Let us first write down the given equation.
80 = 3y + 2y + 4 + 1
80 = 5y + 5
80 - 5 = 5y
75 = 5y
Just reversing both sides we get
5y = 75
Now dividing both sides by 5 we get
y = 15
So the value of y is 15.
I hope the procedure for solving the problem is clear to
you. In future you can solve such problems with ease following the procedure
described.
Answer: a=19
Step-by-step explanation: In order to solve this question, we must first take out the parenthesis, we can do this by taking 1/3 and multiply it by a and 5. This will give us 8=1/3a+5/3, with this we can subtract 5/3 on both sides of the equation, which will then give us, 19/3=1/3a. Last, we can divide 19/3 by 1/3 giving us 19 as our final anwser. a=19
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 .
