Ask question

Ask question

All categories

3 years ago

8

There are 1265 students at Cypress Middle School. 42% of the students are enrolled in PE. Which equation could be used to repres

ent the number of students enrolled in PE?

1 answer:

Dovator [93]3 years ago

7 0

Answer:

0.42* 1265=x

Step-by-step explanation:

There are 1265 students in Cypress Middle School. So we already know one part of the equation. 42% of these students are in PE. So we know that 42% of these 1265 students are in PE. Of these means you probably have to multiply 42% and 1265. 42% = 0.42. So to write an equation we take the 0.42 and 1265 and put it together. Therefore, 0.42 times 1265=x.

You might be interested in

<img src="https://tex.z-dn.net/?f=%20-%2016%281.2%29%5E%7B2%7D%20%2B%2038.4%281.2%29%20%2B%200.96" id="TexFormula1" title=" - 16

Elina [12.6K]

24 is your final answer

6 0

3 years ago

Read 2 more answers

6. All angles in the figure below are right angles. What is the area of the figure? (1 point)

lilavasa [31]

Answer:

Option (4)

Step-by-step explanation:

Area of the figure given in the picture = Area of large rectangle - Area of rectangle A

Area of rectangle A = Length × Width

= [9 - (2 + 4)] × 3

= 3 × 3

= 9

Area of the large rectangle = Length × Width

= 9 × 5

= 45

Therefore, area of the given figure = 45 - 9

= 36

Option (4) will be the correct option.

5 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

The probability that a normal random variable is less than its mean is ____. 1.0 0.5 0.0 Cannot be determined?

Ber [7]

The correct answer for the question that is being presented above is this one: "0.5" <span>The probability that a normal random variable is less than its mean is 0.5. In a normal distribution, 1.0 refers to the one that is stable and is in equilibrium.</span>

7 0

3 years ago

In a triangle , the measures of the angles are x,x + 20 & 2x . What is the value of x ?

Alexus [3.1K]

X+x+20+2x=180
4x+20=180
4x=160
x=40

5 0

3 years ago