The probability of one arrives within the next 10 minutes
when he already been waiting for one jour for a taxi is,
P (X > 70 | X > 60) = P (X > 10) = 1 – P (X ≤ 10)
= 1 – {1 – e ^ -((1 / 10) 10)} = e ^ -1
= 0.3679
The probability of one arrives within the next 10 minutes
when he already been waiting for one hour for a taxi is 0.3679
The answer is C. Because if g equals 4 you multiply 11 times 4 to get 44 then it is a quite simple problem 44 + 5 = 49
Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
==========================================================
Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
==========================================================
Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
==========================================================
Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
==========================================================
Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
==========================================================
Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
Answer:
First of AOD is verticallly opposite to BoC which makes it 55°
Angle at AoC is 55°+55°+105° minus 360 which is 145° hope this helps
3x+7=22
Subtract 7 to both sides
3x=15
Divide by 3
x=5
