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dusya [7]
3 years ago
6

A gambling book recommends the following "winnin strategy" for the game of roulette:

Mathematics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

0.854 (85.4%)

Step-by-step explanation:

Since each bet is independent from others , then for X=gambler’s winnings when she quits

P(X>0) = probability to win in the first bet + probability of losing in the first bet * probability of wining at least one of the 2 additional bets

= 18/38 + (1-18/38) * [18/38*(1-18/38)+(1-18/38)*18/38+18/38*18/38 ] =0.854 (85.4%)

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uysha [10]
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Jack, Jill, and Bill each carried a 48 ounce bucket full of water down the hill.  By the time they reached the bottom, Jack's bu
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Altogether they spilled about 70 ounces

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Step-by-step explanation:

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A bag contains 10 red marbles, 15 yellow marbles, 5 green marbles, and 20 blue marbles. Two marbles are drawn from the bag.
kherson [118]

Answer:

It is the third option: (10C1) (20C1( / 50C2.

Step-by-step explanation:

The number of ways of selecting 2 marbles = the number of combination of any 2 marbles in the 50 marbles in the bag and this is 50C2. It is combinations because the order of drawing the marbles does not matter.

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6 0
3 years ago
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the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger
Reil [10]
<h2>Answer:</h2>

x\geq 3 \ and \ x+1 \geq 4

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

So:

1. The sum of 2 consecutive integers can be written as:

v+(v+1)

2. Nine times the smaller and 5 times the larger can be written as:

9v-5(v+1)

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4

The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

6 0
3 years ago
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