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Nata [24]
3 years ago
12

A colony of bees have produced about 4 oz of honey. Some time later they have about 7oz of honey. If the bees produce at a rate

of 1.5% per hour, how many hours did it take?
Mathematics
1 answer:
yanalaym [24]3 years ago
5 0

It took 50 hours to produce 3 oz of honey.

<u>Step-by-step explanation:</u>

Rate of production= 1.5% per hour

Initial amount of honey = 4 oz

Final amount of honey = 7 oz

Honey produced = 3 oz

Amount produced = (initial amount x rate x time) /100

3 = (4 x 1.5 x t) /100

300 = 6t

t = 300/6

t = 50 hours

It took 50 hours for the bees to produce 3 oz of honey.

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Divide 4 2/3 &amp; 3 1/3. Simplify the answer and write it as a mixed number.
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Answer:

4 2/3 ÷ 3 1/3= 1 2/5

Step-by-step explanation:

First, you turn the mixed terms into an improper fraction like this

4 2/3 → 14/3 because when you multiply 4 times 3 equaling 12 then having 2 then adding that you get 14/3.

3 1/3 → 10/3 because when you multiply 3*3=9 then having 1 and adding that you get 10/3.

Then, you do KCF which stands for <u>Keep Change Flip</u> so for this you would do: 14/3 ÷ 10/3 → 14/3 × 3/10

14 × 3 = 42 and 3 × 10= 30

Now being 42/30 this is considered an improper fraction in which you have to transform it into a mixed number like this:

(For this you need to find the greatest common factor)

42/30 →  42 and 30 greatest common factor is 6 because they are divisible and factor of 6.

Now you divide both the denominator and the numerator y 6 like this:

42 ÷ 6= 7

30 ÷ 6= 5

Now we have 7/5, this is still an improper number so we see how many times 7 can go to 5 which is once. So we have 1 as whole number, now we put the reminder as the numenator of the mix fraction keeping 5 as being the denominator.

Overall, We have our answer 1 2/5

I hope this helps :D  

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