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2 years ago

The function y = 2x + 2 and y = 5/2x + k intersect at the point (2,6). Find the value of k.

Mathematics

2 answers:

antiseptic1488 [7]2 years ago

6 0

Answer:

k = 1

Step-by-step explanation:

We know that:

Given equations: y = 2x + 2 and y = 5x/2 + k
Given intersection point: (2,6)

P.S, I solved this problem graphically. If you have any problems, please write them in the comments.

Explanation:

To find the value of k (y-intercept of 2nd line), we will need to draw a graph. Please check out the uploaded graph. In the graph, it tells us that the y-intercept of 2nd line is 1.

Hence, the value of k is 1.

8_murik_8 [283]2 years ago

5 0

So these are system of two functions. The value of x and y is given. With the help of given x and y values we need to find the values of k.. Let's solve it!~

x = 2
y = 6

Plug in the values and solve for k

$y = 2x + 2 \\ y = \frac{5}{2} x + k$

$6 = 2 \times 2 + 2 \\ 6 = \frac{5}{2} \times 2 + k$

Solve the second system. Since, We have all the values

$6 = \frac{5}{ \cancel2} \times \cancel2 + k$

$6 = 5 + k$

$5 + k = 6$

$k = 6 - 5$

$k = 1$

Hence, The value of k is 1 ...

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Answer:

a) Joint ptobability distribution

$\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}$

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

$P(x,y)=P_x(x)*P_y(y)$

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

$\begin{pmatrix} &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}$

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

$P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56$

We can calculate P(X<= 1) * P(Y<=1)

$P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56$

Both calculations give the same result.

c) Probability of no violations

$P(X+Y=0)=P(0,0)=0.3$

d) P(X + Y <= 1)

$P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53$

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