geno3141

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 #1
avatar+23254 
+5

8 slips of paper each has a 2 on them; 2 slips of paper each has a 4 of them

 

The total value of all the slips when added together is (8)(2) + (2)(4)  =  16 + 8  =  24

 

The expected value of one draw is the average value of these 10 slips of paper is  24 / 10  =  2.4

 

If we add one additional 4, then the total becomes 24 + 4  =  28.

Since there are now 11 slips of paper, the expected value (average) is  28 / 11  =  2.5454...

 

Adding another 4, the total becomes  28 + 4  =  32.

Since there are now 12 slips of paper, the expected value (average) is  32 / 12  =  2.6666...

 

How many 4s  must be added to get an average of 3?

Starting with the original 10 slips of paper and the original sum of 24:

let  n  represent the number of slips of paper (each containing a 4) that we have to add to the original 10.

The sum then becomes:  24 + 4n

The number of slips of paper becomes 10 + n.

The average becomes the sum divided by the number of slips of paper:  (24 + 4n) / (10 + n).

Since we want the average to be 3, we can create this equation:  (24 + 4n) / (10 + n)  =  3

Solving this by multiplying both sides by (10 + n)     --->     24 + 4n  =  3(10 + n)

                                                                                 --->     24 + 4n  =  30 + 3n

                                                                                 --->               n  =  6

We would have to add 6 slips to the original 10.

 

How many  4s  must be added to get an average of at least 3.5.

Using the technique of the previous problem, we get the equation:  (24 + 4n) / (10 + n)  >  3.5

 

Can you take it from here?

Apr 30, 2016
 #2
avatar+23254 
+10

The formula for the sum of the terms is:  S  =  n(a + l) / 2

The formula for the nth (or last) term is:    l  =  a + (n - 1)d

 

Since you want to have a formula for the common difference that does not include the variable  n,  lets solve the top equation for n and then substitute that value into the second equation.

 

S  =  n(a + l)/2     --->     2S  =  n(a + l)     --->     2S / (a + l)  =  n

l  =  a + (n - 1)d     --->     l - a  =  (n - 1)d     --->     l - a  =  ( [ 2S / (a + l) ] - 1 )d

 

Solve the second equation for d:    (l - a) / ( [ 2S / (a + l) ] - 1 )   =   d

 

Rewrite   [ 2S / (a + l) ] - 1  with the common denominator of (a + l)     (where  1  becomes  (a + l) / (a + l) )

     --->     [ 2S / (a + l) ] - (a + l) / (a + l)

Factor out the denominator of (a + l)

     --->     [ 2s - (a + l) ] / (a + l)

 

Look back at the left-hand side of the equation for  d  and replace  ( [ 2S / (a + l) ] - 1 )  with   [ 2s - (a + l) ] / (a + l):

     --->     (l - a) / ( [ 2S / (a + l) ] - 1 )   =   (l - a) / { [ 2s - (a + l) ] / (a + l) }

Multiply both the numerator and denominator of this expression by (a + l):

     --->     (l - a)(a + l)  /  { [ 2s - (a + l) ] / (a + l) · (a + l) }      (the (a + l)s on the right will cancel)

     --->     (l - a)(a + l)  /  [ 2s - (a + l) ]

 

Simplifying and interchanging, this becomes:  ( l2 - a2 ) / [ 2s - (l + a) ]   which equals  d.

Apr 30, 2016