An envelope contains an assortment of photos of the same size. The photos are of children, of pets, or of adults. Billy randomly took a photo from the envelope, recorded what type of picture it was, and then replaced the photo in the envelope. He did this a total of 10 times. The table shows the
outcomes.
Children Pets Adults
5 3 2
Based on these results, what is the probability the next photo taken from the envelope will be a photo of pets?

0.3

Step-by-step explanation:

From the question given above, the following data were obtained:

Children (C) = 5

Pet (P) = 3

Adult (T) = 2

Probability of obtaining a photo of pet, P(P) =?

Next, we shall determine the total outcome i.e the sample space (S).

This is illustrated below:

Sample space (S) = nC + nP + nA

Sample space (S) = 5 + 3 + 2

Sample space (S) = 10

Finally, we shall determine the probability of a photo of pet.

This can be obtained as follow:

Event of obtaining Pet, nP = 3

Sample space (S) = 10

Probability of obtaining a phone of pet, P(P) =?

P(P) = nP/nS

P(P) = 3/10

P(P) = 0.3

Therefore, the probability of obtaining a photo of pet is 0.3

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐