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sp2606 [1]
3 years ago
5

What is the outlier of 29, 2, 28, 30, 26, 31

Mathematics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

2

Step-by-step explanation:

Given the data : 29, 2, 28, 30, 26, 31

Outlier ;

Lower :Q1 - (1.5 * IQR)

Upper : Q3 + (1.5 * IQR)

Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range

Using calculator :

Q1 = 26

Q3 = 30

IQR = (Q3 - Q1) = 30 - 26 = 4

Lower : 26 - (1.5 * 4) = 20

Upper : 30 + (1.5 * 4) = 36

Hence, the number in the given data which falls outside the range is 2

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Is π^2 greater or less than 7.2?
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What on one major characteristic of the ancient herbues?
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A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a
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Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

P_{t}=P_{0}+500t

Where P_{t} = Population after t years

P_{0}= Initial population

t = Time in years

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By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = \frac{5000}{500}

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

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t = \frac{15000}{500}

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For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

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t = \frac{35000}{500}

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b). If the population growth is 5%.

Then the growth will be exponential represented by

T_{n}=T_{0}(1+\frac{r}{100})^{t}

T_{n} = Population after t years

T_{0} = Initial population

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For double once,

Population after t years = 10000

10000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{10000}{5000}

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Take log on both the sides

log(1.05)^{t}=log2

tlog(1.05) = log2

t = \frac{log2}{log1.05}

t = 14.20 years

For double twice,

Population after t years = 20000

20000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{20000}{5000}

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Take log on both the sides

log(1.05)^{t}=log4

tlog(1.05) = log4

t = \frac{log4}{log1.05}

t = 28.413 years

For double thrice

Population after t years = 40000

40000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{40000}{5000}

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Take log on both the sides

log(1.05)^{t}=log8

tlog(1.05) = log8

t = \frac{log8}{log1.05}

t = 42.620 years

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2 years ago
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