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dolphi86 [110]
3 years ago
14

Can someone help me on 11 and 14

Mathematics
1 answer:
sveticcg [70]3 years ago
5 0
For 11, set them equal to each other, since they both equal y.

y=2x+5; y=3x-1
2x+5=3x-1
x=6

and then substitute that x-value to find y.

y=3(6)-1
y=18-1
y=17

for 14 you can substitute the bottom y-value (x+7) into the top equation.

2x=y-10; x+7=y
2x=(x+7)-10
x=-3

and then substitute that x-value to find y.

-3+7=y
y=4
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There is a 99.99998% probability that at least one valve opens.

Step-by-step explanation:

For each valve there are only two possible outcomes. Either it opens on demand, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

In this problem we have that:

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Calculate P(at least one valve opens).

This is P(X \geq 1)

Either no valves open, or at least one does. The sum of the probabilities of these events is decimal 1. So:

P(X = 0) + P(X \geq 1) = 1

P(X \geq 1) = 1 - P(X = 0)

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There is a 99.99998% probability that at least one valve opens.

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