Photos and videos have become an important part of the online social experience, with more than half of Internet users posting photos or videos online that they have taken themselves. 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. A research center finds that P(A) = 0.54, P(B) = 0.36, and P(A or B) = 0.62.
What is the conditional probability that an Internet user posts photos that they have taken themselves, given that they post videos that they have taken themselves? (Round your answer to three decimal places.)

0.778

Step-by-step explanation:

P(an Internet user posts photos that they have taken themselves/they post videos that they have taken themselves)=P(A/B)=?

According to conditional probability definition,

P(A/B)=P(A∩B)/P(B)

P(A)=0.54, P(B)=0.36, P(A or B)=0.62.

P(A or B)= P(A) +P(B) -P(A∩B)

P(A∩B)=P(A)+P(B)-P(A or B)

P(A∩B)=0.54+0.36-0.62

P(A∩B)=0.9-0.62

P(A∩B)=0.28.

P(A/B)=0.28/0.36

P(A/B)=0.778

So, the conditional probability that an Internet user posts photos that they have taken themselves, given that they post videos that they have taken themselves is 0.778.

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

m a t h w a y. c o m

step-by-step explanation: