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adoni [48]
3 years ago
7

Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.

If the 2 times 2 matrix P is the transition matrix for a regular Markov​ chain, then, at​ most, one of the entries of P is equal to 0. Choose the correct answer below. A. This is false. In order for P to be​ regular, the entries of​ P^k must be​ non-negative for some value of k. For​ k=1 the matrix Start 2 By 2 Table 1st Row 1st Column 0 2nd Column 1 2nd Row 1st Column 0 2nd Column 1 EndTable has​ non-negative entries and has two zero entries.​ Thus, it is a regular transition matrix with more than one entry equal to 0. B. This is true. If there is more than one entry equal to​ 0, then the number of entries equal to zero will increase as the power of P increases. C. This is true. If there is more than one entry equal to​ 0, all powers of P will contain 0 entries.​ Hence, there is no power k for which Upper P Superscript k contains all positive entries. That​ is, P will not satisfy the definition of a regular matrix if it has more than one 0. D. This is false. The matrix P must be​ regular, which means that P can only contain positive entries. Since zero is not a positive​ number, there cannot be any entries that equal 0.
Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

C. This is true. If there is more than one entry equal to​ 0, all powers of P will contain 0 entries.​ Hence, there is no power k for which Upper P Superscript k contains all positive entries. That​ is, P will not satisfy the definition of a regular matrix if it has more than one 0

Step-by-step explanation:

The correct option is C as it represents that by considering a matrix P that involves more than one zero and at the same time the powers for all P has received minimum one zero or it included at least one zero

Therefore the statement C verified and hence it is to be considered to be valid

Hence, all the other statements are incorrect

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kupik [55]
Since the perimeter is the sum of the length of all sides we multiply width and length by 2 and then add them together

4 0
3 years ago
Solve the equation x to the power of 2 = 10
Nadusha1986 [10]

Answer:

The answer is √10,−√10 (award brainliest)

Step-by-step explanation:

Take the root of both sides and solve.

Exact Form:  x  = √10,−√10

Decimal Form:  x  =   3.16227766...,−3.16227766...

for more detailed explanation comment down below this post and ask

8 0
3 years ago
A stick is 20cm long.A student makes 5% error in measuring the stick.find two possible values for the student's measurement​
marishachu [46]

Answer:

16cm

24cm

Step-by-step explanation:

8 0
2 years ago
I have an assignment and I am having trouble with it. Can someone please help ASAP???
bezimeni [28]

Answer:

A) Find the sketch in attachment.

In the sketch, we have plotted:

- The length of the arena on the x-axis (90 feet)

- The width of the arena on the y-axis (95 feet)

- The position of the robot at t = 2 sec (10,30) and its position at t = 8 sec (40,75)

The origin (0,0) is the southweast corner of the arena. The system of inequalities to descibe the region of the arena is:

0\leq  x \leq 90\\0\leq y \leq 95

B)

Since the speed of the robot is constant, it covers equal distances (both in the x- and y- axis) in the same time.

Let's look at the x-axis: the robot has covered 10 ft in 2 s and 40 ft in 8 s. There is a direct proportionality between the two variables, x and t:

\frac{10}{2}=\frac{40}{8}

So, this means that at t = 0, the value of x is zero as well.

Also, we notice that the value of y increases by \frac{75-30}{8-2}=7.5 ft/s (7.5 feet every second), so the initial value of y at t = 0 is:

y(t=0)=30-7.5\cdot 2 =15 ft

So, the initial position of the robot was (0,15) (15 feet above the southwest corner)

C)

The speed of the robot is given by

v=\frac{d}{t}

where d is the distance covered in the time interval t.

The distance covered is the one between the two points (10,30) and (40,75), so it is

d=\sqrt{(40-10)^2+(75-30)^2}=54 ft

While the time elapsed is

t=8 sec-2 sec = 6 s

Therefore the speed is

v=\frac{54}{6}=9 ft/s

D)

The equation for the line of the robot is:

y=mx+q

where m is the slope and q is the y-intercept.

The slope of the line is given by:

m=\frac{75-30}{40-10}=1.5

Which means that we can write an equation for the line as

y=mx+q\\y=1.5x+q

where q is the y-intercept. Substituting the point (10,30), we find the value of q:

q=y-1.5x=30-1.5\cdot 10=15

So, the equation of the line is

y=1.5x+15

E)

By prolonging the line above (40,75), we see that the line will hit the north wall. The point at which this happens is the intersection between the lines

y=1.5x+15

and the north wall, which has equation

y=95

By equating the two lines, we find:

1.5x+15=95\\1.5x=80\\x=\frac{80}{15}=53.3 ft

So the coordinates of impact are (53.3, 95).

F)

The distance covered between the time of impact and the initial moment is the distance between the two points, so:

d=\sqrt{(53.5-0)^2+(95-15)^2}=95.7 ft

From part B), we said that the y-coordinate of the robot increases by 15 feet/second.

We also know that the y-position at t = 0 is 15 feet.

This means that the y-position at time t is given by equation:

y(t)=15+7.5t

The time of impact is the time t for which

y = 95 ft

Substituting into the equation and solving for t, we find:

95=15+7.5t\\7.5t=80\\t=10.7 s

G)

The path followed by the robot is sketched in the second graph.

As the robot hits the north wall (at the point (53.3,95), as calculated previously), then it continues perpendicular to the wall, this means along a direction parallel to the y-axis until it hits the south wall.

As we can see from the sketch, the x-coordinate has not changed (53,3), while the y-coordinate is now zero: so, the robot hits the south wall at the point

(53.3, 0)

H)

The perimeter of the triangle is given by the sum of the length of the three sides.

- The length of 1st side was calculated in part F: d_1 = 95.7 ft

- The length of the 2nd side is equal to the width of the arena: d_2=95 ft

- The length of the 3rd side is the distance between the points (0,15) and (53.3,0):

d_3=\sqrt{(0-53.3)^2+(15-0)^2}=55.4 ft

So the perimeter is

d=d_1+d_2+d_3=95.7+95+55.4=246.1 ft

I)

The area of the triangle is given by:

A=\frac{1}{2}bh

where:

b=53.5 ft is the base (the distance between the origin (0,0) and the point (53.3,0)

h=95 ft is the height (the length of the 2nd side)

Therefore, the area is:

A=\frac{1}{2}(53.5)(95)=2541.3 ft^2

J)

The percentage of balls lying within the area of the triangle traced by the robot is proportional to the fraction of the area of the triangle with respect to the total area of the arena, so it is given by:

p=\frac{A}{A'}\cdot 100

where:

A=2541.3 ft^2 is the area of the triangle

A'=90\cdot 95 =8550 ft^2 is the total area of the arena

Therefore substituting, we find:

p=\frac{2541.3}{8550}\cdot 100 =29.7\%

4 0
3 years ago
Solve for x & y solve x = 3 + yy = -2x + 9
sergey [27]

x = 3 + y Eqn(1)

y = -2x + 9​ Eqn(2)

Let us solve the system of equations with the substitution method

x - 3 = y (Subtracting 3 from both sides of the Eqn(1))

Replacing y = x - 3 in Eqn (2), we have:

x - 3 = -2x + 9

x = -2x + 9 + 3 (Adding 3 to both sides of the equation)

x + 2x = 9 + 3 (Adding 2x to both sides of the equation)

3x = 12 ( Adding like terms)

x = 12/3 (Dividing by 3 on both sides of the equation)

x = 4

Replacing x=4 in Eqn(1), we have:

4 = 3 + y

4 - 3 = y (Subtracting 3 from both sides of the equation)

y=1

The answers are:

x= 4 and y=1

6 0
1 year ago
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