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3 years ago

HELP PLEASE ASAP

2 answers:

4 0

The answer to question 1 is b.
The answer to question 2 is b.
The answer to question 3 is c.
The answer to question 4 is c.
The answer to question 5 is b.
The answer to question 6 is a.
The answer to question 7 is b.
The answer to question 8 is a.
The answer to question 9 is c.
The answer to question 10 is a.

charle [14.2K]3 years ago

3 0

To do this, you need to solve either one of the equations for y or x. I'm going to solve for y in the first equation:

6x + 3y = 9

Subtract 6x from both sides:

3y = 9 - 6x

Divide both sides by 3:

y = 3 - 2x

Now that you know what y is equal to, you can substitute it into the second equation:

2x + 3y = 1

2x + 3(3-2x) = 1

Distribute the 3:

2x + 9 - 6x = 1

Combine like terms:

9 - 4x = 1

Subtract 9 from both sides:

-4x = -8

Divide by -4:

x = 2

No substitute this 2 into either one of the equations to get y. I'm going to use the first:

6x + 3y = 9

6(2) + 3y = 9

12 + 3y = 9

Subtract the 12 from both sides:

3y = -3

Divide by 3:

y = -1

ANSWER:

x = 2

y = -1

You can always check by substituting the values into the equations.

Read more on Brainly.com - brainly.com/question/3461332#readmore

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A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p

andreyandreev [35.5K]

Answer:

B) 0.894

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that $\mu = 0.55$

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577$

$P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894$

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3 years ago

300x+290 factor out the gfe of each expression

marta [7]

Answer: 590x.

2 is GFE, hope this helps?

Step-by-step explanation:

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2 years ago

Millie works 8 hours on Saturday and earns £6.50 an hour. Her bus fare to work costs £2.75 and she also spends £4.25 on her lunc

Nimfa-mama [501]

The answer is 7.50 ......

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3 years ago

Suppose that $2000 is invested at a rate of 2.6% , compounded semiannually. Assuming that no withdrawals are made, find the tota

EleoNora [17]

Answer:

$2,589.52

Step-by-step explanation:

$A = P(1 + \dfrac{r}{n})^{nt}$

We start with the compound interest formula above, where

A = future value

P = principal amount invested

r = annual rate of interest written as a decimal

n = number of times interest is compound per year

t = number of years

For this problem, we have

P = 2000

r = 0.026

n = 2

t = 10,

and we find A.

$A = $2000(1 + \dfrac{0.026}{2})^{2 \times 10}$

$A = $2589.52$

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3 years ago

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OleMash [197]

Answer:

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3 years ago

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