Plz help I’ll mark you

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

option (B) is the answer

You might be interested in

Mr. and Mrs. Storey drove 2,400 miles during their vacation. Mr. Storey (y) drove 3 times as many miles as Mrs. Storey (x).

Wittaler [7]

X + y = 2400
y = 3x

x + 3x = 2400
4x = 2400
x = 2400/4
x = 600 <== Mrs. Storey drove 600 miles

y = 3x
y = 3(600)
y = 1800 <== Mr. Storey drove 1800 miles

5 0

3 years ago

For the graph x = 1 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your an

Hitman42 [59]

First, x=1 is a vertical line. A line that is perpendicular would be horizontal so it would have a slope of 0. A line parallel to it would be vertical too so it would have and infinite gradient.

3 0

4 years ago

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)

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