a. The growth rate to 4 decimal places and as a percentage.

b. Estimate the population in 2018 if the trend continues.

necessaryh
2021-02-26
Answered

The population of a rural community was 3050 in 2010 and the population had increased to 3500 in 2015. Assuming exponential growth find the following.

a. The growth rate to 4 decimal places and as a percentage.

b. Estimate the population in 2018 if the trend continues.

a. The growth rate to 4 decimal places and as a percentage.

b. Estimate the population in 2018 if the trend continues.

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Demi-Leigh Barrera

Answered 2021-02-27
Author has **97** answers

a)population in 2010,

population in 2015, p=3500

time in years, t=5

use the exponential growth model

Therefore the growth rate is 0.0279 or 2.79%

b)for 2018,t=8

Therefore, the population is 3801 in 2018

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Solve the following equation.

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Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. The solution is$x=B\otimes$ . (Simplify your answer.)

B. Every real number is a solution.

C. There is no solution.

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. The solution is

B. Every real number is a solution.

C. There is no solution.

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I have some problems, help

Solve the following nonhomogeneous system of linear equations:

$\begin{array}{rl}{x}_{1}+2{x}_{2}+3{x}_{3}+{x}_{4}& =4,\\ 2{x}_{1}+2{x}_{2}+3{x}_{3}+{x}_{4}& =5,\\ 3{x}_{1}+3{x}_{2}+4{x}_{3}+{x}_{4}& =6\end{array}$

Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.

Comment of the teacher

Example. If ${x}_{s}=\left(\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right)$ is a specific solution and $\{\left(\begin{array}{c}1\\ 2\\ 3/2\\ 4\end{array}\right),\left(\begin{array}{c}0\\ -2\\ 2\\ 9\end{array}\right)\}$ is a basis of $\{x:Ax=0\}$ then please enter

[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]

There is an automated system here, checking the solution.

Solution

Augmented matrix:

$\left(\begin{array}{ccccc}1& 2& 3& 1& 4\\ 2& 2& 3& 1& 5\\ 3& 3& 4& 1& 6\end{array}\right)$

I managed to transform the augmented matrix to the diagonal form:

$\left(\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 1& 0& -1& -3\\ 0& 0& 1& 1& 3\end{array}\right)$

So, the specific solution is the last row + 0 at the bottom: ${x}_{s}=\left(\begin{array}{c}1\\ -3\\ 3\\ 0\end{array}\right)$. I checked, and it is really a solution of the original problem.

1-st equation:

$1\cdot 1+2\cdot (-3)+3\cdot 3+1\cdot 0=1-6+9=4$

2-nd equation:

$2\cdot 1+2\cdot (-3)+3\cdot 3+1\cdot 0=2-6+9=5$

3-rd equation:

$3\cdot 1+3\cdot (-3)+4\cdot 3+1\cdot 0=3-9+12=6$

But, why does the basis in the comment consist of 2 columns? I think, it must be only 1: $\left(\begin{array}{c}0\\ -1\\ 1\\ -1\end{array}\right)$ - the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.

When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: "incorrect". Then I wrote this question; but before posting it, I tried it again, and it said: "correct". Probably, I made a typo first time

How can I check, that

$\left(\begin{array}{c}0\\ -1\\ 1\\ -1\end{array}\right)$

is really the basis of the space of solutions, like I did, validating the specific solution?

Solve the following nonhomogeneous system of linear equations:

$\begin{array}{rl}{x}_{1}+2{x}_{2}+3{x}_{3}+{x}_{4}& =4,\\ 2{x}_{1}+2{x}_{2}+3{x}_{3}+{x}_{4}& =5,\\ 3{x}_{1}+3{x}_{2}+4{x}_{3}+{x}_{4}& =6\end{array}$

Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.

Comment of the teacher

Example. If ${x}_{s}=\left(\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right)$ is a specific solution and $\{\left(\begin{array}{c}1\\ 2\\ 3/2\\ 4\end{array}\right),\left(\begin{array}{c}0\\ -2\\ 2\\ 9\end{array}\right)\}$ is a basis of $\{x:Ax=0\}$ then please enter

[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]

There is an automated system here, checking the solution.

Solution

Augmented matrix:

$\left(\begin{array}{ccccc}1& 2& 3& 1& 4\\ 2& 2& 3& 1& 5\\ 3& 3& 4& 1& 6\end{array}\right)$

I managed to transform the augmented matrix to the diagonal form:

$\left(\begin{array}{ccccc}1& 0& 0& 0& 1\\ 0& 1& 0& -1& -3\\ 0& 0& 1& 1& 3\end{array}\right)$

So, the specific solution is the last row + 0 at the bottom: ${x}_{s}=\left(\begin{array}{c}1\\ -3\\ 3\\ 0\end{array}\right)$. I checked, and it is really a solution of the original problem.

1-st equation:

$1\cdot 1+2\cdot (-3)+3\cdot 3+1\cdot 0=1-6+9=4$

2-nd equation:

$2\cdot 1+2\cdot (-3)+3\cdot 3+1\cdot 0=2-6+9=5$

3-rd equation:

$3\cdot 1+3\cdot (-3)+4\cdot 3+1\cdot 0=3-9+12=6$

But, why does the basis in the comment consist of 2 columns? I think, it must be only 1: $\left(\begin{array}{c}0\\ -1\\ 1\\ -1\end{array}\right)$ - the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.

When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: "incorrect". Then I wrote this question; but before posting it, I tried it again, and it said: "correct". Probably, I made a typo first time

How can I check, that

$\left(\begin{array}{c}0\\ -1\\ 1\\ -1\end{array}\right)$

is really the basis of the space of solutions, like I did, validating the specific solution?

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