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Salsk061 [2.6K]
4 years ago
8

If a coin is tossed 5 times, and then a standard six-sided die is rolled 4 times, and finally a group of three cards are drawn f

rom a standard deck of 52 cards without replacement, how many different outcomes are possible?
Mathematics
1 answer:
Step2247 [10]4 years ago
3 0

Answer:  5,499,187,200

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

A coin is tossed 5 times.  

There are two options (heads or tails) so the possible outcomes are: 2⁵

A six-sided die is rolled 4 times.

There are six options so the possible outcomes are: 6⁴

A group of 3 cards are drawn (without replacement).

The first outcome has 52 options, the second has 51 options, and the third has 50 options: 52 x 51 x 50

Now if we want the coin AND the die AND the cards, we have to multiply all of their possible outcomes:

     2⁵ x   6⁴  x  52 x 51 x 50

=   32 x 1296 x 132,600

=  5,499,187,200

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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Year 1 water change:  \displaystyle -2 \frac{1}{3}

Year 2 water change:  \displaystyle -1 \frac{5}{6}

<u>Step 2: Find Total Water Change</u>

  1. [Set up] Add yearly water changes:                                                                \displaystyle -2 \frac{1}{3} + -1 \frac{5}{6}
  2. [Fractions] Convert to improper:                                                                     \displaystyle \frac{-7}{3} - \frac{11}{6}
  3. [Fraction] Rewrite [LCM]:                                                                                  \displaystyle \frac{-14}{6} - \frac{11}{6}
  4. [Order of Operations] Subtract:                                                                       \displaystyle \frac{-25}{6}
  5. [Fraction] Convert to proper:                                                                           \displaystyle -4 \frac{1}{6}
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