C is the answer hopenyou passed!!
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:
5
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
0.3y + y/z
<em>y</em> = 10
<em>z</em> = 5
<u>Step 2: Evaluate</u>
- Substitute in variables: 0.3(10) + 10/5
- Multiply: 3 + 10/5
- Divide: 3 + 2
- Add: 5
Answer:
Step-by-step explanation:
Let s represent the son's age now. Then s+32 is the father's age. In 4 years, we have ...
5(s+4) = (s+32)+4
5s +20 = s +36 . . . . . eliminate parentheses
4s = 16 . . . . . . . . . . . . subtract s+20
s = 4
The son is now 4 years old; the father, 36.
_____
<em>Alternate solution</em>
In 4 years, the ratio of ages is ...
father : son = 5 : 1
The difference of their ages at that time is 5-1 = 4 "ratio units". Since the difference in ages is 32 years, each ratio unit must stand for 32/4 = 8 years. That is, the future age ratio is ...
father : son = 40 : 8
So, now (4 years earlier), the ages must be ...
father: 36; son: 4.