Find x if a = 13 and C = 47.<br> in.<br> 1n.

QUESTION 4 of 5: You are choosing among three apartments. Apartment 1 costs $700 per month, and lets you walk to work. Apartmen

Fittoniya [83]

Apartment 3 because you can transport faster and it only cost $160 more than the bus stop which can cause you stops to et to your destination

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3 years ago

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Six more than the quotient of four and a number xxx.

Shalnov [3]

Answer:

$\large\boxed{\dfrac{4}{x}+6}$

Step-by-step explanation:

Six more than the quotient of four and a number x:

$\dfrac{4}{x}+6$

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3 years ago

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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

7 0

3 years ago

Carlton is solving a linear equation using the steps shown.

mario62 [17]

Answer:

D. Divide each side of the equation by –2.

Step-by-step explanation:

We have been given that Carlton is solving a linear equation using the steps shown.

Equation: $-4x+12=-6x$

Step 1: Add 4x to both sides: $12=-6x+4x$

Step 2: Combine like terms: $12=-2x$

To solve our given equation we will divide both sides of our equation by -2.

$\frac{12}{-2}=\frac{-2x}{2}$

$-6=x$

Therefore, the next step Carlton needs to complete in order to solve the equation is divide each side of the equation by –2.

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3 years ago

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The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am

posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

$N = N_0 * (\frac{1}{2})^{t/p$

$N_0$ is the starting amount of parent element.

$N$ is the end amount of parent element

$t$ is the time elapsed

$p$ is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days$

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2} = 37g$

c.

Now we just gotta do some algebra. Use the original function but this time, replace $N(t)$ with 10g and solve algebraically.

$10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}$

Take the log of both sides.

$log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})$

Use the exponent rule for log laws that, $log(b^x) = x*log(b)$

$log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})$

$\frac{log(\frac{5}{37})}{log(\frac{1}{2})} = \frac{t}{2.8days}$

$2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})} = t$

slap that in your calculator and you get

$t = 8.1 days$

7 0

2 years ago

Find x if a = 13 and C = 47. in. 1n.