Answer:
JK = 11
Step-by-step explanation:
JK ≅ JM
where
JK = 3x - 16
JM = 2x - 7
3x - 16 = 2x - 7
group like terms
3x - 2x = 16 - 7
x = 9
<u>plugin x=9 into JK</u>
JK = 3x - 16
JK = 3(9) - 16
JK = 11
Answer:
5 hours
Step-by-step explanation:
We can write a ratio
20 laps 100 laps
-------------- = --------------
1 hour x hours
Using cross products
20x = 100
Divide by 20
20x/20 = 100/20
x = 5 hours
Answer:
Amy is 1200 seconds fater than Bill.
Step-by-step explanation:
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 .
