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3 years ago

Identify all sets to which the number belongs. –0.249851765...

Mathematics

2 answers:

Ulleksa [173]3 years ago

3 0

Answer:

i. Irrational numbers

ii. Real numbers

Step-by-step explanation:

The given decimal is –0.249851765...

This number does not recur and it does not also terminate.

It belongs to the set of irrational numbers because it cannot be written in the form $\frac{a}{b}$ where $a,b\in R$ and $b\ne0$ .

This numbers also belongs to the set of real numbers.

antoniya [11.8K]3 years ago

3 0

Answer:

irrational

Step-by-step explanation:

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When network cards are communicating, bits can occasionally be corrupted in transmission. Engineers have de- termined that the n

ycow [4]

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

e = 2.71828 is the Euler number

$\mu$ is the mean in the giveninterval.

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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dybincka [34]

Answer: undefined

Step-by-step explanation:

I think im not sure tho

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3 years ago

PLEASE ANSWER ASAP!! I'll mark brainliest!!

vivado [14]

a) she put the numbers in the wrong place, V2 ≠2

b) V32= 4V2

4 0

2 years ago

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Rasek [7]

The answer to the question

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4 years ago

What is - √105 to the nearest integer and nearest tenth?

Vlad [161]

Answer:

[tex]-\sqrt{105}=10.2[/text]

Step-by-step explanation:

We are asked to find the value of [tex]-\sqrt{105}[/text] to the nearest tenth.

Using calculator we find the value if [tex]-\sqrt{105}[/text] which comes out to be 10.24

Hence

[tex]-\sqrt{105}=10.2[/text]

Rounded to nearest tenth decimal .

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4 years ago