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svet-max [94.6K]
3 years ago
15

The cost to rent a moving van is $23 plus as a additional $7 per hour. For $37, what is the maximum amount of time that a moving

van can be rented for?
Mathematics
1 answer:
lapo4ka [179]3 years ago
5 0

Given:

Fixed cost to rent a moving van = $23

Additional cost = $7 per hour.

To find:

For $37, the maximum amount of time that a moving van can be rented for.

Solution:

Let the moving van be rented of t hours.

Total cost = Fixed cost + Additional cost

\text{Total cost}=23+7t

For $37, the total cost should be less than 37.

23+7t\leq 37

7t\leq 37-23

7t\leq 14

Divide both sides by 2.

t\leq 2

Since maximum value of t is 2, therefore the moving van can be rented for maximum 2 hours.

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A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let
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Complete Question:

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has the pmf shown in the table below. Compute E(X) and V(X).

Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left.

NOTE: The pmf is shown in the file attached

Answer:

Expected number of batches, E(X) = 2.0 batches

Variance of batches, V(X) = 0.8 batches²

Expected number of pounds left, E(Y) = 90 lb

Variance of pounds left, V(Y) = 20 lb²

Step-by-step explanation:

The company has 100 batches and it sells to customers in 5 lb batches. The number of pounds left after each customer is filled will be modeled by the mathematical formula: Y = 100 - 5X

The expected value of X, E(X) can be calculated using the formula:

E(X) = \sum x p(x)

E(X) = (1*0.3) + (2*0.5) + (3*0.1) + (4*0.1)

E(X) = 0.3 + 1 + 0.3 + 0.4

E(X) = 2.0

Expected number of batches, E(X) = 2.0 batches

The variance of X, V(X) can be calculated using the formula:

V(X) = E(X^2) - [E(X)]^2\\ E(X^2) = \sum x^2 p(x)\\  E(X^2) = (1^2 *0.3) + (2^2 * 0.5) + (3^2 * 0.1) + (4^2 * 0.1)\\  E(X^2) = 0.3 + 2 + 0.9 + 1.6\\ E(X^2) = 4.8\\V(X) = 4.8 - 2^2\\V(X) = 0.8

Variance of batches, V(X) = 0.8 batches²

Y = 100 - 5X

E(Y) = E(100 - 5X)

E(Y) = 100 - 5E(X)

E(Y) = 100 - 5(2)

E(Y) = 90 lb

Expected number of pounds left, E(Y) = 90 lb

Variance of the number of pounds left:

V(Y) = V(100 - 5X)

V(Y) = V(100) + V(-5X)

V(100) = 0

V(Y) = (-5)² V(X)

V(Y) = 25 * 0.8

V(Y) = 20

Variance of pounds left, V(Y) = 20 lb²

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