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Reil [10]
3 years ago
6

How do you solve 6x+104=16x+114

Mathematics
2 answers:
Luda [366]3 years ago
8 0

Answer:

x = -1

Step-by-step explanation:

6x + 104 = 16x + 114

First subtract 104 from both sides

Now you have:

6x = 16x + 10

Subtract 16x from both sides

-10x = 10

Divide by -10 on both sides and you get your answer

x = -1

defon3 years ago
5 0

Answer:

x = - 1

Step-by-step explanation:

First you wanna subtract 104 from both sides

6x + 104 - 104 = 16x + 114 - 104

you should get 0 on the left and 10 on the right

next you subtract 16x from both sides  

6x - 16x = 16x - 16x + 10

you should get -10x on the left side and get 0 on your right

next you divide both side by -10

-10x ÷ -10 = 10 ÷ -10

the left should be 0 and the right should be -1

that's how i got the answer and I hope it helped !

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Answer:

a) 11.40% probability of 5 bits being in error during the transmission of 1 kb

b) 11.60% probability of 8 bits being in error during the transmission of 2 kb

c) 0.01% probability of no error bits in 3kb

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

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Poisson distribution with mean of 3.2 bits/kb (per kilobyte).

This means that \mu = 3.2kb, in which kb is the number of kilobytes.

(a) What is the probability of 5 bits being in error during the transmission of 1 kb?

This is P(X = 5) when \mu = 3.2

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140

11.40% probability of 5 bits being in error during the transmission of 1 kb

(b) What is the probability of 8 bits being in error during the transmission of 2 kb?

This is P(X = 8) when \mu = 2*3.2 = 6.4

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160

11.60% probability of 8 bits being in error during the transmission of 2 kb

(c) What is the probability of no error bits in 3kb?

This is P(X = 0) when \mu = 3*3.2 = 9.6

Then

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001

0.01% probability of no error bits in 3kb

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3 years ago
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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I need help please i will give u brainliest
IRISSAK [1]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
given the function f(x) = g(x-3)-2, describe transformation of f(x) on a coordinate plane relative to g(x).
Taya2010 [7]

Answer:

f(x) is obtained by translate g(x) 3 units to the right and 2 units down

Step-by-step explanation:

* Lets talk about the transformation of a function

- If the function f(x) translated horizontally to the right  

by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left  

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up  

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∵ f(x) = g(x - 3) - 2

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# x-coordinate is translated 3 units to the right

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  translated 3 units to the right and 2 units down

* f(x) is obtained by translate g(x) 3 units to the right and 2 units down

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