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algol [13]
2 years ago
6

An apartment building has the following distribution of apartments: 1 bedroom 2 bedroom 3 bedroom 1st floor 3 0 1 2nd floor 2 2

2 3rd floor 1 4 1If an apartment is selected at random, what is the probability that it is not on the 2nd floor or has 2 bedrooms?A) none of the given answerB) 1/9C) 14/16D) 1/3
Mathematics
1 answer:
Artyom0805 [142]2 years ago
7 0
1/3 because there are three floors.
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Answer:

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Step-by-step explanation:

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Solve for x. 3(3x - 1) + 2(3 - x) = 0 = 0
BARSIC [14]
If you would like to solve the equation 3 * (3 * x - 1) + 2 * (3 - x) = 0, you can calculate this using the following steps:

3 * (3 * x - 1) + 2 * (3 - x) = 0
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3 years ago
I NEED HELP!!!
andrey2020 [161]

Answer:

<h2>B. The amount of fuel used was lowest at speeds around 60km/h.</h2>

Step-by-step explanation:

The decreasing part is from (20,13) to (60,6), which gives an average rate of

r=\frac{6-13}{60-20}=-\frac{7}{40}=-0.175

Then, we have an increasing part from (60,6) to (120,10), which gives an average rate of

r=\frac{10-6}{120-60}=\frac{4}{60}=0.067

Basically, at 60 km/h the fuel used was 6 liters per 100 kilometers. Notice that the other velocities make the car to use more liters per kilometers.

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2 years ago
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p
Ymorist [56]

Answer:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,7)

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

B\sim Exp(\lambda=\frac{1}{30 min})

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

E(A +B)

Since the two random variables are assumed independent, then we have this

E(A+B) = E(A)+E(B)

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+7}{2}=4 hours

The expected value for the exponential distirbution is given by :

E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx

If we use the substitution y=\lambda x we have this:

E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}

Where X represent the random variable and \lambda the parameter. If we apply this formula to our case we got:

E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

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And in order to find the variance for the random variable A+B we can find the individual variances:

Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2

Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2

We have the following property:

Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)

Since we have independnet variable the Cov(A,B)=0, so then:

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

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3 years ago
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frosja888 [35]
My teacher said A and C
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