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valentinak56 [21]

3 years ago

6

A train travels 360 miles in 4.5 hours, moving at a constant speed. How many hours will it take the train to travel 560 miles at

this speed?

1 answer:

Mashutka [201]3 years ago

8 0

Answer:

The answer would be: 560 miles = 84.5 hours

~ i rlly hope this has helped, have a gr8 day/night my friend!~

Step-by-step explanation:

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Which of the following has the least steep graph

scZoUnD [109]

When a slope is the steepest, the x value would be very small (in terms of fractions and decimals mostly). As the value of x is increased, the further away it starts moving from the y-axis. Therefore, the correct solution would be y=3x-16. Another way to confirm this is by graphing all the equations and comparing.

Hope I helped :)

3 0

3 years ago

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Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

$P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

$P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]$

Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

Transition matrix P raised to the power 5 (at k = 5)

$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]$

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0

4 years ago

Which are samples that are appropriate to survey regarding how the city’s residents feel about building a new youth center? Sele

sasho [114]

Answer:

Its actually 100 residents who live near the proposed site for the youth center

50 children at each city's four elementary schools

100 parents of children at each of the city's four school

Step-by-step explanation:

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3 years ago

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Megan has 25 phone numbers stored in her cell phone. Abby has some phone numbers stored in her cell phone. Together they have a

nikklg [1K]

25+N=61.
N= 36
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3 years ago

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This is my last question i swear pls help

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Answer:

To find a

sin 30=a/6in

1/2=a/6in

2a=6in

a=6in/2

a=3in

To find b

cos 30=b/6in

/3/2=b/6in

2b=/3*6in

b=2/3(two radical three)

I don't get the radical sign so I use / sign on the last 3 steps on the second

8 0

2 years ago

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