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4 years ago

Two lines, A and B, are represented by the equations given below:

2 answers:

4 0

Ok so to be a solution of both, it mus satisfy both equations

therefor only the first choice is correct because (the following is why the other answers are wrong)
2. solutions can lie on axises
3. it mus satisfy both equations
4. it can lie on or between the axises

answer should be the fist one

Cloud [144]4 years ago

3 0

Answer:

the answer is A

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The last one! Because every input has to have a different output! And the last one all have different output compared to the other ones that have the same output with different inputs which does not represent a function

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4 years ago

Use properties of parallel lines to find the value of x.

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3 years ago

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4 years ago

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3 years ago

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A lot of 25 skylight covers are received at your construction site, and before installation are subjected to an acceptance testi

Vladimir79 [104]

Answer:

If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.

Step-by-step explanation:

We have a population of N=25 skylight covers, were K=4 are defective.

We sample n=5 covers, and we will accept the lot if k=2 or fewer are defective.

We will use the hypergeometric distribution to model this probabilities.

First, to be accepted, the sample can have 2, 1 or 0 defective covers, so the probability of being accepted is:

$P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)$

The probability that there are k defective covers in the sample is:

$P(k)=\dfrac{\dbinom{k}{k}\dbinom{N-k}{n-k}}{\dbinom{N}{n}}$

Then, we can calculate the individual probabilities as:

$P(k=0)=\dfrac{\dbinom{4}{0}\cdot \dbinom{25-4}{5-0}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{0}\cdot \dbinom{21}{5}}{\dbinom{25}{5}}\\\\\\P(k=0)=\dfrac{1\cdot 20349}{53130}=0.38$

$P(k=1)=\dfrac{\dbinom{4}{1}\cdot \dbinom{25-4}{5-1}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{1}\cdot \dbinom{21}{4}}{\dbinom{25}{5}}\\\\\\P(k=1)=\dfrac{4\cdot 5985}{53130}=0.45$

$P(k=2)=\dfrac{\dbinom{4}{2}\cdot \dbinom{25-4}{5-2}}{\dbinom{25}{5}}=\dfrac{\dbinom{4}{2}\cdot \dbinom{21}{3}}{\dbinom{25}{5}}\\\\\\P(k=2)=\dfrac{6\cdot 1330}{53130}=0.15$

If we add this probabilities, we have:

$P(accepted)=P(k\leq2)=P(k=0)+P(k=1)+P(k=2)\\\\P(accepted)=0.38+0.45+0.15=0.98$

If the lot has 4 defective covers out of 25 total covers, the probability of accepting the lot is P=0.98.

6 0

3 years ago