Question 1192923
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Part (a)


A = 651 = number of 1-inch nails
B = 243 = number of 2-inch nails
C = 41 = number of 3-inch nails
D = 451 = number of 4-inch nails
E = 333 = number of 5-inch nails


A+B+C+D+E = 651+243+41+451+333 = 1719 nails total


Of that total, D = 451 are four-inch nails.


Answer: 451/1719


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Part (b)


We use the same idea as part (a)


E = 333 = number of five inch nails
A+B+C+D+E = 1719 nails total


It leads to the fraction 333/1719 which reduces to 37/191 after dividing both parts by the GCF 9.


Answer: 37/191


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Part (c)


The question is cut off. Please repost. Thank you.
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