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Maurinko [17]
3 years ago
11

Which of the following factors does not influence the customer when deciding to buy a product

Mathematics
1 answer:
posledela3 years ago
4 0

Answer: The correct answer is C. Weathering

Step-by-step explanation:

It's Weathering because it's very irrelevant and no body talks about weathering when telling the customer to buy a product.

* Hope this helps:) Mark me the brainliest.

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Naomi asked several classmates how much cash they had in their pocket. She recorded the data below:
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She arrived at school 23 minutes later. At what time did. Veronica arrive at school? (3.MD.A.1). 12. ... same way they have measured objects to the nearest inch: ... Students will then use line plots as a tool to record ... line plots your classmates ... Ask students to determine how many of ... Pedro has a dollar bill in his pocket.

Step-by-step explanation:

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Why do cowboys have so much trouble with math
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The Washingtons but a studio apartment for $240,000. They pay a down payment of $60,000. What percent of the purchase price woul
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Eastside middle school ​has 1,500 students. Thirty-two percent of them are in sixth grade. How many sixth-grade students are the
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6 0
3 years ago
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
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