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3 years ago

6x – у = -9 —12х – 4 y = 0

Mathematics

1 answer:

Bumek [7]3 years ago

4 0

Answer:

-15

Step-by-step explanation:

6x-y=-9

6x-(15)=-9

-12x-4y=0

-12x-4(3) =0

-12-12=0

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Simplify 81^5 = 3x x=?

dangina [55]

Um
81^5=3x
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*BRAINLIEST* (Easy Question: Warm UP) Find a positive solution to each equation: x2 = 9/4?

prisoha [69]

Answer:

A

Step-by-step explanation:

Given

x² = 9/4

Solving

Take the square root on both sides
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Option A is the correct option.

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Mike can stitch 7 shirts in 42 hours. he can stitch 1 shirt in how many hours, and in 1 hour he can stitch how much of a shirt.

podryga [215]

Answer:

Mike can stitch one shirt in 6 hours and in 1 hour he can stitch $\frac{1}{6}$ (0.167) shirts.

Step-by-step explanation:

Mike can stitch 7 shirts in 42 hours

to calculate the time that he takes to make 1 shirt, we have to divide 42 by 7.

He can stitch 1 shirt in = $\frac{42}{7}$ = 6 hours

Now we calculate the work done in one hour.

He can stitch in 1 hour = $\frac{1}{6}$ shirts or 0.167 shirts

Mike can stitch one shirt in 6 hours and in 1 hour he can stitch $\frac{1}{6}$ shirts.

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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:

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

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4 years ago

What if there is more than one variable?

Nadya [2.5K]

Answer:

a10 b10

Step-by-step explanation:

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