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3 years ago

13

3 people can build a shed in 8 hours. How long would it take for 5 people to build a shed? Give your answer in minutes.

1 answer:

kolezko [41]3 years ago

7 0

3 people x 8 hours is 24 hours / 5 people = 4 hours and 48 minutes

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Write the number 3.8 in the forum of a/b using integers to show that it is a rational number

Naya [18.7K]

The answer is b. 38 divided by 10 equals 3.8

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3 years ago

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Ms. Clint is comparing the sales of sweaters and gloves at her store for the past ten winter weeks. Select the true statement ba

lianna [129]

Answer:

i would say b

Step-by-step explanation:

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3 years ago

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Enter the lengths of QS and TV<br><br>QS=<br>TV=

kicyunya [14]

3v + 2 = 7v
2 = 4v
1/2 = v

QS = 3v + 2
= 3(1/2)+ 2
= 1.5 + 2
= 3.5

TV = 7v
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3 years ago

Find the approximate perimeter of ABC plotted below.

maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}$

$AB = \sqrt{4 + 4}$

$AB = \sqrt{16}$

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}$

$BC = \sqrt{25 + 49}$

$BC = \sqrt{74}$

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}$

$AC = \sqrt{47 + 25}$

$AC = \sqrt{74}$

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 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

4 years ago