Dmitri buys 23 pound of mixed nuts, 12 pound of chocolate
1 answer:
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Well, the answer will depend on whether the order will count or not (based on Permutations and Combinations). <em>If the order counts</em>, then we would use the formula for Permutations, which is:
Where n is the number of items you have, and r is the number of times you choose from the items.
Which simplifies to
Which simplifies to 15*14*13 (because all the numbers 1-12 in the factorial canceled out), which gets us the answer 2730.
Now, if you wanted to find the number of ways to order the toys without replacement (<em>order doesn't count</em>), you would use the formula:
The variables are still the same, but you are now multiplying by r!.
Simplifies to
Which simplifies to (using the same cancellation method above)
Dividing 2730 by 3! will get us an answer of 455.
Really, it depends on whether they are ordered or not. In this case (since you didn't specify whether the order mattered), the answer would be 455 or 2730.
:)
Where n is the number of items you have, and r is the number of times you choose from the items.
Which simplifies to
Which simplifies to 15*14*13 (because all the numbers 1-12 in the factorial canceled out), which gets us the answer 2730.
Now, if you wanted to find the number of ways to order the toys without replacement (<em>order doesn't count</em>), you would use the formula:
The variables are still the same, but you are now multiplying by r!.
Simplifies to
Which simplifies to (using the same cancellation method above)
Dividing 2730 by 3! will get us an answer of 455.
Really, it depends on whether they are ordered or not. In this case (since you didn't specify whether the order mattered), the answer would be 455 or 2730.
:)
Answer: BC = 5.83
Step-by-step explanation:
Luckily, the triangle is placed on the graph nicely so we can count the legs of the triangle:
AB = 5
AC = 3
BC = ?
To find BC, we can simply use the Pythagorean Theorem:
5^2 + 3^2 = c^2
25 + 9 = c^2
34 = c^2
Now square root to find c, or BC.
c = 5.83 (rounded by nearest hundredth)