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>
Neither because the 0 in 20 has no value so u dont count it.there equal
X + 8x = 81
9x = 81, x = 9
y = 8(9), y = 72
Solution: (9, 72)
Peter's account is 910-40x, and Marla's account is 470-2x. This is a system, so you must find the number that makes both equations end up with the same number. Basically trial and error. Here is what I did:
Let's try the number 10. 910-400=510 and 470-20=450. I need a higher number.
Let's try 13 next. 910-520=390 and 470-26=444. Now the number has to be lower.
Let's try 12. 910-480=430 and 470-24=446. Close, and a little lower.
11.5 is too low, and 11.7 is too high. 11.6 has Equation 1 at 446 and Equation 2 at 446.8. Very close!
It is somewhere around 11.6. I hope that this can give you a start in figuring it out, because I don't believe in giving the complete answer, because then you do not learn anything from it.
