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3 years ago

Don’t really get these very much. Can anyone help ?

Mathematics

1 answer:

storchak [24]3 years ago

8 0

Answer:

23) $g$ has an amplitude greater than the period of $f$ /Both lines have the same period/There are no shifts of any kind between both lines.

24) Both lines have the same amplitude/ $g$ has a period less than period of $f$ /There are no shifts of any kind between both lines.

25) Both lines have the same amplitude/Both lines have the same period/There is an horizontal shift between $f$ and $g$ .

26) Both lines have the same amplitude/Both lines have the same period/There is a vertical shift between $f$ and $g$ .

Step-by-step explanation:

We proceed to describe succintly all differences and simmilarities between $f$ and $g$ in each case:

23) $g$ has an amplitude greater than the period of $f$ /Both lines have the same period/There are no shifts of any kind between both lines.

24) Both lines have the same amplitude/ $g$ has a period less than period of $f$ /There are no shifts of any kind between both lines.

25) Both lines have the same amplitude/Both lines have the same period/There is an horizontal shift between $f$ and $g$ .

26) Both lines have the same amplitude/Both lines have the same period/There is a vertical shift between $f$ and $g$ .

You might be interested in

Linear system word problem

soldi70 [24.7K]

Good job, you got the equations! XD

I'll just help you solve

Multiply the 2nd row by 9
9x+9y=117
9x+27y=207

Subtract the 2nd row from the 1st row
−18y=−90
Divide both sides by −18
y=−90/−18
Two negatives make a positive
y=90/18
y = 5

Substitute 5 into an equation
9x+9(5)=117
9x+45=117
Subtract 45 from both sides
9x=117−45

9x=72
Divide both sides by 9
x = ___

Hope this will help

3 0

3 years ago

Help #2-8 please thank you !

lapo4ka [179]

2. $1.75
3. $1.65
4. $1.05
5. $1.55
6. $1.20
7. $1.50
8. $1.29

3 0

3 years ago

Whats the answer to that question?

gavmur [86]

Answer:

-0.9090... can be written as $\frac{10}{11}$ .

Explanation:

Any repeating decimal can be written as a fraction by dividing the section of the pattern to be repeated by 9's.

We can start by listing out

0.909090... = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now. we let this series be equal to x, that is

$x$ = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now, we'll multiply both sides by 100 .

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + ...

Then, subtract the 1st equation from the second like so:

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

$-x$ = - 9/10 - 0/100 - 9/1000 - 0/10000 - 9/100000 - 0/1000000 - ...

And we end up with this:

$99x=90$

Finally, we divide both sides by 99 in order to isolate x and get the fraction we're looking for.

$x=\frac{90}{99}$

Which can be reduced and simplified to

$x=\frac{10}{11}$

Hope this helps!

4 0

3 years ago

Help! anyone SMART PLEASE

bagirrra123 [75]

Answer choice B is the correct answer the rest don't make sense.

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. Then, the probability of staying in the same state is 90%.

- 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. Then, the probability of staying in the same state is 80%.

- 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. Then, the probability of staying in the same state is 82%.

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