All categories

3 years ago

How do I solve this ?

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

6 0

Answer:

See attachment :D

Step-by-step explanation:

y = 2/3x + 2

This is in slope-intercept form, y = mx+b. m is the slope, 2/3, and b is the y- intercept, 2. So you know the line intersects with the y-axis at (0,2). The slope is 2/3. So you can move up two points, then over three. Because it's a positive slope, you move right. This leaves you at point (0,3). Connect a line between these two points.

The line intersects with points (-3,0) and (0,2).

Hope this is helpful ^^ Good luck with your test :D

You might be interested in

HELP PLEASE........ on these 2

tester [92]

For y=4 draw a horizontal line through the y-axis at point (0,4). Y-axis is the vertical.
For x=-3 draw a vertical line through the x-axis at point (-3,0). X-axis is the horizontal.

8 0

3 years ago

Do you distribute or use exponents first

Temka [501]

My math teacher time me is easier to do it after

3 0

3 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

A right circular cylinder has a volume of 500 cu in. If the base has a radius of 4 in., what's the altitude of the cylinder? Rou

GarryVolchara [31]

Cylinder Volume = π • r² • height
height = Cylinder Volume / (PI * radius^2)
height = 500 cubic inches / 3.14159 * 16
height = 9.9471839432 
height = 9.95 (rounded)

4 0

3 years ago

If 60% of a number is 8, find 30% of that number.

podryga [215]

60/100 = 8/x

Multiply both sides by 100

60 = 800/x

Multiply both sides by x

60x = 800

Divide both sides by 60

x = 13.3333333333333

0.3 • x = answer

Answer = 4

6 0

2 years ago