Aurelio can choose between 3 pairs of pants and 4 shirts. How many different outcomes are in the sample space?

Please I need it for today

DaniilM [7]

Answer:

Which one?

Step-by-step explanation:

7 0

3 years ago

The 7th grade has 220 tudents. There are 3/5 girls. How many boys are there in 7th grade

Helga [31]

Answer:

88

Step-by-step explanation:

Let's see - if there are 3/5 girls, there must be 2/5 boys. Assuming everyone in the class is binary, of course.

So 2/5 of 220 is the number of boys. When you see "of," always think multiplication!

2/5 of 220

2/5 * 220 = 88 boys in 7th grade

3 0

3 years ago

In winter, the price of apples suddenly went up by $0.75 per pound. Sam bought 3 pounds of apples at the new price for a total o

Vilka [71]

Answer:

look down

Step-by-step explanation:

sam bought 3 pounds so 0.75 multiplied with 3 is 2.25 then we subtract 5.88 by 2.25 to get the original price for 3 pounds which is 3.63. Last step we divide 3.63 by 3 to get the original price for 1 pound and the answer should be 1.21 per pound.

The final equation should be $5.88- $2.25÷3= $1.21

8 0

3 years ago

Read 2 more answers

A bag contains ten identical blue marbles and ten identical green marbles. In how many distinguishable ways can five of these ma

ivann1987 [24]

Answer: 11 different rows.

Step-by-step explanation:

As the marbles are identical, we do not really care for permutations (as we can really not difference them)

If we have 5 blue marbles, we have only one combination.

B-B-B-B-B

If we have 4 blue marbles, we have 3 combinations:

B-B-B-B-G, G-B-B-B-B, B-B-G-B-B

This is because the blue marbles need to be next to another blue one, so from here we can do the same analysis.

If we have 3 of them, we have 3 combinations.

B-B-B-G-G, G-B-B-B-G, G-G-B-B-B

If we have 2 of them, we have 4 combinations

B-B-G-G-G, G-B-B-G-G, G-G-B-B-G, G-G-G-B-B

then we have 1 + 3 + 3 + 4 = 11 combinations

7 0

3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

a.

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

6 0

3 years ago