The expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
Solution:
Given expressions are 15t – 2t and 2t – 15t.
To determine 15t – 2t is equivalent to 2t – 15t or not.
Substitute t = 2 in above two expressions.
15t – 2t = 15(2) – 2(2)
= 30 – 4
= 26
2t – 15t = 2(2) – 15(2)
= 4 – 30
= –26
The values of the expressions are different when t = 2.
So, 15t – 2t is not equivalent to 2t – 15t.
Hence the expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
You would put 4-1 over 4-1 so the slope would be 3/3 which is one
so slope=1
The table shows a linear function. x -5,-4,-3,-2,-1, 0, 1, 2, 3. on top row f(x) -11, -3, 5, 13, 21, 29, 37, 45, 53. on bottom r
NARA [144]
Wouldn’t the inputs be 4 and 5? I could wrong though
I think You have to divide 159 by 11, and that would make you get 14.
We are given the following information concerning the three production machines;

Also, we are given the percentage of defective output as follows;

Therefore, if an item is selected randomly, the probability that the item is defective would be;

ANSWER:
The probability that the item is defective would be 0.0415