All categories

4 years ago

What is missing to make this true? 2x^2+4x prove it with the distributive property

1 answer:

5 0

Take you the common factors. those are 2 and x. So if you divided 2x^2 by 2x you end with only one x. and if you divided 4x by 2x you end with 2. The final answer should look like 2x(x+2)=2x^2+4x

You might be interested in

If f(x)=7x-5, find f-|(16) ?<br>

svet-max [94.6K]

Answer:

F-16 cuz of the absolute sign

3 0

3 years ago

The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Ro

QveST [7]

Answer:

a) There is a 16.0623% probability that 5 messages are received in 1 hour.

b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.

c) There is a 22.4042% probability that 2 messages are received in 0.5 hours.

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 time interval.

In this problem, we have a mean of 6 messages per hour.

(a) What is the probability that 5 messages are received in 1 hour?

Find the value of P when $x = 5$ and $\mu = 6$

$P(X = 5) = \frac{e^{-6}*(6)^{5}}{(5)!} = 0.160623$

There is a 16.0623% probability that 5 messages are received in 1 hour.

(b) What is the probability that 10 messages are received in 1.5 hours?

The mean is 6 messages in one hour.

For 1.5 hours, the mean is 6*1.5 = 9 messages.

We have to find the value of P when $x = 10$ and $\mu = 9$ .

$P(X = 10) = \frac{e^{-9}*(9)^{10}}{(10)!} = 0.115880$

There is a 11.5880% probability that 10 messages are received in 1.5 hours.

(c) What is the probability that less than 2 messages are received in 1/2 hour?

The mean is 6 messages in one hour.

For 0.5 hours, the mean is 6*0.5 = 3 messages.

We have to find the value of P when $x = 2$ and $\mu = 3$ .

$P(X = 10) = \frac{e^{-3}*(3)^{2}}{(2)!} = 0.224042$

There is a 22.4042% probability that 2 messages are received in 0.5 hours.

6 0

3 years ago

Question 7 of 10

Vsevolod [243]

Answer:

g(x) = –3(x + 6)2 + 48.

Step-by-step explanation:

The graph of f(x) = x2 is shifted right 6 units. The graph of f(x) = x2 is shifted down 48 units. The graph of f(x) = x2 is reflected over the y-axis.

6 0

3 years ago

Read 2 more answers

Emily can rake her lawn in 5 hours. Andrew can rake the same lawn in 8 hours. Rounded to the nearest hour, how long will the job

Dmitry_Shevchenko [17]

Emily can rake 1/5 of the lawn in an hr.
Andrew can rake 1/8 of the lawn in an hr.

Together, they rake 1 yardful. x represents time in hours.
(1/5)x+ (1/8)x= 1

Multiply all values by 40
8x+5x=40
13x=40
x= 40/13
x=3.0769...

Final answer: 3 hours

8 0

3 years ago

You have a job and you make $11.45 per hour. If your total check for the week was $171.75, how many hours did you work this week

zaharov [31]

Answer: you have worked 15 hours

Step-by-step explanation: i multiplied $11.45 and 15 which got me to $171.75 your welcome ;)

6 0

3 years ago