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2 years ago

14

Determine whether the graph of the equation is increasing or decreasing. Write the correct answer before the number. Show your s

olutions.
5y = x - 2

1 answer:

KATRIN_1 [288]2 years ago

5 0

Answer:

Increasing

Step-by-step explanation:

We are given the function:

$\displaystyle \large{5y = x - 2}$

Since y is the function of x; we isolate y by dividing both sides by 5.

$\displaystyle \large{ \frac{5y}{5} = \frac{x - 2}{5} }$

Thus:

$\displaystyle \large{ y = \frac{x - 2}{5} }$

Simplify the expression, separating the fraction.

$\displaystyle \large{ y = \frac{x}{5} - \frac{2}{5} }$

Familiar with this equation? This function is a linear function.

Now to the increasing and decreasing part. There are several ways to find whether if the graph is increasing.

By substituting values
Graph Visualization (Or look at the graph)

I will demonstrate the first method. Start from substituting negative to positive, if we keep substituting higher numbers and we get higher y-value then the graph is increasing.

If we substitute higher numbers but we get lower y-value then the graph is decreasing.

<u>S</u><u>u</u><u>b</u><u>s</u><u>t</u><u>i</u><u>t</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u>

y = -1/5 - 2/5 = -3/5

y = 0-2/5 = -2/5

y = 1/5-2/5 = -1/5

y = 2/5-2/5 = 0

y = 3/5-2/5 = 1/5

y = 4/5-2/5 = 2/5

y = 5/5-2/5 = 3/5

y = 6/5-2/5 = 4/5

...

As so on, as we see, when x keeps increasing, y increases too.

Therefore, the graph is increasing.

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Answer:

a) $P(X=3) = 0.1$

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And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

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And replacing we got:

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And the variance would be:

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And the deviation:

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We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

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Answer:

Step-by-step explanation:

Assuming that the differential equation is

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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,

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