In this region, are the events "silver" and "black" mutually exclusive?

Nuetrik [128]

Answer:

y(s) = $\frac{5s-53}{s^{2} - 10s + 26}$

we will compare the denominator to the form $(s-a)^{2} +\beta ^{2}$

$s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$

comparing coefficients of terms in s

$s^{2} :$ 1

s: -2a = -10

a = -2/-10

a = 1/5

constant: $a^{2}+\beta ^{2} = 26$

$(\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09$

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = $A(s-a)+B((s-a)^{2} +\beta ^{2} )$

we insert the values of a and β

$\\5s-53$ = $A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )$

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

the constants are

a = 0.2

β = 5.09

A = 5.4

B = -2

Step-by-step explanation:

since the denominator has a complex root we compare with the standard form $s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$
Expand and compare coefficients to obtain the values of a and β as shown above
substitute the values gotten into the function
Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...

3 0

3 years ago

Afina-wow [57]

Answer:tytgh

Step-by-step explanation:

8 0

2 years ago

Natali [406]

Answer: