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Kazeer [188]
3 years ago
12

a school district collected 14,621 pounds of recyclables during last year. This year was 3,943 more pounds of recyclables than i

t collected last year. How many pounds of recyclables did the school district collected last year?
Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0

Answer:

14,621 lbs.

Step-by-step explanation:

It says at the beginning of the question that the school district collected 14,621 lbs. of recycling. The part where it says that the school district collected 3,943 lbs. of recycling more than last year is un-needed information.

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If ln(4x+y)=2x-3,then dy/dx=
Viktor [21]

\ln(4x+y)=2x-3

Differentiate both sides wrt x:

\dfrac{\mathrm d(\ln(4x+y))}{\mathrm dx}=\dfrac{\mathrm d(2x-3)}{\mathrm dx}

By the chain rule, we get

\dfrac1{4x+y}\dfrac{\mathrm d(4x+y)}{\mathrm dx}=2

\dfrac{4+\frac{\mathrm dy}{\mathrm dx}}{4x+y}=2

Solve for \frac{\mathrm dy}{\mathrm dx}:

4+\dfrac{\mathrm dy}{\mathrm dx}=8x+2y

\boxed{\dfrac{\mathrm dy}{\mathrm dx}=8x+2y-4}

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2 years ago
Please help me i'm being timed
Naddika [18.5K]

Answer:

B. A linear, partial variation

Step-by-step explanation:

We know that speed = distance / time. From the table we have a linear function, and it's indirect, or partial.

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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
3 years ago
0.0000981 in scientific notation
Pachacha [2.7K]
9.81 times 10 to the -5th power
4 0
3 years ago
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SOMEONE HELP ME PLEASE
faust18 [17]

Answer:

The answer is either 1, 3, or 5

Step-by-step explanation:

There are 6 sides on a singular die and the die is rolled twice the first time it rolls onto 4 and the second onto an odd number. The only odd numbers on a 6 sided die are 1, 3, and 5.

5 0
3 years ago
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