Solve for the system of liner equations by substitution. <br> X+4y= -1 <br> -3x-14=y

ted took a trip to china. upon leaving he decided to convert all his Yuan back into dollars. how many dollars did he receive if

romanna [79]

Ted would receive 2 dollars

5 0

2 years ago

I NEED HELP WITH ME MATH

ruslelena [56]

ok so lets start of with the fact that the whole thing is 21 units.

So we do 21-9=12 so then we now know that the rectangle is 12 so then we do 12*8 units which is= 96 . Now the triangles. so the first one we know that its 8*3 and times it by 1/2 because two triangles is equal to a rectangle so the first triangle is 12 and now the second. its 9-3= 6 then its 6*8*1/2 which is equal to 24 so now the final answer is

96+12+24 which is equal to 132 so i'm guessing it 132 square units

5 0

2 years ago

Find the lateral surface area of this

Allushta [10]

Answer:226.2

Step-by-step explanation

Lateral= 2πrh

H=12

R=3

2*π*3*12=226.19

Round to nearest 10th is 226.2

6 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: