Find The value of each variable

Pls help fast maths

-Dominant- [34]

Answer:

the answer of question number 4 is : (x+4)

6 0

2 years ago

Abby takes a 75-centimeter by 40-centimeter rectangle of fabric and cuts from one corner of

Grace [21]

answer: 195cm

the photo should explain

5 0

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:

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

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

4b+13/2 equivalent expressions

k0ka [10]

The equivalent expression of 4b+13/2 is (8b+13)/2.

Given an expression be 4b+13/2.

We are required to find the equivalent of the expression 4b+13/2.

Expression is combination of numbers, symbols, fractions, coefficients, determinants, indeterminants. In an expression we do not find any equal to sign. It shows some relationship. In this expression b is the only variable that exists. If we have been given some value of th expression at some point then we can get the value of variable.

The expression is 4b+13/2.

Taking LCM first.

The LCM of any two numbers is basically the value that is evenly divisible by the two given numbers.

=(8b+13)/2

Hence the equivalent expression of 4b+13/2 is (8b+13)/2.

Learn more about expressions at brainly.com/question/723406

#SPJ1

5 0

2 years ago

A baseball team had 75 players show up for tryouts last year and this year had 90 players show up for tryouts. Find the percent

olga_2 [115]

Answer: 33.3%

Step-by-step explanation:

First find the difference between the players in try-outs over the years.

= 90 - 75

= 25

Then use this difference to find the percent increase:

= Difference / Players last years * 100%

= 25 / 75 * 100%

= 33.3%

3 0

3 years ago