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3 years ago

9

Please help me with my homework.

1 answer:

QveST [7]3 years ago

8 0

Answer:

47.6 maybe?

Step-by-step explanation:

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Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of tim

Debora [2.8K]

Answer:

The Probability of exactly one tag being lost, in terms of π is $2\pi(1-\pi)-\pi^2(1-\pi)^2$

Step-by-step explanation:

Using the tree diagram attached to the bottom of this answer, you can see that the probability of only one tag being lost is the union of the probability of the left tag being lost when the right one is not lost and the probability of the right tag being lost when the left one is not lost.

Probability of losing only the right tag:

$P(C2|\frac{}{C1})=P(C2)*\frac{}{P(C1)}=\pi*(1-\pi)$

Probabilty of losing only the left tag:

$P(C1|\frac{}{C2})=P(C1)*\frac{}{P(C2)}=\pi*(1-\pi)$

Now, to unite those two probabilities, we use basic probability properties:

$P(C2|\frac{}{C1})$ ∪ $P(C1|\frac{}{C2})=P(C2|\frac{}{C1})+P(C1|\frac{}{C2})-(P(C1|\frac{}{C2})$ ∩ $P(C2|\frac{}{C1}))$

Since the events are independent:

$P(C1|\frac{}{C2})$ ∩ $P(C2|\frac{}{C1})=P(C1|\frac{}{C2})*P(C2|\frac{}{C1})$

So, the union becomes:

$P(C2|\frac{}{C1})$ ∪ $P(C1|\frac{}{C2})=P(C1|\frac{}{C2})+P(C2|\frac{}{C1})-P(C1|\frac{}{C2})*P(C2|\frac{}{C1})$

Replacing:

$=\pi(1-\pi)+\pi(1-\pi)-\pi^2(1-\pi)^2=2\pi(1-\pi)-\pi^2(1-\pi)^2$

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There is 27 campers. This is nine times as many as the number of counselors. how many counselors are there?

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June, Gavyn and Alex share some sweets in the ratio 2:2:1. June gets 14 sweets. How many did Alex get?

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3 years ago

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So do you know ughhh

Sauron [17]

The answer is 6 i hope this helped you

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Answer:

don't click the link/search the link to the other answer.

Step-by-step explanation:

trust me it's bad

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