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4 years ago

If x=√987678948729 and y=15927085, then what is (x÷9.5357)π + y/9743268?

1 answer:

8 0

First, we determine the value of x by evaluating the equation given,
x = sqrt (987678948729)
x = 993,820.38

With the equation given above,
(x/9.5357)π +y/9743268
Substituting,
(993,820.38/9.5357)π + 15927085/9743268

Simplifying,
327,254 + 1.6347
Further simplifying,
327,255.6347

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-5/8(-3/8 minus one fourth)

Ne4ueva [31]

The value of $\frac{-5}{8} (\frac{-3}{8}-\frac{1}{4} )$ = 25/64

<h3>What is fraction?</h3>

A fraction simply tells us how many parts of a whole we have. You can recognize a fraction by the slash that is written between the two numbers.

$\frac{-5}{8} (\frac{-3}{8}-\frac{1}{4} )$

= -5/8 (-3/8 -2/8)

= -5/8 (-5/8)

= 25/64

Thus, $\frac{-5}{8} (\frac{-3}{8}-\frac{1}{4} )$ = 25/64

Learn more about fraction here:

brainly.com/question/10354322

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3 0

2 years ago

A $20 T-shirt was reduced by 20%. What is the new price of the T-shirt?

Sloan [31]

Answer: $16

Step-by-step explanation:

reduce by 20% means cutting from 100% and leftover with 80% of original

100%-20%=80%

original price: $20×100%=$20

after reduction: $20×(100%-20%)=$16

4 0

4 years ago

Read 2 more answers

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)

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