Answer:
The answer is 0.5
hope this help :)
Step-by-step explanation:
Answer:
133/143
Step-by-step explanation:
Let S be the sample space
Let E be the event of selecting three committee partners with at least one junior partner.
Partners in the law firm include:
Senior partners = 6
Junior partners = 7
Total partners = 13
n(S) = number of ways of selecting 3 partners from 13 = 13C3
n(S) = 13C3 = 13!/(10!3!) = (13x12x11)/(3x2x1) = 286
To get n(E) i.e least 1 junior partner in the selected committee, we may have:
(2 senior and 1 junior) or ( 1 senior and 2 junior) or (3 junior).
Therefore, the required number of way is given below:
= (6C2 x 7C1) + (6C1 x 7C2) + 7C3
= [(6x5)/2 x 7] + [6 x (7x6)/2] + [(7x6x5)/(3x2)]
= 105 + 126 + 35
n(E) = 266
Therefore, the probability P(E) that at least one of the junior partners is on the committee is given below:
P(E) = n(E) /n(S)
P(E) = 266/286
P(E) = 133/143
Answer:
There were 210 downloads of the standard version.
Step-by-step explanation:
This question can be solved using a system of equations.
I am going to say that:
x is the number of downloads of the standard version.
y is the number of downloads of the high-quality version.
The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.2 MB. The total size downloaded for the two versions was 4074 MB.
This means that:

Yesterday, the high-quality version was downloaded four times as often as the standard version.
This means that 
How many downloads of the standard version were there?
This is x.

Since 




There were 210 downloads of the standard version.
Answer:
5 is the hight
Step-by-step explanation: