I found the table that contains the following data:
25 times:
Column No. 1 2 3 4 5 Total
New York 3 2 2 1 2 10
Connecticut 1 1 1 2 0 5
Pennsylvania 1 1 1 1 1 5
Rhode Island 0 1 1 1 2 5
New York: 10/25 or 2/5
Connecticut: 5/25 or 1/5
Pennsylvania: 5/25 or 1/5
Rhode Island: 5/25 or 1/5
300 times * 2/5 = 120 times for New York
The best estimate of the number of times Amber would pull out a New York quarter is 120.
The means is the average of a data set
Distributive Property of each given equation:
<span>96 divided by 6
</span> => 2(40 + 8) / 6
=> (80) + (16) / 6
=> 96 / 6
=> 16
<span>85 divided by 5
</span> =>(2(30 + 15) - 5 )/ 5
=> ((60 + 30) - 5) / 5
=> (90 - 5 ) / 5
=> 85 / 5
=> 17
<span>85 divided by 6
</span> =>(2(30 + 15) - 5 )/ 6
=> ((60 + 30) - 5) / 6
=> (90 - 5 ) / 6
=> 85 / 6
=> 14.166…
<span>168 divided by 7
</span> =>3(50 + 6) / 7
=> (150 + 18) / 7
=> (168 ) / 7
=> 24
<span>104 divided by 4
</span> =>2(50 + 2) / 4
=> (100 + 4) / 4
=> 104 / 4
<span> => 26 </span>
<span>171 divided by 9
=> 3 (50 + 7) / 9
=> (150 + 21) / 9
=> (171) / 9
=> 19
</span>
<span>102 divided by 9
=> 2 (50 + 1) / 9
=> ((2 x 50) + (2 x 1)) / 9
=> (100 + 2 ) / 9
=> 102 / 9
=> 11.333……
</span>
<span>102 divided by 6
=> 2 (50 + 1 ) / 6
=> ((2 x 50) + (2 x 1)) / 6
=> (100 + 2) / 6
=> 102 / 6
=> 17
</span>
210 divided by 5
=> 2 (100 + 5) / 5
=> ( (2 x 100) + (2 x 5) ) /5
<span> => (200 + 10) / 5 </span>
=> 210 /5
<span> => 42
</span>
These are linear equations that can be solved simultaneously;
5m+3n=41
3m-6n=9 (multiplying the first equation by 3 and the second by 5)
15m+9n=123
15m-30n= 45 (subtracting the two equations)
39n = 78
n = 2,
and for m, 3m = 9+6(2)
= 21
m = 7
Therefore, n=2 and m=7
