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3 years ago

Create a model to show that 2(2x+3y)=4x+6y

1 answer:

4 0

You would first do 2 multiplied by 2x and that would give you 4x, then you would do 2 multiplied by 3y which would give you 6y, you would put 4x under 2x and put 6y under 3y and bring down your addition symbol

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Please help me with this. i don't understand it.

Vika [28.1K]

Answer:

A. 11160 cubes B. 1395cm^3

Step-by-step explanation:

Volume= length x width x height

Its easiest to solve B. first and use it to solve A.

B. 7.5 x 12 x 15.5 = 1395cm^3 (volume of prism)

A. .5 x .5 x .5 = 1/8cm^3 (which is the volume of the cube)

1395 ÷ 1/8 = 11160

4 0

3 years ago

Read 2 more answers

12 guests are coming to my party. They each get 4/5 cup of juice. How many cups of juice do I need to purchase

dusya [7]

You would need to purchase 10 cups of juice

3 0

3 years ago

Read 2 more answers

An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli

oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0

3 years ago

Which question(s) could be answered by collecting data that has variability? Click on each statistical question. To change an an

Viefleur [7K]

Answer:

how old are the cats that live in Kyle's neighborhood?

Step-by-step explanation:

Because the question is statistical which mean you will get more then one answers

7 0

2 years ago

This is a System Of Equations and I'd like it to get solved by the Elimination or Substitution Method.x+2y=83x+5y=8

Sonbull [250]

Answer:

x=133 y=-25

Step-by-step explanation:

I'll do both ways for you. So let's start with Substitution:

With the sub method, you have to set both equations equal to each other by setting them equal to the same variable. Since there is no coefficient in front of both x's in both equations, that variable will be easiest to solve for.

x + 2y = 83 & x + 5y = 8

Solve for x.

x = 83 - 2y & x = 8 - 5y

Once you have solved for x, set each equation equal to one another and solve for y now.

83 - 2y = 8 - 5y

Isolate all variables to one side:

83 = 8 - 3y

Now subtract the 8 to fully isolate the y variable:

75 = -3y

Divide by -3:

-25 = y Now that you have your first variable, plug it into one of the original equations and solve for x.

x + 2(-25) = 83

x - 50 = 83

x = 133

Now for the Elimination method. For this method you need to get rid of a variable by either subtracting/adding each equation together. Again, since you can subtract and x from both equations, you will be left with only the y variable to solve:

Put each equation on top of one another and subtract:

x + 2y = 83

- (x + 5y = 8)

The x's will cancel out:

(x - x) + (2y - 5y) = (83 - 8)

Simplify:

-3y = 75

Solve for y

y = -25

Then, plug y = -25 into one of the original equations:

x + 5(-25) = 8

Solve for x:

x - 125 = 8

x = 133

Hope this helps!

8 0

2 years ago