Factor completely: 9x^4+12x^3-45x^2-60x

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 p

djverab [1.8K]

Answer:

The probability that a given user is transmitting is 0.1 = 10%.

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$ .

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Each user transmits only 10 percent of the time.

This means that $p = 0.1$

Find the probability that a given user is transmitting.

The probability that a given user is transmitting is 0.1 = 10%.

Suppose there are 120 users.

This means that $n = 120$

Find the probability that at any given time, exactly n users are transmitting simultaneously.

This is $P(X = n)$ .

This n is different from the n of total number of users(120 in this case) from the standard binomial formula. This is the number of successes, which is the equivalent of x. So

$P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

Find the probability that there are 21 or more users transmitting simultaneously.

Now we use the binomial approximation to the normal. We have that:

$\mu = E(X) = np = 120*0.1 = 12$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.1*0.9} = 3.2863$

Using continuity correction, this probability is $P(X \geq 21 - 0.5) = P(X \geq 20.5)$ , which is 1 subtracted by the pvalue of Z when X = 20.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.5 - 12}{3.2863}$

$Z = 2.59$

$Z = 2.59$ has a pvalue of 0.9952

1 - 0.9952 = 0.0048

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.

7 0

3 years ago

The area of a square is 81square centimeters. First, find the length of a side. Then, find the length of the diagonal?

brilliants [131]

Part 1:

The length of a side is 9 cm.

The square root of 81 is 9. We have to find the square root bc we know it is a square.

Part 2:

The hypotenuse (or diagonal) is c.

Using the Pythagorean Theorem
$a^{2} + {b}^{2} = {c}^{2}$
you would know that 9 is a and b.

${9}^{2} + {9}^{2} = {c}^{2}$
We know that
${9}^{2} = 81$
so you add 81 to 81.

$81 + 81 = 162$
81 and 81 make 162.

So you know ...

${c}^{2} = 162$
You take the square root of 162.

$\sqrt{162} = 12.73$
well almost.

You can write your answer as

$about \: 12.73 cm$
or
$\sqrt{162} cm$

Hope I helped!

3 0

3 years ago

1. Solve for x.

Allushta [10]

Answer:

i am able to do this only

hope this might help you