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3 years ago

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The profits in hundreds of dollars, P(c), that a company can make from a product is modeled by a function of the price, c, they

charge for the product: P(c) = –20c2 + 320c + 5,120. What is the maximum profit the company can make from the product?

2 answers:

postnew [5]3 years ago

7 0

Answer:

The correct answer is B. 640,000

Step-by-step explanation:

Jobisdone [24]3 years ago

4 0

The maximum profit is the y-coordinate of the vertex of the parabola represented by the equation.

$P(c)=-20c^2+320c+5120 \\ \\
a=-20 \\
b=320 \\ \\
\hbox{the vertex } (h,k): \\
h=\frac{-b}{2a}=\frac{-320}{2 \times (-20)}=\frac{-320}{-40}=8 \\
k=f(h)=f(8)=-20 \times 8^2+320 \times 8+5120= \\
=-1280+2560+5120=6400$

The maximum value is 6400, but the profit is given in hundreds of dollars, so multiply the value by 100.

The maximum profit the company can make is $640,000.

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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

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Answer:

$-\frac{21}{2}$

Step-by-step explanation:

We can solve this system just by summing each side of the equation:

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5.2.14. For the negative binomial pdf p (k; p, r) = k+r−1 (1 − p)kpr, find the maximum likelihood k estimator for p if r is know

Volgvan

Answer:

$\hat p = \frac{r}{\bar x +r}$

Step-by-step explanation:

A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".

And the probability mass function is given by:

$P(X=x) = (x+r-1 C k)p^r (1-p)^{x}$

Where r represent the number successes after the k failures and p is the probability of a success on any given trial.

Solution to the problem

For this case the likehoof function is given by:

$L(\theta , x_i) = \prod_{i=1}^n f(\theta ,x_i)$

If we replace the mass function we got:

$L(p, x_i) = \prod_{i=1}^n (x_i +r-1 C k) p^r (1-p)^{x_i}$

When we take the derivate of the likehood function we got:

$l(p,x_i) = \sum_{i=1}^n [log (x_i +r-1 C k) + r log(p) + x_i log(1-p)]$

And in order to estimate the likehood estimator for p we need to take the derivate from the last expression and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\frac{x_i}{1-p}$

And we can separete the sum and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}$

Now we need to find the critical point setting equal to zero this derivate and we got:

$\frac{dl(p,x_i)}{dp} = \sum_{i=1}^n \frac{r}{p} -\sum_{i=1}^n \frac{x_i}{1-p}=0$

$\sum_{i=1}^n \frac{r}{p} =\sum_{i=1}^n \frac{x_i}{1-p}$

For the left and right part of the expression we just have this using the properties for a sum and taking in count that p is a fixed value:

$\frac{nr}{p}= \frac{\sum_{i=1}^n x_i}{1-p}$

Now we need to solve the value of $\hat p$ from the last equation like this:

$nr(1-p) = p \sum_{i=1}^n x_i$

$nr -nrp =p \sum_{i=1}^n x_i$

$p \sum_{i=1}^n x_i +nrp = nr$

$p[\sum_{i=1}^n x_i +nr]= nr$

And if we solve for $\hat p$ we got:

$\hat p = \frac{nr}{\sum_{i=1}^n x_i +nr}$

And if we divide numerator and denominator by n we got:

$\hat p = \frac{r}{\bar x +r}$

Since $\bar x = \frac{\sum_{i=1}^n x_i}{n}$

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