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3 years ago

8

Hannah sells computers. The amount of her monthly sales for the first half of the year are listed below:

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

Mean = $16,500

Mode = $14,000

Median = $16,000

Step-by-step explanation:

Mean: We add all the numbers together and divide by 6 but this isn't the mean as we have to factor out and get it to the correct number. Therefore we will do the following.

14000+18000+13000+19000+21000+14000 / 6 Cancel the common factors

factor 2 out of everything

2(7000+9000+6500+9500+10500+7000) / 6

we get rid of the 2 and the 6 by factoring them out. so we now get

7000+9000+6500+9500+10500+7000 / 3

Now we add the numbers together and divide by 3

49500 / 3 = $16,500

Mode: This is easy as there is only 1 number that appears more then 1 time and therefore is 14,000

Median = we arrange the numbers from smallest to largest and find the center number. since there are an even number we will take the 2 center numbers and add them together and divide by 2.

13000, 14000, 14000, 18000, 19000, 21000

So we will take 14000 and 18000 and add them together then divide by 2.

14000 + 18000 = 32000/2 = $16,000

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

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4 years ago