Answer:
A. 1/5k - 2/3j and -2/3j +1/5k
Step-by-step explanation:
A. 1/5k - 2/3j and -2/3j +1/5k
B. 1/5k - 2/3j and -1/5k +2/3j
There is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
Not equivalent
C. 1/5k - 2/3j and 1/5j - 2/3k
There is a change in the variables
1/5k changed to 1/5j
-2/3j changed to -2/3k
D. 1/5k - 2/3j and 2/3j - 1/5k
The is a change in the signs of each term
1/5k changed to -1/5k
-2/3j changed to +2/3j
The only equivalent expression is
A. 1/5k - 2/3j and -2/3j +1/5k
= 2a - 1
2() = 2(2a - 1) <em>multiplied both sides by 2 </em>
ab = 4a - 2 <em>distributed the 2 on the right side</em>
ab - 4a = -2 <em>subtracted 4a from both sides</em>
a(b - 4) = -2 factored out "a" from the left side
a = 
<em>divided (b - 4) on both sides</em>
Answer: a =
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
