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2 years ago

Solve the system graphically 1. x-2y=6 and 3x+2y=-6 2. x-y=0 and 2x+3y=-5

Mathematics

1 answer:

zysi [14]2 years ago

3 0

Step-by-step explanation:

1)The given equations are:

x − 2y = 6 ...(i)

3x − 6y = 0 ...(ii)

Putting x = 0 in equation (i) we get

=> 0 - 2y = 6

=> y = -3

x = 0, y = -3

Putting y = 0 in equation (i) we get

⇒x-2×0=6

⇒x=6

x = 6, y = 0

Use the following table to draw the graph

x 0 6

y -3 0

Plotting the two points A(0, -3) and B(6,0) equaion (1) can be drawn

Graph of the equation ..(ii)

3x - 6y = 0 ...(ii)

Putting x = 0 in equation (ii) we get

⇒3×0-6y=0

=> y = 0

x = 0, y = 0

Putting x = 2 in equation (2) we get

⇒3×2-6y=0

=> y = 1

x = 2, y = 1

Use the following table to draw the graph.

x 0 2

y 0 1

Draw the graph by plotting the two points O(0,0) and D(2,1) from table

We see that the two lines are parallel, so they won’t intersect

Hence there is no solution

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Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the conference is 0.10.

(d) The probability that John stays home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (M), Roberto (R), John (J), Dominique (D) and Clarice (C) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

${5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10$

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

The probability of the event E is:

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of Roberto and John is, RJ.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the conference as follows:

$P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10$

Thus, the probability that Roberto and John attend the conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the conference as follows:

$P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40$

Thus, the probability that Clarice attends the conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays home as follows:

$P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60$

Thus, the probability that John stays home is 0.60.

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