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lapo4ka [179]
3 years ago
15

The Mendes family bought a new house 5 years ago for $100,000. The house is now worth $198,000. Assuming a steady rate of growth

, what was the yearly rate of appreciation?
Mathematics
1 answer:
Andrei [34K]3 years ago
5 0
(new amount-old amount)=(198,000)-(100,000)=98% increase

98/5 years=19,600
Yr 1=$100,000+19,600=$119,600
Yr 2=$119,600+19,600=$139,200
Yr 3=$139,200+19,600=$158,800
Yr 4=$158,800+19,600=$178,400
Yr 5=$178,400+19,600=$198,000

The mortgage grew at a steady rate of $19,600 over the 5 years...

HOPE THIS HELPS :)

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
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Answer:

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Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

5 0
3 years ago
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