This is a problem of conditional probability that can be calculated by the formula:
P(B | A) = P(A ∩ B) / P(A)
We know that:
- between 1 and 50 there are 41 two-digit numbers, therefore
P(A) = 41/50 = 0.82
- between 1 and 50 there are 8 multiples of six, therefore
P(B) = 8/50 = 0.16
- <span>between 1 and 50 there are 7 two-digits mutiples of six, therefore
P(A ∩ B) = 7/50 = 0.14
Now, we can calculate:
</span>P(B | A) = P(A <span>∩ B) / P(A)
= 0.14 / 0.82
= 0.17
Therefore, the probability of getting a multiple of 6 if we draw a two-digit number is 17%.</span>
The 12th decades every other day every three days and four days all have 12 in common I think
X = smallest odd
x + 2 = second odd
x + 4 = third odd
x + x + 2 + x + 4 = 3
3x + 6 = 3
3x = -3
x = -1
The numbers are -1, 1, 3. Hope this helps!
Answer:
-0.7
Step-by-step explanation:
-2n+1.8=3.2
-2n=3.2-1.8
-2n=1.4
n=1.4/-2
n=-0.7
