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1 year ago

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ſ x - 5y = -9( 4x + 4y = - 12Step 1 of 2: Determine if the point (-4,1) lies on both of the lines in the system of equations by

substituting theordered pair into both equations.

1 answer:

Phoenix [80]1 year ago

3 0

Given:

$\begin{gathered} x-5y=-9\ldots\ldots(1) \\ 4x+4y=-12\ldots\text{.}\mathrm{}(2) \end{gathered}$

The given point is (-4,1)

Let's check the ordered pair (-4,1) in the first equation

$\begin{gathered} x-5y=-9 \\ (-4)-5(1)=-9 \\ -4-5=-9 \\ -9=-9 \end{gathered}$

Hence, (-4,1) is a solution of the first equation x-5y=-9.

Now, let's check (-4,1) in the second equation.

$\begin{gathered} 4x+4y=-12 \\ 4(-4)+4(1)=-12 \\ -16+4=-12 \\ -12=-12 \end{gathered}$

So, (-4,1) is a solution of the second equation 4x+4y=-12.

Hence, (-4,1) is a solution of both equation in the system, then it is a solution to the overall system.

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2 years ago

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3 years ago

An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. (a) Use a bin

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

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Step-by-step explanation:

Binomial 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 combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

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In which

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$\mu$ 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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