Answer:
0.1-0.6
Step-by-step explanation:
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6
Answer:
(- 1/2 , √3/2)
Step-by-step explanation:
t=20π/3
t' = 20π/3 - 6π = 2π/3 (120°) ... 2nd quadrant
If we start from (1,0) of unit circle, the coordinate of terminal point (x,y)
OF/OE = - cos 60° = x / 1 = - 1/2
FE/OE = sin 60° = y/1 = √3/2
The sample space is:
(1, 1); (1, 2) - sum of 3; (1, 3); (1, 4); (1, 5) - sum of 6; (1, 6);
(2, 1) - sum of 3; (2, 2); (2, 3); (2, 4) - sum of 6; (2, 5); (2, 6);
(3, 1); (3, 2); (3, 3) - sum of 6; (3, 4); (3, 5); (3, 6) - sum of 9;
(4, 1); (4, 2) - sum of 6; (4, 3); (4, 4); (4, 5) - sum of 9; (4, 6);
(5, 1) - sum of 6; (5, 2); (5, 3); (5, 4) - sum of 9; (5, 5); (5, 6);
(6, 1): (6, 2); (6, 3) - sum of 9; (6, 4); (6, 5); (6, 6)
The answer is a I believe
