Is it true or false?? please help me.

The rate of growth of profit (in millions) from an invention is approximated by Upper P prime (x )equals x e Superscript negati

Alecsey [184]

Answer:

The function is $P(x) = \frac{1}{2} e^{-x^2} +0.016$

Step-by-step explanation:

From the question we are told that

The rate of growth is $P'(x) = xe^{x} - x^2$

The total profit is $P(2) =$ $25,000

The time taken to make the profit is $x = 2 \ years$

From the question

$P'(x) = xe^{-x^2}$ is the rate of growth

Now here x represent the time taken

Now the total profit is mathematically represented as

$P(x) = \int\limits {P'(x)} \, = \int\limits {xe^{-x^2}} \,$

So using substitution method

We have that

$u = - x^2$

$du = 2xdx$

So

$p(x) = \int\limits {\frac{1}{2} e^{-u}} \, du$

$p(x) = {\frac{1}{2} [ e^{-u}} +c ]$

$p(x) = {\frac{1}{2} e^{-x^2}} + \frac{1}{2} c$ recall $u = - x^2$ and let $\frac{1}{2} c = Z$

At x = 2 years

$P(x) =$ $25,000

So

Since the profit rate is in million

$P(x) =$ $25,000 = $\frac{25000}{1000000} =$ $0.025 millon dollars

So

$0.025 = {\frac{1}{2} e^{-2^2}} + Z$

=> $Z = 0.025 - {\frac{1}{2} e^{-2^2}}$

$Z = 0.016$

So the profit function becomes

$P(x) = \frac{1}{2} e^{-x^2} +0.016$

5 0

3 years ago

Solve for x. 10 m 42

kramer

Answer:

The measures of a pair of same-side exterior angles are 10w and 5v. ... Find the measure of angles 1-7 given that lines m and n are parallel and t is transversal. ... Solve for x. 120= 15x +5+ 22x +4. 120 = 37% +. 4. (X=3]. 6. Use the following figure, given that ... Find the measure of angle 1. m24 = 123, mZ1 = 2x, mZ2 = x +42.

Step-by-step explanation:

5 0

2 years ago

A flowerbed surrounds the base of a tree. It’s enclosed by stones to form a circle that measures 25.12 feet. What is the radius

Ymorist [56]

2.83 if rounded. 2.82842712475 would be the actual.

Hope this helped!

5 0

3 years ago

A research study investigated differences between male and female students. Based on the study results, we can assume the popula

garri49 [273]

Using the normal distribution and the central limit theorem, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of $\mu = 3.5$ .

The standard deviation is of $\sigma = 0.5$ .

Sample of 100, hence $n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05$

The interval that contains 95.44% of the sample means for male students is between Z = -2 and Z = 2, as the subtraction of their p-values is 0.9544, hence:

Z = -2:

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

By the Central Limit Theorem

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