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26 Nov 2017, 09:05
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A toy company expects the ratio of broken toys to the total number of toys in all future shipments to be equal to the corresponding ratio for shipments S1, S2, S3, and S4 combined, as shown in the table above. What is the expected number of defective chips in a shipment of 630,000 toys?
A. 20
B. 24
C. 29
D. 30
E. 33
Attachment:
2017-11-26_2003.png [ 11.81 KiB | Viewed 2929 times ]
26 Nov 2017, 11:21
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Bunuel
So, all future shipments ratio of broken to total will be the same as the ratio of shipments combined above: (broken/total)*future = (20/420,000) * 630,000 = 30 (D)
27 Nov 2017, 01:43
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Because we are given the broken toys for each shipment 2+6+8+4 = 20
and the total # number of shipments = 50,000 + 135,000 + 165,000 + 70,000 = 420,000
the ratio is 20:420,000, simplifying it is becomes 1:21,000
Given the total number of shipment is 630,000 then to get how many defects given the ratio \(\frac{630,000}{21,000}= 30\)
Or we can do it this way also 10:210,000 and \(\frac{630,000}{210,000} = 3\) so \(3*10 = 30\)
and the total # number of shipments = 50,000 + 135,000 + 165,000 + 70,000 = 420,000
the ratio is 20:420,000, simplifying it is becomes 1:21,000
Given the total number of shipment is 630,000 then to get how many defects given the ratio \(\frac{630,000}{21,000}= 30\)
Or we can do it this way also 10:210,000 and \(\frac{630,000}{210,000} = 3\) so \(3*10 = 30\)
