Answer:
x= -2
Step-by-step explanation: If you feel skeptical about my answer, then you should plug -2 back into the equation to check if it's correct. And it's correct for me when I did it.
Answer:
Find annual profit: $75,000/6 = $12,500
ROI = Annual Profit/ initial investment
ROI = $12,500/$15,000 or 83.3%
hope that help
You have to make a cos graph that starts its minimum and of -2, has an amplitude of 10, a period of 10 and a maximum of 18.
I decided to use a cos graph since cos graphs start at their minimum or maximum unlike a sine graph that starts halfway between the minimum and maximum. You also know the amplitude has to be 10 since 18+2=20 and 20/2=10. We were also told that the water wheel completels a rotation every 10 minutes which means the period is 10 minutes.
lets start of with a regular cos(x) graph. This starts on its maximum instead of minimum so we have to multiply it by -1 to get -cos(x) which does start on its minimum.
-cos(x) has an amplitude of 1 instead of 10, to fix that we multiply it by 10 to get -10cos(x) which has an amplitude of 10.
-10cos(x) has a period of 2π instead of 10, to fix this we multiply the x by 2π/10 to get -10cos((π/5)x) which now has a period of 10.
-10cos((π/5)x) has a minimum of -10 and maximum of 10 instead of a minimum of -2 and maximum of 18, to fix this we add 8 to -10cos((π/5)x) to get -10cos((π/5)x)+8 which does have a minimum of -2 and maximum of 18.
Therefore the answer is y=-10cos((π/5)x)+8. x being time in minutes and and y being the height in feet.
I hope this helps. Let me know if anything is unclear.
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 .
