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2 years ago

Complete the table Y=6x-4

Mathematics

1 answer:

viktelen [127]2 years ago

4 0

Answer:

With Two points is enough to describe this graph

$\left[\begin{array}{ccc}\frac{4}{6} &0\\0&-4\\\end{array}\right]$

Step-by-step explanation:

Interception with Y axis (y=0), thus we have the following equation:

$0=6*x+4$

$x=4/6$

The point for this is (4/6 , 0)

Interception with x axis (x=0), thus we have the following equation:

$Y=4*0-4$

$y=-4$

The point for this is (0, -4)

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adelina 88 [10]

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3 years ago

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What is the sum of 55 66

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3 years ago

Consider a communication source that transmits packets containing digitized speech. After each transmission, the receiver sends

Greeley [361]

Answer:

The probability mass function is

$F(x) = [ \left \ 8 } \atop {x}} \right. ] * p^x * q^{n - x }$

Step-by-step explanation:

From the question we are told that

The maximum transmission time is $n = 8$

Generally the number of times a packet is transmitted follows a binomial distribution given that the the results of successive transmissions are independent of one another and that the probability of any particular transmission being successful is p

Generally the probability mass function is mathematically represented as

$F(x) = [ \left \ 8 } \atop {x}} \right. ] * p^x * q^{n - x }$

Here x can take values x = 0 , 1 , 2 , 3 , ... , 7

8 0

3 years ago

Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)

xenn [34]

Answer:

if the sequence is:

12, 13, 13, 14, 14 etc, and each term keeps growing up, the sequence obviusly diverges.

Now, if the sequence is

1/2, 1/3, 1/3, 1/4, 1/4, 1/5 , 1/5

so the terms after the first one repeat, we could group the terms with the same denominator and get:

1/2, 2/3, 2/4, 2/5..... etc.

So the terms after the first one are aₙ = 2/n.

Now, a criteria to see if a sequence converges if seing if:

$\lim_{n \to \infty} a_n = 0$

and here we have;

$\lim_{n \to \infty} 2/n$

that obviusly tends to zero, so we can conclude that this sequence converges.

then the limit is:

There exist a n' such that for any n > n' then IL -aₙI < ε

where L is the limit

I2/n - 0I = I2/nI < ε

then this is true if n > 2/ε = n'

3 0

3 years ago