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Natalija [7]
3 years ago
15

Solve the problem.

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

<h2>560</h2>

Step-by-step explanation:

You must use a combination:

_nC_k=\dfrac{n!}{k!(n-k)!}

We have <em>n = 16</em>, <em>k = 3</em>.

Substitute:

_{16}C_3=\dfrac{16!}{3!(16-3)!}=\dfrac{13!\cdot14\cdot15\cdot16}{2\cdot3\cdot13!}\qquad\text{cancel}\ 13!\\\\=\dfrac{14\cdot15\cdot16}{2\cdot3}=\dfrac{7\cdot5\cdot16}{1}=560

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Lizzie has 18 dimes and 12 quarters

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that LIzzie has 30 coins

number of dimes + number of quarters = 30

d + q = 30 ---- eqn 1

Also given that the coins total $ 4.80

number of dimes x  value of 1 dime + number of quarters x  value of 1 quarter = 4.80

d \times 0.10 + q \times 0.25 = 4.80

0.1d + 0.25q = 4.8 ------ eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 30 - q ---- eqn 3

Substitute eqn 3 in eqn 2

0.1(30 - q) + 0.25q = 4.8

3 - 0.1q + 0.25q = 4.8

0.15q = 1.8

<h3>q = 12</h3>

From eqn 3,

d = 30 - 12

<h3>d = 18</h3>

Thus she has 18 dimes and 12 quarters

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