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Lorico [155]
3 years ago
6

Kimberly and Mike have an equal amount of money. After Kimberly spent $50 and Mike spent $25, Mike's money is 50% more than Kimb

erly's. How much did Kimberly and Mike have at first?
Mathematics
1 answer:
Paul [167]3 years ago
7 0

Answer:

Kimberly and Mike had at first $100 each

Step-by-step explanation:

Assume that Kimberly and Mike each have $x

∵ Kimberly has $x

∵ Kimberly spent $50

- Find the reminder with Kimberly by subtracting 50 from x

∴ The money left with Kimberly = x - 50

∵ Mike has $x

∵ Mike spent $25

- Find the reminder with Mike by subtracting 25 from x

∴ The money left with Mike = x - 25

∵ Mike's money is 50% more than Kimberly's money

- That means Mike's money is more than Kimberly's money by

   50% of Kimberly's money

∴ x - 25 = (x - 50) + 50%(x - 50)

∵ 50% = \frac{50}{100} = 0.5

∴ x - 25 = (x - 50) + 0.5(x - 50)

- Simplify the right hand side

∴ x - 25 = x - 50 + 0.5x - 25

- Add the like terms in the right hand side

∴ x - 25 = 1.5x - 75

- Add 75 to both sides

∴ x + 50 = 1.5x

- Subtract x from both sides

∴ 50 = 0.5x

- Divide both sides by 0.5

∴ 100 = x

∵ x represents the amount of money that Kimberly and

   Mike each had at first

∴ Kimberly and Mike had at first $100 each

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Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
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Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

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Assuming the population has a normal distribution.

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<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

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P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

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Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

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Step-by-step explanation:

here's your solution

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