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3 years ago

Which expression is undefined?

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

Answer: C

Work: So, after you simplify that equation, it's 8/0. You Cannot divide zero like that. You can divide 0/8, but you can't with 8/0. I hope this wasn't to confusing, and <em>Happy Holidays! :)</em>

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Photo and Video Sharing. Photos and videos have become an important part of the online social experience, with more than half of

love history [14]

Answer: 0.923

Step-by-step explanation:

Let A be the event an Internet user posts photos that they have taken themselves, and B be the event an Internet user posts videos that they have taken themselves.

Pew Research Center finds that

P(A)=0.52 P(b)=0.26, and P(A or B)=0.54.

To find : P(A|B)

Since , $\text{P(A or B)=P(A+P(B)-P(A and B)}$

i.e. $\text{P(A and B)=P(A)+P(B)-P(A or B)}$

$\text{P(A and B)=}0.52+0.26-0.54=0.24$

Now, using conditional probability formula ,

$P(A|B)=\dfrac{\text{P(A and B)}}{\text{P(B)}}\\\\=\dfrac{0.24}{0.26}=0.923076923077\approx0.923$

Hence, the conditional probability that an Internet user posts photos that they have taken themselves, given that they post videos that they have taken themselves = 0.923

5 0

4 years ago

There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one

diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

$P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02$

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01$

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05$

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C$

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0

4 years ago

Which values are solutions of the inequality 5 < y -8<br><br>8 , 9. , 13. , 15. , 20

Sladkaya [172]

Answer:

5 is greater than y-8 since 5 is positive it + in the begging +5 while - 8 is less than 0 because it has - in the begging

6 0

3 years ago

First try was incorrect

Veronika [31]

Answer:

Circumference = 2(pi)r

C = 2*3.14*4.0= 25.12 round 25.10 m

Step-by-step explanation:

7 0

3 years ago

I really need help it’s an emergency this is due in a few hours

dedylja [7]

The answer is the first choice!!

5 0

3 years ago

Read 2 more answers