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3 years ago

What is the answer to this problem?

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

the simplified answer to the problem would be x/1

Step-by-step explanation:

-1/x and +1/x cancel eachother out so the answer is x/1 :)

AURORKA [14]3 years ago

5 0

The answer to the problem is x - 1

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A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with t

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Answer:

a) P=0.2

b) P=0.42

c) P=0.7

d) Marginal probability distribution of x:

$p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+0.04+0.02=0.16\\\\p_x(1)=p(1,0)+p(1,1)+p(1,2)=0.08+0.2+0.06=0.34\\\\p_x(2)=p(2,0)+p(2,1)+p(2,2)=0.06+0.14+0.30=0.50$

Marginal probability distribution of y:

$p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+0.08+0.06=0.24\\\\p_y(1)=p(0,1)+p(1,1)+p(2,1)=0.04+0.2+0.14=0.38\\\\p_y(2)=p(0,2)+p(1,2)+p(2,2)=0.02+0.06+0.3=0.38$

e) The two variables are dependant.

Step-by-step explanation:

The joint pmf of X and Y appears in the accompanying tabulation (attached).

a) Is the proability that one hose being used in the self service island and one hose being use in the full service island.

$P(X=1 \& Y=1)=P(1,1)=0.2$

(according to the tabulation)

$P(X\leq1 \& Y\leq1)=P(0,0)+P(1,0)+P(0,1)+P(1,1)\\\\P(X\leq1 \& Y\leq1)=0.10+0.08+0.04+0.20\\\\P(X\leq1 \& Y\leq1)=0.42$

c) This event means that at least one hose is in use in each of the island.

$P(X\neq0 \& Y\neq0)=P(1,1)+P(1,2)+P(2,1)+P(2,2)\\\\P(X\neq0 \& Y\neq0)=0.20+0.06+0.14+0.30=0.70$

Marginal probability distribution of x:

$p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+0.04+0.02=0.16\\\\p_x(1)=p(1,0)+p(1,1)+p(1,2)=0.08+0.2+0.06=0.34\\\\p_x(2)=p(2,0)+p(2,1)+p(2,2)=0.06+0.14+0.30=0.50$

Marginal probability distribution of y:

$p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+0.08+0.06=0.24\\\\p_y(1)=p(0,1)+p(1,1)+p(2,1)=0.04+0.2+0.14=0.38\\\\p_y(2)=p(0,2)+p(1,2)+p(2,2)=0.02+0.06+0.3=0.38$

Compute P(X≤1)

$P(X\leq1)=p_x(0)+p_x(1)=0.16+0.34=0.50$