A division of Ditton Industries manufactures the Futura model microwave oven. The daily cost (in dollars) of producing these mic
1 answer:
Answer:
The costs associated with the production are:
- For the 111st oven, it costs $114.07
- The 221st oven comes to $122.71
- The 311st oven amounts to $140.59
The marginal costs for x equal to 110, 220, and 310 are outlined as follows:
Step-by-step explanation:
The daily cost for manufacturing the microwave ovens can be expressed through the given function:
(a) To determine the actual cost of the 111st oven, subtract the total cost of producing the first 110 ovens from that of the first 111 ovens:
For the 221st oven:
For the 311st oven:
(b) If C(x) represents the total cost function, then the marginal cost function is its derivative, expressed as C'(x).
We need to calculate the derivative of the total cost function:
The marginal costs when x equals 110, 220, and 310 are represented by
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Response:
Detailed explanation:
Greetings!
You have the variable
X: Area eligible for painting with a can of spray paint (feet²)
This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²
As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.
a.
P(X>27)
The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².
The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:
X[bar]~N(μ;δ²/n)
To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999d.No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.
I trust this information is helpful!