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3 years ago

What is the volume of this prism?

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

It’s b, I’m positive

Olenka [21]3 years ago

5 0

I’m pretty sure the correct answer the teacher is looking for is B

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How far can you run in 5/7 of an hour if you run at a speed of 8 2/5 mph?

Mazyrski [523]

Answer: you can run 6 miles

explanation: 8 2/5 divided by seven = 1.2 1.2x5 = 6

5 0

4 years ago

SOMEONE PLEASEEEE HELPPPP?!?!?!?!!?

Zarrin [17]

Answer:

Step-by-step explanation:

Top/Bottom = 12

Sides = 15

Front/Back = 20

20 + 15 + 12 = 47

47 / 1/2 = 23.5

8 0

3 years ago

Find the area of the following figure.

Zigmanuir [339]

Answer:

312

Step-by-step explanation:

Triangles

1/2 12 *10 = 6*10 = 60

there are two triangles meaning we have to multiply 60 by 2 = 120

Rectangle

12*16 = 192

Add the two and you have the area 120+192=312

5 0

3 years ago

How do you simplify the exponent

BigorU [14]

You just apply the ^4 to everything inside. It multiplies with the powers. of the numbers inside. 9^(1*4) * g^(1*4) * s^(4*4)

5 0

4 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:

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

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