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masha68 [24]
3 years ago
9

To transmit information on the internet, large files are broken into packets of smaller sizes. Each packet has 1,500 bytes of in

formation. An equation relating packets to bytes of information is given by b=1,500p where p represents the number of packets and b represents the number of bytes of information.
How many packets would be needed to transmit 30,000 bytes of information?
How much information could be transmitted in 30,000 packets?
Each byte contains 8 bits of information. Write an equation to represent the relationship between the number of packets and the number of bits.
Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

Step-by-step explanation:

Jjhu

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Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.

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A is the intersection of angle bisectors, so is the incenter of triangle EFG. It is also the intersection of the perpendicular bisectors of the sides of triangle EFG, so is the circumcenter.

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For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df
rjkz [21]

Answer:

a) For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

b) For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

t_{\alpha/2} =\pm 2.35

Part b

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

t_{\alpha/2} =\pm 2.62

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Answer:

63°

Step-by-step explanation:

TSU = 180 - 117 = 63

6 0
3 years ago
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