Ask question

Ask question

All categories

3 years ago

9

There are three monomials such that the greatest common factor of the first and second monomial is 2xy and the greatest common f

actor of the second and third monomials is 2x^2y

1 answer:

Fynjy0 [20]3 years ago

3 0

Hi, I'm an employer of Brainly. Someone will answer you in the next 10 minutes. Thank you for your patience! Have a great day.

You might be interested in

Fast I need to know what 20/22= x/11

sergij07 [2.7K]

Answer:

To Find => The value of x. in the equation below

=>

$20 \div 22 = x \div 11$

Step-by-step explanation:

We will cross Multiply the digits

=>20 \div 22 = x \div 20

=>We'll cross Multiply

=>

$20 \div 22 = x \div 11 \\ = > 22x = 20 \times 11 \\ = > 22x = 220 \\ = > x = 220 \div 22 \\ = > x = 10$

<em><u>Hence,</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u> </u></em><em><u>is </u></em><em><u>the</u></em><em><u> </u></em><em><u>required</u></em><em><u> </u></em><em><u>answer</u></em>

4 0

3 years ago

Number 6 help. <br> .......................

hichkok12 [17]

Since it says only write a expression, you can do 45xM= ?/?

6 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

7 pens and 2 pencils cost 22

Romashka [77]

Pen : $3

Pencil : $0.50

4 0

3 years ago

There are exactly 8 boys for every 2 girls. There is 46 girls how many boys are there?

expeople1 [14]

It would be 23 girls if you divide 46 by 2 since your trying to figure out the amount of boys for 46 girls.

8 0

3 years ago

Read 2 more answers