The student activities department of a community college plans to rent buses and vans for a spring-break trip. Each bus has 40 r
egular seats and 1 handicapped seat; each van has 8 regular seats and 3 handicapped seats. The rental cost is $350 for each van and $975 for each bus. If 320 regular and 36 handicapped seats are required for the trip, how many vehicles of each type should be rented to minimize cost
2 answers:
Answer:
6 buses and 10 vans
Step-by-step explanation:
It is convenient to solve this graphically. The problem setup is ...
minimize 975b + 350v subject to ...
40b +8v ≥ 320 . . . . . regular seats required
1b +3v ≥ 36 . . . . . . . .handicapped seats required
where b and v are the numbers of buses and vans, respectively.
The constraint lines intersect at (b, v) = (6, 10). 6 buses and 10 vans will provide the required number of seats at minimum cost.
_____
At any other point in the feasible region, the cost line (green) will be farther from the origin, hence the cost will be greater.
In the attached, x represents buses; y represents vans.
Answer:
10 vans and 6 buses.
Step-by-step explanation:
Minimize cost = 350*x + 975*y
subject to
40*y + 8*x >= 320 (regular seats needed)
y + 3*x >= 36 (handicap seats needed)
x >= 0
y >= 0
where
x, number of vans rented
y, number of buses rented
This is a linear programing problem. In this kind of problem, it is known that the optimal solution is on a vertix of the feasible region. In the figure attached, feasible region can be seen. For the vertices costs are
vertex cost
(0, 36) $35100
(10, 6) $9350
(40, 0) $14000
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Answer:
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income
Step-by-step explanation:
We use the binomial aproxiation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
. So
In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?
We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.
X = 34
has a pvalue of 0.9993
X = 21
has a pvalue of 0.5239
0.9993 - 0.5239 = 0.4754
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income