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ioda
3 years ago
14

how many different four letter permutations can be formed using four letters out of the first 12 in the alphabet?

Mathematics
1 answer:
Verizon [17]3 years ago
3 0

Answer:

11,800 different four letter permutations can be formed using four letters out of the first 12 in the alphabet

Step-by-step explanation:

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

P_{(n,x)} = \frac{n!}{(n-x)!}

In this question:

Permutations of four letters from a set of 12 letters. So

P_{(12,4)} = \frac{12!}{(12-4)!} = 11800

11,800 different four letter permutations can be formed using four letters out of the first 12 in the alphabet

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Let's look at an example.

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Before we can add, the denominators must be the same.

To get the denominators to be the same, we can...

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  • multiply top and bottom of 1/8 by 6 to get 6/48

At this point, both fractions involve the denominator 48. We can add the fractions like so

8/48  + 6/48 = (8+6)/48 = 14/48

Add the numerators while keeping the denominator the same the entire time.

The last step is to reduce if possible. In this case, we can reduce. This is because 14 and 48 have the factor 2 in common. Divide each part by 2.

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The fraction 14/48 reduces to 7/24

Overall, 1/6 + 1/8 = 7/24

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2 years ago
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2 years ago
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Which of the following inequalities matches the graph? (1 point)
pickupchik [31]

Answer:

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Step-by-step explanation:

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3 years ago
The perimeters of two similar figures are 15 in. and 24 in. What is the ratio of the areas of the two figures? State your answer
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Answer:

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Step-by-step explanation:

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\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2

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