Simplify: (2 x^5 5 y^3)^3

daser333 [38]

Answer:

The monopolist's net profit function would be:

$N(y)=198\,y\,-\,2.5\,y^2$

Step-by-step explanation:

Recall that perfect price discrimination means that the monopolist would be able to get the maximum price that consumers are willing to pay for his products.

Therefore, if the demand curve is given by the function:

$P(y)=200-2y$

P stands for the price the consumers are willing to pay for the commodity and "y" stands for the quantity of units demanded at that price.

Then, the total income function (I) for the monopolist would be the product of the price the customers are willing to pay (that is function P) times the number of units that are sold at that price (y):

$I(y)=y*P(y)\\I(y)=y\,(200-2y)\\I(y)= 200y-2y^2$

Therefore, the net profit (N) for the monopolist would be the difference between the Income and Cost functions (Income minus Cost):

$N(y)=I(y)-C(y)\\N(y)=(200\,y-2y^2)-(2y+0.5y^2)\\N(y)=198\,y\,-\,2.5\,y^2$

5 0

2 years ago

What's 16 over 8 in a mixed number

olya-2409 [2.1K]

16/8= 2
2 is a whole number

4 0

3 years ago

Read 2 more answers

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .