What's a very complex equation that equals 3

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago

The graph of the even function F(X) Has 5 x-intercepts if (6,0) is the one of the intercepts what set of points can be the other

alex41 [277]

f(x) being even means

f(x) = f(-x)

So the zeros come in positive and negative pairs. If there are an odd number of intercepts like there are here, it's because one of them is x=0 which is its own negation.

Given zero x=6 we know x=-6 is also a zero.

So we know three zeros, and know the other two zeros are a positive and negative pair.

The only choice with (-6,0) and (0,0) is A.

Choice A

3 0

3 years ago

Read 2 more answers

You pick one card from a standard deck of 52 cards. What is the probability that the card will be a two?

Natali [406]

Answer:

4 out of 52 or simplified 2 out of 26

Step-by-step explanation:

There are 4 suites of cards, spades, hearts, clubs, diamonds and there is one card for two in each suite, so we therefore know that the probability of picking a two out of a 52 card deck will be four out of 52 or the simplified answer would be 2 out of 26. Hope this helps, have a great day!

3 0

4 years ago

businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message

KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let X = number of text messages receive or send in an hour.

The random variable X follows a Poisson distribution with parameter λ.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: $\lambda=\frac{62.7}{24}= 2.6125$ .

The probability of a random variable can be computed using the formula:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...$

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

$P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180$

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667$

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0

3 years ago

PLEASE HELP DUE SOON MORE POINTS IF CORRECT!!

SVETLANKA909090 [29]

Answer:

y=4x+35

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers