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Kruka [31]
4 years ago
7

If 5,000 beef cattle were slaughtered in one day at Tyson Foods and the average animal weight was 1,350 lbs. How many pounds of

animals were processed at that facility during 5 days of production?
Mathematics
1 answer:
krok68 [10]4 years ago
4 0

Answer:

33,750,000 pounds of animals were processed during 5 days.

Step-by-step explanation:

If the average animal weight is 1350 lbs and 5,000 beef are slaughtered in one day, the total amount for one day would be: (1350)(5000) = 6,750,000 lbs.

Now we need to multiply this number by five since we want to know the total of pounds of animals processed during 5 days.

6,750,000 (5) = 33,750,000 pounds of animals.

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All of the following fractions are equal to 1/2 except?<br> 9/18<br> 20/40<br> 4/6<br> 5/10
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3 years ago
The mass, m grams, of a radioactive substance, present at time t days after first being observed, is given by the formula m=24e^
Reika [66]

Answer:

(i) The value of<em> m</em> when t = 30 is 13.2

(ii) The value of <em>t</em> when the mass is half of its value at t=0 is 34.7

(iii) The rate of the mass when t=50 is -0.18            

Step-by-step explanation:

(i) The <em>m</em> value when t = 30 is:

m = 24e^{-0.02t} = 24e^{-0.02*30} = 13.2

Then, the value of<em> m</em> when t = 30 is 13.2

(ii) The value of the mass when t=0 is:

m_{0} = 24e^{-0.02t} = 24e^{-0.02*0} = 24    

Now, the value of <em>t </em>is:

ln(\frac{m_{0}/2}{24}) = -0.02t

t = -\frac{ln(\frac{24}{2*24})}{0.02} = 34.7

Hence, the value of <em>t</em> when the mass is half of its value at t=0 is 34.7

(iii) Finally, the rate at which the mass is decreasing when t=50 is:

\frac{dm}{dt} = \frac{d}{dt}(24e^{-0.02t}) = 24(e^{-0.02t})*(-0.02) = -0.48*                            (e^{-0.02*50}) = -0.18

Therefore, the rate of the mass when t=50 is -0.18.

I hope it helps you!                  

3 0
3 years ago
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