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3 years ago

Which of the following is the most appropriate unit to describe the rate at which data is transferred using the internet?

2 answers:

4 0

Megabits per second, because the independent quality is time in seconds and dependent quality is amount of data in megabits.

sveticcg [70]3 years ago

3 0

The correct answer is:

Mbps, or megabits per second.

Explanation:

The standard measurement unit of data sent or received over the Internet is megabits per second. This means "millions of bits per second."

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Round to the nearest hundred 16,537

mart [117]

Answer:

16,500

Step-by-step explanation:

thats it

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3 years ago

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Paul [167]

The area of rectangle is lengthXwidth. A rectangular prism is essentially a rectangle with a height. So you multiply the area of rectangle with a height to get volume of prism.

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2 years ago

36 cans of soda coast $8.90 how much does 12 cans cost

maw [93]

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2 years ago

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A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are si

nordsb [41]

Answer:

Step-by-step explanation:

Let GS denote the good service and SP denote the signal problem.

A subway has good service 70% of the time, that is, $P(GS)=0.7$ and a subway runs less frequently 30% of the time because of the signal problems, that is, $P(SP)=0.3$ .

If there are signal problems, the amount of time T in minutes that have to wait at the platform is described by the probability density function given below:

$P_{T|SP}(t)=0.1e^{0.1t}$

If there is good service, the amount of time T in minutes that have to wait at the platform is described the probability density function given below:

$P_{T|GOOD}(t)=0.3e^{0.3t}$

(a)

The probability that you wait at least 1 minute if there is good service P(T ≥ 1| GS) is obtained as follows:

$P(T\geq 1|GS)=\int\limits^{\infty}_1 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_1 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_1\\\\=-(0-e^{-0.3})\\=0.74$

(b)

The probability that you wait at least 1 minute if there is signal problems P(T ≥ 1| SP) is obtained as follows:

$P(T\geq 1|SP)=\int\limits^{\infty}_1 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_1 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.3})]\\\\=-(e^{-0.1t})\limits^{\infty}_1\\\\=-(e^{\infty}-e^{-0.1})\\=-(0-0.904)\\=0.904$

(c)

After 1 minute of waiting on the platform, the train is having signal problems follows an

exponential distribution with parameter $\lambda= 0.1$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is obtained using the result given below:

$P(SP|T\geq 1)=\frac{P(T\geq 1|SP)P(SP)}{P(T\geq 1)}$

$P(T\geq 1|GS)=0.74, P(T\geq 1|SP)=0.904$

Now calculate the $P(T \geq 1)$ as follows:

$P(T \geq 1)=P(T\geq 1|SP)P(SP)+P(T\geq 1|GS)P(GS)\\=(0.904)(0.3)+(0.74)(0.7)=0.7892$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is calculated as follows:

$P(SP|T\geq 1)= \frac{0.904 \times 0.3}{0.7892}
= 0.3436
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.3436$ .

(d)

After 5 minutes of waiting on the platform, the train is having signal problems follows an exponential distribution with parameter $\lambda= 0.1$ .

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is obtained using the result given below:

$P(SP|T\geq 5)=\frac{P(T\geq 5|SP)P(SP)}{P(T\geq 5)}$

First, calculate the $P(T\geq 5|SP)$ as follows:

$P(T\geq 5|SP)=\int\limits^{\infty}_5 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_5 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.1})]\\\\=-(e^{-0.1t})\limits^{\infty}_5\\\\=-(e^{\infty}-e^{-0.5})\\=-(0-0.6065)\\=0.6065$

Now, calculate the $P (T\geq5|GS )$ as follows:

$P(T\geq 5|GS)=\int\limits^{\infty}_5 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_5 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_5\\\\=-(0-e^{-1.5})\\=0.2231$