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Arlecino [84]
3 years ago
6

Word problem.. i need help. Due at midnight

Mathematics
1 answer:
aniked [119]3 years ago
8 0

The Brady family received 1 letter on July 28.

Step-by-step explanation:

Step 1:

Assume a = the number of letters, b = the number of bills, c = the number of ads, and d = the number of magazines.

From the question, they got 15 pieces of mail that day so a+b+c+d=15.

They got 5 more magazines than bills so d = b + 5.

The got the same number of letters as ads so a=c.

They got 3 more bills than ads so b= c + 3.

Step 2:

As we have only one equation to solve the values, we need to make every term in terms of a single term.

We make everything in terms of c.

So a=c, b = c+3, c=c, d = b+5 = (c+3)+5=c+8.

Substituting all this in the equation, we get

a + b+c+d =(c)+(c+3)+(c)+(c+8) = 15, 4c+11=15.

4c=4, c = 1.

Step 3:

Substituting c = 1 in other terms, we get

a=c=1,

b=c+3=1+3=4,

c=1, and

d = c+8 = 1+8=9.

As a = 1, they received 1 letter that day.

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

Explanation:

If a set has n elements inside it, then it has 2^n different subsets.

Consider a set like {a,b,c}. It has n = 3 elements.

It has 2^n = 2^3 = 8 subsets

Those 8 subsets are...

  1. {a,b,c} ..... any set is a subset of itself
  2. {a,b}
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As you can see, each subset consists of items that are selected from the original set {a,b,c}. We can't have any repeat letters.

In that list, items 2 through 4 represent subsets with exactly two things inside it. Items 5 through 7 are known as singletons as they only have one item inside each set. The empty set can be written with the symbol \varnothing or you could have a pair of braces with nothing inside them. The empty set is a subset of any set.

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

Note how I included {a,b,c} as a subset. Any set is a subset of itself.

A proper subset will ignore the original set and only look at smaller subsets. If we say B is a proper subset of A, then set B will have fewer items compared to set A. Without the "proper" in there, it's possible that A = B.

So all we've done really is kick out one set which drops 2^n to (2^n)-1 when counting the number of proper subsets. I'm using parenthesis to indicate the "-1" is not part of the exponent. If you wrote this on your paper, then you would likely write 2^{n} - 1

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

For this problem, n = 11

This means there are 2^n = 2^11 = 2048 subsets and (2^n)-1 = 2048-1 = 2047 proper subsets.

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