Solve each system by substitution. Check your answers. 4x+2y=7 , y = 5x
1 answer:
By substitution, the solution of the system of equations, 4x + 2y = 7 and y = 5x, is (1/2 , 5/2).
A system of linear equations is a set of two or more equations which includes common variables. To solve system of equations, we must find the value of the unknown variables used in the equations that must satisfy both equations.
There are three methods that can be used to solve system of linear equations.
1. Elimination
2. Substitution
3. Graphing
Using substitution method, given two linear equations in x and y,
4x + 2y = 7 (equation 1)
y = 5x (equation 2)
Since the second equation is already expressed in y in terms of x, substitute the second equation to the first equation.
4x + 2y = 7 (equation 1)
4x + 2(5x) = 7
4x + 10x = 7
Combining all terms containing the variable x on one side and the constants on the other side of the equality, and solving for x.
4x + 10x = 7
14x = 7
x = 1/2
Substitute the value of x in the second equation and solve for y.
y = 5x
y = 5(1/2)
y = 5/2
Hence, the solution of the given system of equations is (1/2 , 5/2).
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Probability that is 0.523
Given that a point (x, y, z) is in space which is randomly selected as
These three equations represents a cube in space.
Length of each side of the cube = 1 - -1 = 2
Now represents a sphere with center (0, 0, 0) and of radius less than or equal to 1.
So we can imagine that this sphere is inside the cube.
So the probability that = Volume of the sphere/Volume of the cube
Volume of sphere with radius 1 = cubic unit
Volume of the cube of length 2 = cubic units
So the probability that = 4.186/8 = 0.523
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Answer:
a) See the code and the figure attached below
b) See the code and the figure attached below
c) We see significant differences on the boxplots. For Catalyst A the variation is less than the variation for Cat B. For both categories we see that we don't have a symmetric distributions. The maximum for Cat A is a little higher than the maximum for Cat B. The median for Cat A is a little higher than Cat B. And the third quartile for Cat A is lower than the third quartile for Cat B.
Step-by-step explanation:
Data given
Catalyst A: 4.4,3.4,2.6,3.8,4.9,4.6,5.2,4.7,4.1,2.6,6.9,0.8,3.6,2.9,2.6,4.0,4.3 .9,4.8,4.5,4.4,3.1,5.7,4.5
Catalyst B: .4,1.1,2.9,5.5,6.4,5.0,5.8,2.5,3.7,3.8,3.1,1.6, 3.5,5.9,6.7,5.2, 6.3, 2.6,4.3, 3.8
Solution to the problem
We are going to use R in order to do the plots required, we put the code and the output its attached to the solution.
Part a
R Code:
> catA<-c(4.4,3.4,2.6,3.8,4.9,4.6,5.2,4.7,4.1,2.6,6.9,0.8,3.6,2.9,2.6,4.0,4.3,0.9,4.8,4.5,4.4,3.1,5.7,4.5 )
> length(catA)
[1] 24
> catB<-c(.4,1.1,2.9,5.5,6.4,5.0,5.8,2.5,3.7,3.8,3.1,1.6,3.5,5.9,6.7,5.2,6.3,2.6,4.3,3.8 )
> length(catB)
[1] 20
> hist(catA)
> hist(catB)
Part b
R code:
> par(mfrow=c(1,2))
> boxplot(catA,main="Boxplot for CatA")
> boxplot(catB,main="Boxplot for CatB")
The result is attached.
Part c
We see significant differences on the boxplots. For Catalyst A the variation is less than the variation for Cat B. For both categories we see that we don't have a symmetric distributions. The maximum for Cat A is a little higher than the maximum for Cat B. The median for Cat A is a little higher than Cat B. And the third quartile for Cat A is lower than the third quartile for Cat B.
