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horrorfan [7]
2 years ago
13

If you were writing a two-column proof proving that ANQP APON, which of the following statements would have

Mathematics
2 answers:
larisa [96]2 years ago
8 0
The answer is D: NP NP
Vlada [557]2 years ago
4 0

Answer:

NP NP

Step-by-step explanation:

this is the only solution that works for the question

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At a clothing store, jackets are on sale for 30% off. Sales tax is 5%. Which expression will calculate the total cost of a jacke
VLD [36.1K]

Answer: B.x(0.75)

Step-by-step explanation:

If the jacket has a cost of x, but has 30\% off and we have to add the 5\% in tax, we can write all in the following expression:

x-30\% x+5\%x

Where:

x is the cost

30\% x=\frac{30}{100}x=0.3 x is the percent off

5\%x=\frac{5}{100}x=0.05x is the tax

Then:

x-0.3 x+0.05 x=0.75 x

This can be also rewritten as:

x(0.75)

Hence, the correct option is B.

3 0
3 years ago
John and Jane are married. The probability that John watches a certain television show is .7. The probability that Jane watches
ikadub [295]

a) The probability  that both John and Jane watch the show is 0.12.

b.The probability that Jane watches the show, given that John does is  0.1714.

c)They do watch show independent of each other.

Step-by-step explanation:

Here, as given in the question:

Probability that John watches a certain show  = 0.7   ⇒ P(J) = 0.7

Probability that Jane watches a certain show  = 0.3   ⇒ P(Je) = 0.3

Probability that John watches a certain show given Jane watches it to

 = 0.4   ⇒ P(J/Je) = 0.4

a. )  Here, we need to find the probability of  both Jane and John watching a show , P(J∩Je)

Now, by BAYES THEOREM:

P(J/Je)  = \frac{P(J \cap Je)}{P(Je)}

\implies  0.4  = \frac{P(J \cap Je)}{0.3} \\\implies P(J \cap Je) = 0.4 \times 0.3  = 0.12

Hence the probability  that both John and Jane watch the show is 0.12.

b.) The probability that Jane watches the show, given that John does is P(J/Je)

By BAYES Theorem:

P(Je/J)  = \frac{P(J \cap Je)}{P(J)}\\\implies P(Je/J)  = \frac{0.12}{0.7}  = 0.1714

Hence, the probability that Jane watches the show, given that John does is  0.1714.

c) As we can see P(Jane ∩ John) =  0.12

So, the probability of both of them seeing the show TOGETHER is 0.12.

Hence, they do watch show independent of each other and the probability of doing that is 1.0.12  = 0.88

7 0
3 years ago
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