16
y = 16 - 10
y = 6
so endpoint R has coordinates of (8,6) <===
SOLUTION
The mean is 4min
standard deviation is 1min
the z score is

where

then we have

The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
No bucease it has a negative on the 11 and the other one don’t
We can solve this by finding one of the numbers first
Let the smaller number be y.
Since these 2 numbers are consecutive odd numbers, the larger number should be 2 more than the smaller one (y+2).
Therefore,
y + (y+2) = 284
2y +2 = 284
shift +2 to the other side and turn it into -2
2y = 284 - 2
2y = 282
shift x2 to the other side and turn it into /2
y = 282/2
y = 141
Now we got the smaller number, which is 141, we can also find the larger number by adding 2 to it. 141+2 = 143,
Therefore, your answer is 141 and 143