Explain how finding 7x20 is similar to similar to finding 7x2000.then find each product

2 answers:

8 0

They are both the same because each answer just has more zeros.7x20=140 7x2000=14000 P.S help from 7x2 =14 just add zeros

notka56 [123]4 years ago

7 0

They are both similar because you just have to multiply 7 x 2 and then add the zeros. 7 x 20 is 140 and 7 x2000 is 14000

You might be interested in

A number and its absolutely value are equal. If you subtract 2 from the number and its absolute value are not equal. What do you

Naily [24]

$n-the\ number\\\\|n|=n\ then\ n\geq0\\\\|n-2|\neq n-2\ what's\ a\ number\\\\I\ think,\ it's\ n=1\\\\because\ |1|=1\ and\ |n-2|=|-1|=1\ \ \ :correct\ \ :)\\\\3\ is\ not,\ because:\ |3|=3\ and\ |3-2|=|1|=1\to3\neq1$

5 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)

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