r/q= sin y
I got sin y,
or option one
Hope this helps ;3
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:
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Step-by-step explanation:
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You would just do 5.7-8.1, which is -3.6 degrees Celcius. Hope I helped!
Answer:
Answer to fourth part is :
Angle ABC= Angle DBC
So for fifth part reason answer is ASA congruency.
Step-by-step explanation:
Here we are given that CB bisects angle ABD and angle ACD.
So we have ,
Angle ABC= Angle DBC
Answer to fourth part is :
Angle ABC= Angle DBC
Now here we have two angles and one side equal in two triangles.
So we can say that ASA congruency fits in the best here.
So for fifth part reason answer is ASA congruency.
