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3 years ago

WILL GIVE BRAINLIEST!!!! 50 POINTS!!!

Mathematics

2 answers:

m_a_m_a [10]3 years ago

5 0

Answer:

The answer is C.

Norma-Jean [14]3 years ago

5 0

Hello,

What is the value of the x variable in the solution to the following system of equations?

3x + 2y = 8

2x + 3y = 2

Lets use 4:

3(4) + 2y = 8

12 + 2y = 8

12 + 2(-2) = 8

12 + -4 = 8

--------------------

2x + 3y = 2

2(4) + 3y = 2

8 + 3y = 2

8 + 3(-2) = 2

8 + (-6) = 2

So option A is true! (But remember that there can be more than one solution.)

Lets use -2:

3x + 2y = 8

3(-2) + 2y = 8

-6 + 2y = 8

-6 + 2(7) = 8

-6 + 14 = 8

-------------------

2x + 3y = 2

2(-2) + 3y = 2

-4 + 3y = 2

-4 + 3(2) = 2

-4 + 6 = 2

So option B is also true! (But again there can be more than one solution)

Answer:

So we have proved that there can be more than 1 solution for these set of equations. So the best answer is C: x can be any number as there are infinitely many solutions to this system.

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3 years ago

Read 2 more answers

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08

kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

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.

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 with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So $\mu = 10*0.08 = 0.8$

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

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

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a $p = 0.4493$ probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.