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damaskus [11]
2 years ago
15

At the bright-as-day light bulb factory, 5 out of each 136 bulbs produced are defective. If the daily production is 2720 bulbs,

how many are defective?
Mathematics
1 answer:
lions [1.4K]2 years ago
4 0

Answer:

100 lightbulbs

Step-by-step explanation:

Basically find the percentage of lightbulbs that are bad. 5/136. So about 3. 6 percent. I'm going to use a more exact form of this percent for my calculations though. Now use the decimal for of this (0.036....) and multiply it by 2720. Using my exact decimal, the answer just so happened to be exactly 100. So there will be 100 defective lightbulbs per day. (Teachers are a stickler for units, so don't forget them if it's for a teacher)

Hope this helps!

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puteri [66]
  • Let x be the percentage of p.
  • Therefore,
  • m\% \:  \: of \:  \: n = x\% \:  \: of \:  \: p \\  =  >  \frac{mn}{100}  =  \frac{px}{100}  \\  =  > mn =  \frac{px}{100}  \times 100 \\  =  > mn = px \\  =  >  \frac{mn}{p}  = x

<u>Answer:</u>

<u>\frac{mn}{p}</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0
2 years ago
A city which has a population of 250,000 has been experiencing a population decline of 5.5% every year. What will the population
nadezda [96]

Answer:

126250

Step-by-step explanation:

5.5% of 250000 is 13750 so you multiply 13750 by 9 which = 123750 then you subtract 123750 from 250000 so you get your answer

4 0
2 years ago
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White raven [17]

Hi!

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4 0
3 years ago
The population of the town increased from 175000 to 262500 in a decade. What is the average percent increase of population per y
Alenkinab [10]

Answer:

5%

Step-by-step explanation:

As given,

The population of the town increased from 175000 to 262500 in a decade.

⇒ In 10 years , population increased from 175000 to 262500

⇒% increase = \frac{262500 - 175000}{175000}× 100% = \frac{87500}{175000}×100% = 0.5×100% = 50%

Now,

Average % increase = \frac{percent increase}{Total years} = \frac{50}{10}% = 5%

∴ we get

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