A dozen eggs cost 3.84 at this rate how much would 50 eggs cost

Mathematics

1 answer:

Zina [86]2 years ago

5 0

Answer:

50 eggs would cost 16$

Step-by-step explanation:

A dozen eggs is 12. So for 4 dozen eggs, you get 48. 48 eggs is 15.36$. If you divide the 3.84 to 12, you will get the individual price for each egg. The individual price for each egg is .32 cents. You need to add 2 more eggs to the 4 dozen to get 50 eggs. You times .32 to 2 and you get .64. After that you add .64 to the price of the 4 dozen eggs and you will get 16$

Simple version:

dozen eggs =12, cost 3.84$

4 dozen eggs= 48, cost 15.36$

3.84$ divided by 12 = .32 cents (price of the individual egg)

Add two of the individual eggs together and add it to 15.36$

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Suppose we want to choose 4 colors, without replacement, from 18 distinct colors.

amm1812

Answers:

(a) 73440
(b) 3060

=====================================================

Explanation for part (a)

Consider four slots labeled A,B,C,D.

We have

18 choices for slot A
17 choices for slot B
16 choices for slot C
15 choices for slot D

We started at 18 and counted down until we filled the four slots. The countdown is because we cannot repeat a color. Multiply out those values to get the answer: 18*17*16*15 = 73440

You can use the nPr permutation formula as an alternative method

$_nP_r = \frac{n!}{(n-r)!}$

in this case n = 18 and r = 4. The exclamation marks indicate factorial.

-----------------------------------

Explanation for part (b)

Let's say we had color labels A,B,C,... all the way up to R which is the 18th letter in the English alphabet. From those 18 letters, we can only pick four. Let's say we pick D,E,F,G.

Focusing solely on the set {D,E,F,G}, we only have one set and the order doesn't matter. So {D,E,F,G} is the same as {D,E,G,F}.

There are 4*3*2*1 = 24 ways to arrange those four items meaning that we'll need to divide the result of part (a) by 24 to get the result of part (b)

73440/24 = 3060

You could use the nCr combination formula to get the same result

$_nC_r = \frac{n!}{r!(n-r)!}$