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2 years ago

11

robert plant 7 trees behind westwood school. he planted 6 times as many trees in front of the school. how many trees did he plan

t?

2 answers:

Sati [7]2 years ago

4 0

49 in total or 42 trees in the front

Verdich [7]2 years ago

3 0

42 in front and 49 in total.

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The pet store had 6 puppies selling for $104 each and 12 kittens selling for $24 each. Today on 2 puppies and 8 kittens were lef

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You multiply 2*104 that is for puppies and then you do 8 by 24 cause that is for kittens and you get 208:192

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There are nine teddy bears in the diagram below. How many TOTAL squares are<br> there?

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Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

</span><span> $P(D)= \frac{1}{3}$ </span><span>

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$ </span>

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2 years ago

If a histogram of a sample of men's ages is skewed, what do you expect to see in the normal quantile plot?

Scorpion4ik [409]

Answer: Points are not following a straight-line pattern

Step-by-step explanation:

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2 years ago