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3 years ago

8

A town has a population of 5000 and grows at 4% every year. To the nearest year, how long will it be until the population will r

each 11600?

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

About 21.457 years

Step-by-step explanation:

You can solve this by setting up a simple equation.

$5000\cdot 1.04^x \geq 11600$

$1.04^x \geq 2.32$

log base 1.04 2.32=x

$x\approx 21.457$ years

Hope this helps!

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4 years ago

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4 years ago

The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied y

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$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

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Step-by-step explanation:

For this case we can create the following table

Interval Frequency (f) Midpoint(xi) xi *f xi^2* f

0-499 9 249.5 2245.5 560252.3

500-999 13 749.5 9743.5 7302753

1000-1499 33 1249.5 41233.5 51521258.25

1500-1999 115 1749.5 201193.5 361986278.8

2000-2499 125 2249.5 281187.5 632531281.3

2500-2999 81 2749.5 222709.5 612339770.3

3000-3499 47 3249.5 152726.5 496284761.8

3500-3999 45 3749.5 168727.5 632643761.3

4000-4499 22 4249.5 93489 397281505.5

4500-4999 10 4749.5 47495 225577502.5

Total 500 1220750 3408029125

$\sum f_i = 500 , \sum x_i f_i = 1220750, \sum x^2_i f_i = 3408029125$

For this case we can calculate the sample mean with this formula:

$\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5$

And for the sample variance we can use the following formula:

$s^2= \frac{\sum x^2_i f_i - \frac{(\sum x_i f_i)^2}{n}}{n-1}$

And if we replace we got:

$s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7$

And the deviation is just the square root of the sample variance and for this case is:

$s= \sqrt{856849.7}=925.662$

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