Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
I'm having trouble with this math problem, can someone please help!?
A freight version of Boeing's 747 airliner has 2.7×104 cubic feet of cargo space. A CD in its jewel case has a volume of 10.8 cubic inches. Assume that each CD holds 650 megabytes of information.
(a) If you packed the aircraft with as many CDs as it could hold, how much information could you store in megabytes?
(b) A "56K" modem typically transmits information at 45 kilobits per second. In computer terminology, a kilobit is 1024 bits, and a megabyte is 1,048,576 bytes. One byte equals eight bits of information. How long would it take to transmit the amount of information contained in one 747 planeful of CDs through a modem?
(c) A T3 high-speed internet connection transmits information at 45.0 megabits per second. As with bytes and bits, one megabyte equals eight megabits of information. Assume that it takes 12 hours to load, fly and unload the 747. How many T3 connections would it take to transmit this information in 12 hours?
1 Answer
- 1 decade agoFavorite Answer
Let's assume that the 747 can hold each CD, in its case, without space loss (for instance, think of packing oranges into a box...they don't exactly stack well)
2.7 * 10^4 cu.ft. = Volume of 747
10.8 * 10^1 cu.in = Volume of CD and case
1 cubic foot is equivalent to 1728 cubic inches
(2.7 * 10^4 * 1.728 * 10^3) / (10.8) =
0.432 * 10^7 CDs
4.32 * 10^6 CDs
4,320,000 CDs on 1 jet
a)
Each CD has 650 MB, so:
4.32 * 10^6 * 650 =
2808 * 10^6 MB =
2.808 * 10^9 MB
2,808,000,000 Megabytes of storage
b)
First, let's convert Megabytes to bits
1,048,576 (bytes/MB) * 8 (bits/byte) = 8,388,608 (bits/MB)
Now, the plane stores 2.808 * 10^9 MB
2.808 * 10^9 (MB) * 8.388608 * 10^6 (bits/MB) =
23.555211264 * 10^15 bits
A modem transfers 45 * 1024 bits per second
23.555211264 * 10^15 / (45 * 1024) =
511180800000 =
5.111808 * 10^11 seconds
(roughly 16200 years)
c)
Okay, we have to transmit 2,808,000,000 Megabytes in 12 hours (43200 seconds)
2.808 * 10^9 MB =
2.808 * 10^9 * 1024^2 Bytes =
2.808 * 10^9 * 1024^2 * 8 Bits =
23555211.264 * 10^9 Bits =
2.3555211264 * 10^16 Bits
We have to transmit
2.3555211264 * 10^16 Bits in 43200 Seconds
(2.3555211264 / 4.32) * (10^16 / 10^4) b/s =
0.54525952 * 10^12 b/s =
5,4525952 * 10^11 b/s
1 T3 line can transfer 45 Megabits per second
45 * 8 * 1,048,576 =
377487360 =
3.7748736 * 10^8 b/s
(5.4525952 / 3.7748736) * (10^11 / 10^8) =
1.4444444444444444444444444444444 * 10^3 =
1444.444.....
It would take 1445 T3 connections to transmit all of the information stored in the CDs in a 12 hour period
