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denis-greek [22]
3 years ago
12

Sammy has 2 pairs of jeans, 3 pairs of black pants, and 1 skirt. She also has 2 pink shirts, 4 striped shirts, and 2 hooded swea

tshirts. What's the probability Sammy would wear black pants and a striped shirt?
Mathematics
1 answer:
Semenov [28]3 years ago
8 0

I'm not too sure if this a a question for school but i think it really depends on what Sammy wanna wear. I would also say she may wear black pants and a striped shirt because of the amount she has of them. All the other clothes have only two pairs, So because she has more than two pairs for black pants and striped, it means that she likes that outfit of the black pants and striped shirt so she will most likely wear them.

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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:

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

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

3 0
3 years ago
Rewrite the expression 2(x + 5) using the Distributive Property.​
ehidna [41]

Answer:

2x + 10

Step-by-step explanation:

To expand (using the distributive property) <u>multiply</u> the number outside the bracket i.e. in this case '2', with the <u>values inside the brackets</u>.

So multiply '2' and 'x' and '2' and 5' and add or subtract on basis of whether the second value is positive or negative.

So

2(x + 5)

= (2*x)+(2*5)

=2x+10

<em>extention note:</em> <u>be careful</u> when the symbol within the equation within the brackets is a subtraction because it implies that the second value would instead be a negative number and should be treated as such.

an example

2(x-5)

= (2*x)+(2*-5)

=2x -10

Anyhow, I hope this helped!

5 0
3 years ago
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