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3 years ago

Simplify: -8c + w + 6c - 5w I will give Brainliest and Thanks

Mathematics

2 answers:

Gekata [30.6K]3 years ago

5 0

Answer:

The answer is -2c - 4w

Step-by-step explanation:

You have to combine the c's and the w's and you get -2c and -4w

artcher [175]3 years ago

4 0

Step 1 : Combine like terms

−8 + + 6 - 5

2 + − 5

Step 2: Combine the like terms again

2 + − 5

−2c − 4w

Solution: −2c − 4w

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Corrine picked 1/4 gallon of blackberries. She poured the berries equally into 4 containers. What fraction of a gallon is in eac

Eduardwww [97]

Answer:

1/16

Step-by-step explanation:

In the starting, Corrine had 1/4 gallon of blackberries. Now, she poured the berries equally into 4 containers. Therefore, the amount of berries in each container is:1/4 : 4=1/16

Thus, each container has 1/16 of a gallon of blackberries.

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2 years ago

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The events committee buys 100 flowers for a school dance the flowers are a combination of carnations an roses. Each carnation co

Natali5045456 [20]

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We don't know the price of the roses, so there is not enough information to answer.

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3 years ago

Suppose a certain population satisfies the logistic equation given by dP

Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

$\frac{dP}{dt}=10P-P^2$

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

$\frac{dP}{10P-P^2}=dt$

Integrate both sides.

$\int \frac{dP}{10P-P^2}=\int dt$ .... (1)

Using partial fraction

$\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}$

$A=\frac{1}{10},B=\frac{1}{10}$

Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

$\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt$

On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

$\frac{1}{10}(\ln \frac{P}{10-P})=t+C$

We have P(0)=1

Substitute t=0 and P=1 in above equation.

$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

$\frac{1}{10}(\ln \frac{1}{9})=C$

The required equation is

$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

Multiply both sides by 10.

$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

$e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}$

$\frac{P}{10-P}=\frac{1}{9}e^{10t}$

Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

The population when t = 3 is

$P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}$

Using calculator,

$P=9.999\approx 10$

Therefore, the population when t = 3 is 10.

8 0

2 years ago

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

3 years ago

Y=-2x - 2<br> y=7x - 20<br> Solve each by substitution

valina [46]

Answer:

(2, -6)

Step-by-step explanation:

Because each equation is equal to y, you can plug the y equal to each other. Because one y is equal to -2x-2 and the other is 7x-20, you can set them equal to each other. So you should have

-2x - 2 = 7x - 20

The next step is to solve for x, which mean that x is 2, then plug 2 into one of the original equations ( it doesn't matter which one) and solve for y, and when you do that, you will get -6

6 0

2 years ago