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2 years ago

Find the distance between the points (4,-2), (4,-5)

1 answer:

6 0

$\text{Given that,}\\\\(x_1,y_1) = (4,-2)~~ \text{and}~~ (x_2 ,y_2) = (4,-5)\\\\\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2 -y_1)^2}\\\\~~~~~~~~~~~~=\sqrt{(4-4)^2 + (-5+2)^2}\\\\~~~~~~~~~~~~=\sqrt{0^2+(-3)^2}\\\\~~~~~~~~~~~~=\sqrt{0+9}\\\\~~~~~~~~~~~~=\sqrt 9\\\\~~~~~~~~~~~~=3$

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8 Opposite sides are congruent. Parallelogram Rhombus Rectangle Square

miv72 [106K]

Answer:

Parrelelogram

Step-by-step explanation:

Conggruent means sam angle but diiferent length and a parrelelogram has s standardized size deviation where the symmetry monsizes the ooppiosite size

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2 years ago

how three different fractional number representations of a rational number are related to determining the sign of of quotient of

lakkis [162]

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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. 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

Compute the permutations and combinations. 6 P 4 15 30 360

3241004551 [841]

The answer is c.
6 P 4 = 360

7 0

2 years ago

Read 2 more answers

Can i have help on this q plz

lukranit [14]

Answer:

No, they will need an additional £34 to stay the planned 10 nights.

Step-by-step explanation:

Attached screenshot of a spreadsheet.

The cost for 10 nights for the 2 people is £100.

The tent site rental (over 10m^2) for 10 nights is £150.

Savings for the two over the 9 weeks is £216.

They will need an additional £34 to camp as planned.

I suggest fibbing about the extra 0.5m^2 needed for the tent site. That will reduce the bill by £30, so they are only £4 short. Time to make a couple of friends.

3 0

2 years ago