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denis-greek [22]
4 years ago
6

Solve by elimination 5x+4y = 12 3x - 3y = 18 Please help solve and provide steps.

Mathematics
2 answers:
kramer4 years ago
6 0

5x + 4y = 12   ⇒   3(5x + 4y = 12)   ⇒   15x + 12y = 36

3x - 3y = 18   ⇒    4(3x - 3y = 18)   ⇒   <u> 12x - 12y</u> =  <u>72  </u>

                                                              27x         = 108

                                                              \frac{27x}{27}   =  \frac{108}{27}

                                                                   x        =   4

5x + 4y = 12   ⇒   5(4) + 4y = 12   ⇒  20 + 4y = 12   ⇒   4y = -8   ⇒   y = -2

Answer: x = 4, y = -2


SOVA2 [1]4 years ago
4 0
First you solve for y.
5x+ y= 12 so Y= 12-5x
And plug it into the next equation
3x- 3(12-5x) = 18
Distribute
3x-36+15x=18
Add 36 to both sides
3x+ 15x= 54
18x= 54 ... divide
x= 3
And you can plug in x to solve for Y
5(3) + y = 12
15+ y = 12 ...subtract 15 from both sides
y= -3
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The probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

<h3>What is chain rule in probability?</h3>

For two events A and B, by chain rule, we have:

P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)

where P(A|B) is probability of occurrence of A given that B already occurred.

For this case, the table given is:

Entries                          4-wheel drive        No 4-wheel drive        Total

Third Row Seats                  18                                12                        30

No Third Row Seats             7                                28                        35

Total                                     25                               40                        65

Let we take

A = event that a randomly selected car has no-4 wheel drive

B = event that a randomly selected car has third row seats

The total ways a car can be selected = 65 = n(S)

Total ways A can happen = n(A) = 40 (from the table).

Similarly, n(B) = 30

P(A) = n(A)/n(S) = 40/65

P(B) = n(B)/n(S) = 30/65

P(A∩B) = n(A∩B)/n(S) = 12/65

As by chain rule, we have:

P(A∩B) = P(A)P(B|A) = P(B)P(A|B)

We need P( A randomly selected car has three seats given that the selected car is with no 4-wheel drive)

which is symbolically P( B | A)

Thus, we use: P(A∩B) = P(A)P(B|A)

or

12/65 = (30/65)(P(B|A))\\\dfrac{12/65}{30/65} = P(B|A)\\\\P(B|A) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4

Thus, the probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

Learn more about probability here:

brainly.com/question/1210781

3 0
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