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TO TRANSMIT INFORMATION ON THE INTERNET, LARGE FILES ARE BROKEN INTO PACKETS OF SMALLER SIZES. EACH PACKET HAS 1,500 BYTES OF IN

borishaifa [10]

Answer:

A. It would be needed 20 packets to transmit 30,000 bytes of information.

B. 45'000,000 (45 million) of bytes could be transmitted in 30,000 packets.

Step-by-step explanation:

1. Let's review the data given to us for solving the question:

Size of each packet of information = 1,500 bytes

Number of packets = P

Number of bytes of information = B

2. Let's answer the questions:

A. How many packets would be needed to transmit 30,000 bytes of information?

P = 30,000 ÷ 1,500

P = 20

It would be needed 20 packets to transmit 30,000 bytes of information

B. How much information could be transmitted in 30,000 packets?

B = 30,000 * 1,500

B = 45'000,000

45'000,000 (45 million) of bytes could be transmitted in 30,000 packets.

6 0

3 years ago

Sont answer if u don't have a answer

Readme [11.4K]

Answer:

no association because there is no specific pattern

7 0

3 years ago

Read 2 more answers

Is (x + 5) a factor of f(x) = x3 − 4x2 + 3x + 7?

Vaselesa [24]

Answer:

Step-by-step explanation:

Gcfygfvfvftf

6 0

3 years ago

A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of

Ivenika [448]

The equation for the height of the rocket at time t given

$h= -16t^2+192t$

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

$h= -16t^2+192t$

$560 = -16t^2+192t$

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

$560 = -16(t^2 - 12t)$

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

$560/-16 = -16(t^2-12t)/-16$

$-35 = t^2 -12t$

Now we will move -35 to the righ side by adding 35 to both sides.

$-35+35 = t^2-12t+35$

$0 = t^2 -12t+35$

$t^2-12t+35 = 0$

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

$t^2-12t+35 =0$

$(t-5)(t-7) =0$

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

$h= -16t^2+192t$

$0= -16t^2 + 192 t$

$0 = -16(t^2-12t)$

$-16(t^2-12t) = 0$

We will move -16 to the other side by dividing it to both sides.

$-16(t^2-12t)/-16 = 0/-16$

$t^2-12t = 0$

We will take out the common factor t from the left side. By taking out t we will get,

$t(t-12) = 0$

We will use zero product property now. By using that we will get,

$t = 0$

ans also $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0

3 years ago

Please help me !!!!!

romanna [79]

Answer:

3/8

Step-by-step explanation:

If 1 and 3 are odd numbers then all you have to do is count the number of 1 and 3's(3). then count the number of numbers (8).

3 0

3 years ago