The CFO for the Shelton Corporation has $1.2 million to allocate to the following budget requests from five departments: Dept. 1 Dept. 2 Dept. 3 Dept. 4 Dept. 5
$450,000 $310,000 $275,000 $187,500 $135,000
Because the total budget requests exceed the available $1.2 million, not all the requests can be satisfied. Suppose the CFO considers the requests for departments 2 and 3 to be twice as important as those from departments 4 and 5, and the request from department 1 to be twice as important as those from departments 2 and 3. Fur- ther suppose the CFO wants to make sure each department receives at least 70% of the requested amount.
a. Formulate a GP model for this problem.
b. Implement your model and solve it. What is the optimal solution?
c. Suppose the CFO is willing to allocate more than $1.2 million to these budgets but regards exceeding the $1.2 million figure as being twice as undesirable as not meeting the budget request of department 1. What is the optimal solution?
d. Suppose the CFO regards all deviations from the original budget amounts (including the $1.2 million available) to be equally undesirable. What solution minimizes the maximum percentage deviation from the budgeted amounts?

it will be a cylinder. with a diameter of 8 and 4 single pieces high

step-by-step explanation:

days of the week

step-by-step explanation:

javier earned more money then kyle, so that won't always be a constant term, but the days of the week do not change so that term is always constant.

edit: i meant

a long term investor

step-by-step explanation:

because he is a long term investor

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.