Help with this question please

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

8 0

4 years ago

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term

Klio2033 [76]

Answer:

see explanation

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = $a_{2}$ - $a_{1}$ = $a_{3}$ - $a_{2}$

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

--------------------------------------------------------

The n th term of an arithmetic sequence is

$a_{n}$ = $a_{1}$ + (n - 1)d

$a_{1}$ = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

$a_{8}$ = 4 + (7 × 5) = 4 + 35 = 39

4 0

3 years ago

ITO

Setler [38]

Using the speed - distance relationship, the time left before the appointed time is 27 minutes.

Recall :

Distance = speed × time

Let time = t

At 10mph :

Distance = 10 × (t + 3) = 10t + 30 - - - (1)

At 12 mph :

Distance = 12 × (t - 2) = 12t - 24 - - - - (2)

Equate (1) and (2) :

10t + 30 = 12t - 24

Collect like terms

10t - 12t = - 24 - 30

-2t = - 54

Divide both sides by - 2

t = 54 / 2

t = 27

Hence, the time left before the appointed time is 27 minutes.

Learn more : brainly.com/question/25669152

8 0

2 years ago

Read 2 more answers

Isaac creates a scatter plot showing the association between the height and weight of his male

Black_prince [1.1K]

Answer:

Step-by-step explanation:

Given

w is the classmate's weight in pounds

h is the height

The general formula for the expression is y = MX+c where

m is the slope

c is the intercept

M is the change in y to change in x

Comparing with the given equation

w = 6.5h - 275, we will see that the slope of the equation is 6.5 and can be expressed as the change in the classmate's height to the classmate's weight

slope = dh/dw = 6.5

8 0

3 years ago

I need help with my homework please answer this correctly

Alla [95]

98,561

By following the rules for each digit, your answer should be 98,561

5 0

3 years ago