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2 years ago

X=-4y 3x+2y=20 Substitution

Mathematics

2 answers:

nadezda [96]2 years ago

7 0

Answer: y= -2; x = 8

Step-by-step explanation:

X=-4y.......equation 1

3x+2y=20......equation 2

Substitute the value of x into equation 2

3 (-4y) + 2y = 20

-12y + 2y = 20

-10y = 20

Divide both sides by -10

y = 20/-10

y = -2

Substitute the value of y into equation 1 to get the value of x

x = - 4(-2)

x = 8

I hope this helps greatly, please mark as brainliest answer.

inysia [295]2 years ago

6 0

Answer:

y=-2

x=8

Step-by-step explanation:

Substitute x=-4y into the other equation:

3(-4y)+2y =20

Multiply out the parenthesis:

-12y +2y = 20

-10y =20

Divide on both sides by -10:

y = 20/-10 =-2

Now, knowing y,you can substite it in the previous equation for x:

x = -4(-2)= 8

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2 years ago

A university found that of its students withdraw without completing the introductory statistics course. Assume that students reg

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Answer:

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.

a. Compute the probability that 2 or fewer will withdraw (to 4 decimals).

= 0.0355

b. Compute the probability that exactly 4 will withdraw (to 4 decimals).

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c. Compute the probability that more than 3 will withdraw (to 4 decimals).

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d. Compute the expected number of withdrawals.

= 6

Step-by-step explanation:

This is a binomial problem and the formula for binomial is:

$P(X = x) = nCx p^{x} q^{n - x}$

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 2) = 20C2(0.3)^2(0.7)^{18}\\P(X = 2) =190 * 0.09 * 0.001628413597\\P(X = 2) = 0.027845872524$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 1) = 20C1(0.3)^1(0.7)^{19}\\P(X = 1) =20 * 0.3 * 0.001139889518\\P(X = 1) = 0.006839337111$

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 0) = 20C0(0.3)^0(0.7)^{20}\\P(X = 0) =1 * 1 * 0.000797922662\\P(X = 0) = 0.000797922662$

Finally, the probability that 2 or fewer students will withdraw is

$P(X = 2) + P(X = 1) + P(X = 0) \\= 0.027845872524 + 0.006839337111 + 0.000797922662\\= 0.035483132297\\= 0.0355$

b) Compute the probability that exactly 4 will withdraw.

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 4) = 20C4(0.3)^4(0.7)^{16}\\P(X = 4) = 4845 * 0.0081 * 0.003323293056\\P(X = 4) = 0.130420974373\\P(X = 4) = 0.1304$

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

$P(X = x) = nCx p^{x} q^{n - x}\\P(X = 3) = 20C3(0.3)^3(0.7)^{17}\\P(X = 3) = 1140 * 0.027 * 0.002326305139\\P(X = 3) = 0.071603672205\\P(X = 3) = 0.0716$

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1 - 0.1071=0.8929

d) Compute the expected number of withdrawals.

E(X) = 3/10 * 20 = 6

Expected number of withdrawals is the 30% of 20 which is 6.

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