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xxTIMURxx [149]
2 years ago
12

The property taxes on a business office were $2160. what was the tax rate if the business office was valued at $270,000?

Mathematics
1 answer:
ahrayia [7]2 years ago
7 0

Answer:

Use the graphing tool to graph the linear equation.

Step-by-step explanation:

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Y = 2/3x - 3 as a function
Ksivusya [100]

Answer:

∫(x) = \frac{2}{3}x - 3

6 0
3 years ago
Read 2 more answers
PLZ HELP ME I BEG YOU !!!!!!!!!!!!!!
Zigmanuir [339]
3x200= 600
1000-600=400
400 divided by 5 = 80
answer is 80 adults

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3 0
3 years ago
Whats the next number 0 3 8 15 24 35
exis [7]
I believe it is 48. 
0+3=3
3+5=8
8+7=15
15+9=24
24+11=35
35+13=48
5 0
3 years ago
In a population of 10,000, there are 5000 nonsmokers, 2500 smokers of one pack or less per day, and 2500 smokers of more than on
Kazeer [188]

Answer:

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

Step-by-step explanation:

We have to write the transition matrix M for the population.

We have three states (nonsmokers, smokers of one pack and smokers of more than one pack), so we will have a 3x3 transition matrix.

We can write the transition matrix, in which the rows are the actual state and the columns are the future state.

- There is an 8% probability that a nonsmoker will begin smoking a pack or less per day, and a 2% probability that a nonsmoker will begin smoking more than a pack per day. <em>Then, the probability of staying in the same state is 90%.</em>

-  For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. <em>Then, the probability of staying in the same state is 80%.</em>

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. <em>Then, the probability of staying in the same state is 82%.</em>

<em />

The transition matrix becomes:

\begin{vmatrix} &NS&P1&PM\\NS&  0.90&0.08&0.02 \\  P1&0.10&0.80 &0.10 \\  PM& 0.08 &0.10&0.82 \end{vmatrix}

The actual state matrix is

\left[\begin{array}{ccc}5,000&2,500&2,500\end{array}\right]

We can calculate the next month state by multupling the actual state matrix and the transition matrix:

\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

\left( \left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] \right)^{12} =\left[\begin{array}{ccc}4792.63&3005.44&2201.93\end{array}\right]

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

3 0
4 years ago
Given 4 &lt; x &lt; 9 and 5 - y&lt; 4 , where x and y are integers. Find the maximum value of x-y. ​
djyliett [7]

Answer:

You want to maximize x and minimize y

Maximum y = 8

Minimum y = 10

8-10= -2

7 0
2 years ago
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