Answer:
24
Step-by-step explanation:
percent of 280 is 67.2
280×y/100=67.2
280y/100=67.2
2.8y=67.2
divide by 2.8
y=24
verify
280×24/100=67.2
6720/100=67.2
67.2=67.2
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

.
The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by

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

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

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

Therefore,

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>

</span><span>

</span><span>
Therefore,

</span>
Answer:
78
Step-by-step explanation:
Since the two angles at the bottom are congruent, you would the two equations equal to each other. So it would be:
5x-19=2x+11
Once you simplify, you will find that x= 10.
Then you plug in x for both equations, so 5(10)-19= 31 and 2(10)+11= 31
So you add up the sides: 31+31+16= 78
Answer: (-10)^2
Step-by-step explanation:
(b / 2)^2 = (-10)^2
Hope this helps! Please give brainliest!