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max2010maxim [7]
3 years ago
7

Solve for x -3x+2=-11 please

Mathematics
2 answers:
Svetach [21]3 years ago
6 0

Answer:

x = 6.5

Step-by-step explanation:

x - 3x + 2 =  - 11  \\  - 2x + 2 =  - 11 \\  - 2x =  - 11 - 2 \\   \frac{ - 2x}{ - 2} =  \frac{ - 13}{ - 2}  \\  x =  + 6.5

larisa [96]3 years ago
4 0

Answer:

6.5

Step-by-step explanation:

We need to solve out for x , the given Equation is ,

: \implies x - 3x + 2 = -11

: \implies -2x = -11 -2

: \implies -2x = -13

: \implies x = -13/-2

: \implies x = 6.5

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b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

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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 combinations 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.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

\mu = np

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that p = \frac{1}{2500} = 0.0004

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is P(X = 4) when n = 7000. So

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

P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

\mu = np = 7000*0.0004 = 2.8. So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

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3 years ago
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