Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or vice versa). Because of this, the computer sending a message repeats each bit three times, so a 0 is sent as 000 and a 1 as 111. The receiving computer "decodes" each triplet by majority rule: whichever number, 0 or 1, appears more often in a triplet is declared to be the intended bit. For example, both 000 and 100 are decoded as 0, while 101 and 011 are decoded as 1. Suppose that 6% of bits are switched (0 to 1, or 1 to 0) during transmission between two particular computers, and that these errors occur independently during transmission.(a) Find the probability that a triplet is decoded incorrectly by the receiving computer.(b) Using your answer to part (a), explain how using triplets reduces communication errors.(c) How does your answer to part (a) change if each bit is repeated five times (instead of three)?(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Answer: D

Step-by-step explanation:

Answer:

271 degrees

Step-by-step explanation:

143-(-128)=143+128=271 degrees

Answer:

b + 2 = 6

Step-by-step explanation:

This may not be the right answer, but its what I think it is.

Answer:

2*2*3*5

Step-by-step explanation:

Answer:

∠EBF = 51°

∠DBE = 17°

∠ABF = 141°

∠EBA = 90°

∠DBC = 107°

∠DBF = 68°

Step-by-step explanation:

Hope this helps

Answer:

1.

a. m<EBF = 90 - 39 = 51

b. m<EBA = 90

c. m<DBE = 90 - 73 = 17

d. m<DBC = 90 + 17 = 107

e. m<ABF = 90 + 51 = 141

f. m<DBF = 17 + 51 = 68

2.

<ABD, <DBE, <EBF, <FBC, <DBF

3.

<ABF, <DBC

4.

<ABE, <EBC

Answer:

17.7

Step-by-step explanation:

The tenths place is the very first value after the decimal. The number 17.68694 rounds up to 17.7, because .68 is closer to .7 than it is to .6.

Answer:

486 inches is 13.5 yards and 162 inches is 4.5 yards please mark brainlist thank you

Answer:

37

Step-by-step explanation:

Sorry if this is wong I haven't done these in awhile, but this is how I got it.

f(X)=3x+7

f(10)=3(10) +7

f(10)=30 +7

f(10)=37

Answer:

f(10)=37

Step-by-step explanation:

f(x)=3x+7

(for f(10) we sub 10 for x)

f(10)=3×10+7

=30+7

=37

Answer:

I'll assume that x2 = x^2 and y2 = y^2 and the answer is sqrt{35}.

Step-by-step explanation:

This problem invovles completing the square, so if I square (x+y), I get

x^2+2xy+y^2, and since xy = 6, 2xy = 12, 23+12 = 35, and your answer is sqrt{35}.

Answer:

1) [tex]f'(t)=4t,\ g'(t)=3t^2+4[/tex]

2) [tex]p(t) =2t^5+8t^3[/tex]

[tex]p'(t)=10t^4+24t^2[/tex]

3) False

4)[tex]q(t) =\dfrac{1}{2}t+2t^{-1}[/tex]

[tex]q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]

5) False

Step-by-step explanation:

Given that:

[tex]f(t) = 2t^2[/tex] and [tex]g(t) = t^3 + 4t[/tex]

Formula:

[tex]1. \dfrac{d}{dx}x^n=nx^{n-1}[/tex]

[tex]2. \dfrac{d}{dx}C.f(x)=C.f'(x)\ \{\text{C is a constant}\}[/tex]

1) Using above formula:

[tex]f'(t)=2\times 2 t^{2-1}=4t[/tex]

[tex]g'(t)=3t^{3-1}+4\times 1 t^{1-1}=3t^2+4[/tex]

2) [tex]p(t) =2t^2(t^3+4t)[/tex]

Rewriting the formula by distributing the [tex]2t^2[/tex] term:

[tex]p(t) =2t^2.t^3+2t^2.4t=2t^5+8t^3[/tex]

[tex]p'(t) = 10t^4+24t^2[/tex]

3) By using answers of part (1):

[tex]f'(t).g'(t)=12t^3+16t[/tex]

[tex]p'(t) = 10t^4+24t^2[/tex]

Therefore it is False that [tex]p'(t) = f'(t).g'(t)[/tex]

4) [tex]q(t)=\dfrac{t^3+4t}{2t^2}[/tex]

Writing by distributing:

[tex]q(t)=\dfrac{t^3}{2t^2}+\dfrac{4t}{2t^2}\\\Rightarrow q(t) =\dfrac{t}{2}+\dfrac{2}{t}\\\Rightarrow q(t) =\dfrac{1}{2}t+2t^{-1}[/tex]

Using the formula:

[tex]q'(t)=\dfrac{1}{2}t^{1-1}+2\dfrac{-1}{t^2}\\\Rightarrow q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]

(5)By using answers in part (1):

[tex]\dfrac{g'(t)}{f'(t)}=\dfrac{3t^2+4}{4t}=\dfrac{3}{4}t+\dfrac{1}t[/tex]

[tex]q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]

Therefore, it is False that:

[tex]q'(t)=\dfrac{g'(t)}{f'(t)}[/tex]

Answer:

Directly proportional relationships always pass through the origin (0,0). There are other linear relationships that do not pass through the origin.

The graph of a proportional relationship must pass through the origin (0,0). Because of the value of k, the ordered pair (0, 0) satisfies this equation.

The relation between lines and points is described by a graph, which is a mathematical description of a network. A graph is made up of some points and connecting lines. It doesn't matter how long the lines are or where the points are located. A node is specified for each element in a graph.

A ray is used to describe any proportional relationship using an equation of the form y = kx. The ordered pair (0, 0) satisfies this equation regardless of the value of k, hence the ray or line must traverse the origin.

Thus, the graph of a proportional relationship must pass through the origin (0,0). Because of the value of k, the ordered pair (0, 0) satisfies this equation.

Learn more about the graph here,

https://brainly.com/question/17267403

#SPJ2

Answer: 600

Step-by-step explanation: 555

The five in the tens place is equal to 5

To round to the nearest tens place the number has to be 5 or more than 5

Answer:

i can't graph it but i'll give you the points to graph

Step-by-step explanation:

(-2, -5) (-1,-2) (0,1) (1,4) (2,7)

start at the point in bold

Answer:

Step-by-step explanation:

A coefficient is basically the number infront of a variable or term.

E.g. the coefficient in 7x is 7

The coefficient in 23y is 23

The coefficient in 43a is 43.

So for your question: 7a + 23b + 2c + 16,

The coefficients are 7, 23 and 2

To see if these ratios make a proportion let us first change them into fractions

= 4:10 = 4/10 = 4 ÷ 2 / 10 ÷ 2 = 2/5

= 6:15 = 6/15 = 6 ÷ 3 / 15 ÷ 3 = 2/5

= Since both these ratios are getting simplified into a same fraction , they are proportional .

Therefore , 4 : 10 :: 6 : 15 ( 4 : 10 is proportional to 6 : 15 )

Answer:

List some of the given data:

1 L = 33.8 oz.

A) Which one has more volume?

In a six-pack, each can has 12 oz.

Then in total, this is:

6*12oz = 72oz.

In one liter we have 33.8 oz.

Then in two liters, we have two times that:

2L = 2*33.8 oz = 67.7 oz.

Then the sixpack has more volume.

B) Which one has more value?

Here we must divide the volume by the price:

Six-pack:

Volume  = 72 oz.

Price = $1.37

Ratio = 72oz/$1.37 = 52.55 oz/$.

2L bottle:

Volume = 67.7 oz

Price = $1.31

Ratio = 67.7oz/$1.31 = 51.68 oz/$.

You can see that the ratio is larger in the case of the six-pack, this means that you get more ounces for each dollar.

Answer:

28

Step-by-step explanation:

Hope this helps

Multiplying a matrix by a scalar is equivalent to multiplying each entry in the matrix by that scalar:

[tex]A=\begin{bmatrix}0&-3\\-5&5\end{bmatrix}\implies3A=\begin{bmatrix}0&-9\\-15&15\end{bmatrix}[/tex]

[tex]B=\begin{bmatrix}-2&-6\\2&5\end{bmatrix}\implies-6B=\begin{bmatrix}12&36\\-12&-30\end{bmatrix}[/tex]

Now combine them:

[tex]3A-6B=3A+(-6B)=\begin{bmatrix}0+12&-9+36\\-15+(-12)&15+(-30)\end{bmatrix}=\begin{bmatrix}12&27\\-27&-15\end{bmatrix}[/tex]

Answer:

3

Step-by-step explanation:

he pattern is continued by adding 3 to the last number each time, like this: 1, 8, 27, 64, 125, 216, 343, 512, 729

Answer:

-56.9034

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

ill give u the answer if u do 73 divided by 2628

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

Answer:

4^2 + 5 = 21, 5^2 + 3 = 28, 7^2 + 3 = 52, 3^2 + 43 = 52, 7^2 + 19 = 68, 3^2 + 11 = 20, 1^2+19 = 20. I can't find a prime number and a perfect square that sums to 13.

Explanation:

these are primarily derived by finding the closest perfect square to a number and finding the difference to make the sum work, or finding the simplest prime number that will reduce the number which is a multiple of a perfect square.

Answer:

[tex] \cos (u + v) = -\dfrac{56}{65} [/tex]

Good job!

Step-by-step explanation:

[tex] \cos (u + v) = \cos u \cos v - \sin u \sin v [/tex]

[tex] \sin u = -\dfrac{3}{5} [/tex]; u is in QIII.

That makes [tex] \cos u = -\dfrac{4}{5} [/tex]

[tex] \sin v = -\dfrac{12}{13} [/tex]; v is in QIV.

That makes [tex] \cos v = \dfrac{5}{13} [/tex]

[tex] \cos (u + v) = (-\dfrac{4}{5})(\dfrac{5}{13}) - (-\dfrac{3}{5})(-\dfrac{12}{13}) [/tex]

[tex] \cos (u + v) = -\dfrac{20}{65} - \dfrac{36}{65} [/tex]

[tex] \cos (u + v) = -\dfrac{56}{65} [/tex]

You are correct.

Answer:

Step-by-step explanation:

-0.8615

Answer:

a) C) The system has no solutions only when h=3 and k is any real number.

b) D) The system has a unique solution when [tex]h=(-\infty,3)U(3,\infty)[/tex] and k is any real number.

c) The system has may solutions when h=3 and k=15

Step-by-step explanation:

a) In order to determine when the system will have no solution, we can start by solving the equation by substitution. We can solve the first equation for x1:

[tex]x_{1}+hx_{2}=3[/tex]

so

[tex]x_{1}=3-hx_{2}[/tex]

Next we can substitute this into the second equation so we get:

[tex]5(3-hx_{2})+15x_{2}=k[/tex]

We distribute the 5 into the first parenthesis so we get:

[tex]15-5hx_{2}+15x_{2}=k[/tex]

and group like terms:

[tex]-5hx_{2}+15x_{2}=k-15[/tex]

we factor x2 so we get:

[tex]x_{2}(-5h+15)=k-15[/tex]

and solve for x2:

[tex]x_{2}=\frac{k-15}{-5h+15}[/tex]

this final answer is important because it tells us what value the system of equations is not valid for. That answer will not ve vallid if the denominator is zero, so we can set the denominator equal to zero and solve for h, so we get:

[tex]-5h+15= 0[/tex]

and solve for h:

[tex]-5h= -15[/tex]

[tex]h=\frac{-15}{-5}[/tex]

[tex]h= 3[/tex]

so it doesn't really matter what value k gets since all that matters is that the denominator of the answer isn't zero.

b)

For part b we need to know when the system of equations will have infinitely many answers. Generally, this will happen when both equations are basically the same, so we need to make sure to simplify the second equation so it looks like the first equation, compare them and determine the respective coefficients.

So we take the second equation and factor it:

[tex]5x_{1}+15x_{2}=k[/tex]

we start by factoring a 5 from the left side of the equation so we get:

[tex]5(x_{1}+3x_{2})=k[/tex]

Next, we divide both sides of the equation into 5 so we get:

[tex]x_{1}+3x_{2}=\frac{k}{5}[/tex]

we now compare it to the first equation:

[tex]x_{1}+hx_{2}=3[/tex]

[tex]x_{1}+3x_{2}=\frac{k}{5}[/tex]

In this case, every coefficient of the two equations must be the same for us to get infinitely many answers, so we can see that h=3 and [tex]\frac{k}{5}=3[/tex]

when taking the second condition and solving for k we get that:

[tex]k=3(5)[/tex]

so

k=15

Anything else than the specific combination h=3 and k=15 will give us unique solutions, so for b, the answer is:

D) The system has a unique solution when and k is any real number.

c)

We have already solved part c on the previous part of the problem, so the answer is:

The system has many solutions when h=3 and k=15