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2 years ago

7

Write the standard form of the line that passes through the given points. Include your work in your final answer. Type your answ

er in the box provided to submit your solution.
(7, -3) and (4, -8)

1 answer:

I am Lyosha [343]2 years ago

6 0

Answer:

|x y 1|

|7 -3 1| = 0

|4 -8 1|

x*(-3*1 - (-8)*1) - y*(7*1 - 4*1) + 1*(7*-8 - 4*-3) = 0

5x - 3y - 44 = 0

Step-by-step explanation:

You might be interested in

The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,

uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

$\frac{500-100}{100} = A}$ and $A=4$ .

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$ .

Finally, as we want the value of $t$ such that $P(t)=200$ , we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$ .

This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$ .

So, the population of rats will be 200 after 25 months.

6 0

3 years ago

A woman walks 5 rounds along the border of a square field of area 49². Find the total distance she walks in m.

Luba_88 [7]

You need to estimate the square root of 200 ft². Rewrite 200 as 2 x 100, then square root of (2*100) is:

10*sqrt(2). The sqrt(2) = 1.414, so the length of each side of the square will be 10*1.414 = 14.14 ft or about 14 ft.

4 0

2 years ago

Read 2 more answers

Please help! Show all work.

Morgarella [4.7K]

First, let's work with the denominator alone:

4⁻⁴ = 1 / 4⁴ = 1/256 .

The original problem is the reciprocal of that = 256.

8 0

3 years ago

Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me

Alexus [3.1K]

Answer:

a) 18.94% probability that the sample mean amount purchased is at least 12 gallons

b) 81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c) The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we can apply the theorem, with mean $\mu$ and standard deviation $s = \sqrt{n}*\sigma$

In this problem, we have that:

$\mu = 11.5, \sigma = 4$

a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?

Here we have $n = 50, s = \frac{4}{\sqrt{50}} = 0.5657$

This probability is 1 subtracted by the pvalue of Z when X = 12.

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

By the Central Limit theorem

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

$Z = \frac{12 - 11.5}{0.5657}$

$Z = 0.88$

$Z = 0.88$ has a pvalue of 0.8106.

1 - 0.8106 = 0.1894

18.94% probability that the sample mean amount purchased is at least 12 gallons

b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.

For sums, so $mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28$

This probability is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 575}{28.28}$

$Z = 0.88$

$Z = 0.88$ has a pvalue of 0.8106.

81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

$1.645 = \frac{X- 575}{28.28}$

$X - 575 = 28.28*1.645$

$X = 621.5$

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

5 0

3 years ago

Tell weather the given value is a solution of the inequality yes or no

Hitman42 [59]

Yes it is a solution, 8-3 is less than or equal to 9

3 0

3 years ago