Answer:
a) 504
b) 56
c) 0.111
Step-by-step explanation:
Data provided in the question:
There are nine golf balls numbered from 1 to 9 in a bag
Three balls are randomly selected without replacement
a) 3-digit numbers that can be formed
= 
n = 9
r = 3
= ⁹P₃
= 
= 9 × 8 × 7
= 504
b) 3-digit numbers start with the digit 1
= _ _ _
in the above 3 blanks first digit is fixed i.e 1
we and we have 8 choices left for the last 2 digits
Thus,
n = 8
r = 2
Therefore,
= 1 × ⁸P₂
= 1 × 
= 1 × 8 × 7
= 56
c) Probability that the 3-digit number formed is less than 200
Now,
The number of 3-digit number formed is less than 200 will be the 3-digit numbers start with the digit 1 i.e part b)
and total 3-digit numbers that can be formed is part a)
therefore,
Probability that the 3-digit number formed is less than 200
= 56 ÷ 504
= 0.111
Answer:
19,17,15,13
Step-by-step explanation:
25-6=19
25-8=17
25-10=15
25-12=13
Answer:
The approximate solution to the system is (1.2, 4.4)
x = 1.2 and y = 4.4
Step-by-step explanation:
The solution of the system of linear equations equation y = –0.25x + 4.7 and y = 4.9x – 1.64 is shown in the attached graph. The red line represents the equation y = –0.25x + 4.7 and the blue line represents the equation = 4.9x – 1.64.
The solution of the system of equations is their point of intersection shown on the graph.
The point of intersection is (1.231, 4.392). To the nearest tenth, it is (1.2, 4,4). So x = 1.2 and y = 4.4.
So the approximate solution to the system is (1.2, 4.4)
Answer:
Step-by-step explanation:
- The product of 5 and the cube of x
- the difference of 6 and x^3
I understand as increase as sum so
Answer:
The probability of friend A winning with HH = 1/4.
Step-by-step explanation:
The probability of an event, A is P(A) given by the relationship;
P(A) = (The number of required outcome)/(The number of possible outcomes)
The parameters given are;
The condition of friend A winning = Coin toss sequence HH shows up
The condition of friend B winning = Coin toss sequence TH shows up
The number of possible outcomes = TT, TH, HH, HT = 4
(TH and HT are taken as different for the game to be fair)
The number of required outcome = HH = 1
Therefore;
The probability of friend A winning with HH = 1/4.
