The probability of choosing a number that is not a multiple of 2 is P = 0.44
<h3 /><h3>How to find the probability?</h3>
We need to count the number of options for each digit.
- For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
- For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
- For the third digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}.
The total number of combinations is the product between the numbers of options:
C = 8*9*9 = 648
If we want our number to not be a multiple of 2 then it must end in a odd digit, the combinations that meet that condition are:
- For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
- For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
- For the third digit, we have 4 options {1, 3, 5, 7}.
C = 8*9*4 = 288
Then the probability of selecting a 3 digit number that is not a multiple of 2 is:
P = 288/648 = 0.44
If you want to learn more about probability, you can read:
brainly.com/question/251701
Answer:
Step-by-step explanation:
eq of a line through (x1,y1) and slope m is
y-y1=m(x-x11)
y+3=3(x+4)
y+3=3x+12
y=3x+12-3
y=3x+9
1) (f + g)(2) = 7 + 3 = 10 The answer is C
2) (f - g)(4) = 11 - 15 = -4 The answer is A
3) f(1) = 2(1) + 3 = 5 g(1) = 1² - 1 = 0 The answer is D
4) (f xg ) (1) = 7/3 The answer is B
For the first 60 positive integers, a = 1, n = 60, l = 60.
Sn = n/2(a + l)
s = 60/2(1 + 60) = 30(61)
For the next 60 positive integer, a = 61, n = 60, l = 120
Sum = 60/2(61 + 120) = 30(61 + 120) = 30(61) + 30(120) = s + 3600
Sum of first 120 positive integers = s + s + 3600 = 2s + 3600
The midpoint formula is basically (averaging the x coordinates, averaging the y coordinates).
Point A: (3, 7)
Point B: (2, -1)
Midpoint x: (3 + 2) / 2 = 5 / 2
Mindpoint y: (7 - 1) / 2 = 3
Therefore, the midpoint of the segment is choice C (5/2, 3)
