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3 years ago

4) Solve each pair of simultaneous equations.

1 answer:

7 0

4) a. x+y=1–(1)
y=2x-8—(2)
(2) into (1)
x+(2x-8)=1
3x-8=1
3x=1+8
x=9/3
x=3—(3)
(3) into (2)
y=2(3)-8
y=-2

ans x=3, y=-2

b. x+y=19—(1)
y=5x+1—(2)

(2) into (1)

x+(5x+1)=19
6x+1=19
6x=19-1
x=18/6
x=3—(3)

(3) into (2)

y=5(3)+1
y=16

ans x=3, y=16

c.x+y=-2—(1)
y=x-10—(2)

(2) into (1)

x+(x-10)=-2
2x-10=-2
2x=-2+10
x=8/2
x=4–(3)

(3) into (2)
y=4-10
y=-6

ans x=4, y=-6

5) 3x=y—(1)
x=y-16–(2)

(2) into (1)

3(y-16)=y
3y-48=y
2y=48
y=48/2
y=24–(3)

(3) into (2)

x=24-16
x=8

ans x=8, y=24

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4 years ago

Exercise 12.1.2: The probability of an event under the uniform distribution - random permutations. About A class with n kids lin

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Answer:

a) $P_a=\frac{1}{n}$

b) $P_b=\frac{1}{n(1-n)}$

c) $P_c=\frac{2}{n}$

Step-by-step explanation:

The question is incomplete:

(a) What is the probability that Celia is first in line? (b) What is the probability that Celia is first in line and Felicity is last in line? (c) What is the probability that Celia and Felicity are next to each other in the line?

a) The probability that Celia is first in line, if there are n kids and all of them have the same chance, is 1/n.

$P_a=\frac{1}{n}$

b) The probability that Celia is first in line and Felicity is last in line is

$P_b=\frac{1}{n} \frac{1}{n-1} =\frac{1}{n(1-n)}$ .

It is deducted like that:

If Celia is placed in the first line (what has a probablity of 1/n), there are left (n-1) kids. Then, the probability of placing Felicity in the last place is 1/(n-1).

Both probabilities multiplied give 1/(n*(n-1)).

c) The probability that Celia and Felicity are next to each other in the line is

$P_c=\frac{2(n-1)!}{n!}=\frac{2}{n}$

There are n! combinations for kids lines, where (n-1)! permutations have Celia before Felicity and other (n-1)! permutations have Felicity before Celia.

Then, we have 2(n-1)! permutations that have Celia and Felicity next to each other, over n! permutations possible.

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3 years ago

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