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mash [69]
1 year ago
12

There were 340,000 cattle placed on feed. Write an equivalent ratio that could be used to find how many of these cattle were bet

ween 700 and 799 pounds. How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?

Mathematics
2 answers:
BartSMP [9]1 year ago
6 0
2/5 = x/340,000

2 x 340,000 = 680,000

680,000 / 5 = 136,000

x = 136,000

136,000 cows were between 700 and 799 pounds


lutik1710 [3]1 year ago
3 0

There were 340,000 cattle placed on feed

How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?

Given the fraction of total cattle for 700 - 799 pounds is 2/5

Let x be the number of cattle between 700 - 799 pounds

We make a proportion using the fraction

\frac{2}{5} = \frac{x}{340,000}

Cross multiply it and solve for x

340000* 2 = 5x

680000 = 5x

Divide by 5 on both sides

So x= 136,000

There were 136,000 cattle between 700 and 799 pounds

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Alright, lets get started.

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Suppose they are working today t hrs, and his department will complete 43 packages per hour today.

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This 43 t packages includes those 44 too , which they are short of yesterday due to picnic.

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Cross multiplying

39 t = 43 t - 44

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Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

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(Br) Brown = 30%

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(R) Red = 20%

(G) Green =10%  =0.1

(O) Orange = 10%

(T) Tan = 10%

 

After 1995

(Br) Brown = 13%

(Y) Yellow = 14%  =0.14

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Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

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Since the draws from the 1994 and 1996 bag are independent,

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The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

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