I have a series of 2 questions.

Mathematics

1 answer:

Margarita [4]3 years ago

3 0

1)
I:y=3x-4
II:9x-3y=14

substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14

this is obviously not equal so there is no solution, the lines are parallel

2)
I:y=4x+6
II:5x-y=6

substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54

the solution is (12,54)

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Answer:

a) The probability that at least 5 ties are too tight is P=0.0432.

b) The probability that at most 12 ties are too tight is P=1.

Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

$P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432$

The probability that at least 5 ties are too tight is P=0.0432.

a) We can express the probability that at most 12 ties are too tight as:

$P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1$