Answer: 40:80:80
Step-by-step explanation:
The ratio is 1:2:2. This means, for every 5, the money is split in a ratio of 1:2:2. Let’s calculate how many 5s there are in 200. 200 / 5 = 40. So, the money will be split in 1:2:2 every 5 dollars 40 times.
Excellent. So, we can easily find out the ratio by multiplying the ratio 1:2:2 by 40. 1*40:2*40:2*40 = 40:80:80. Let’s add these up, and see if they result in 200. 40 + 80 + 80 = 200. So, our answer is correct.
Hope this helps!
First off, your chances of red are not really 50-50. You are overlooking the 0 slot or the 00 slot which are green. So, chances of red are 18 in 37 (0 slot) or 38 (0 and 00 slots). With a betting machine, the odds does not change no trouble what has occurred before. Think through the simplest circumstance, a coin toss. If I toss heads 10 times one after the other, the chances of tails about to happen on the next toss are still on a 50-50. A betting machine has no ability, no plan, and no past.
Chances (0 slot) that you success on red are 18 out of 37 (18 red slots), but likelihoods of losing are 19 out of 37 (18 black plus 0). For the wheel with both a 0 and 0-0 slot, the odds are poorer. You chances of red are 18 out of 38 (18 red slots win), and down are 20 out of 38 (18 black plus 0 and 00). It does not really matter on how long you play there, the probabilities would always continue the same on every spin. The lengthier you play, the more thoroughly you will tie the chances with a total net loss of that portion of a percent in accord of the house. 18 winning red slots and either 19 or 20 losing slots.
Answer:
3/5
Step-by-step explanation:
I took the test
Answer:
ok so u =?? 15 and 17= 2 but cant be cs they 2 4 6 8 10 12 14 16 not 15 or 17so 3
Step-by-step explanation:
You have the following data set: 10, 14, 12, 16, 13, 15, 20, 16, 10, 14. Based on the values in the given data set, which of the
kolbaska11 [484]
Trequency distribution table:
10.....2
14.....2
12.....1
16.....2
13.....1
15.....1
20....1
Answer:
The number 11 is NOT INCLUDED in the frequency distribution
