Answer:
Following are the code to the given question:
Explanation:
<html>
<body>
<p align="center" >VOCATIONAL SUBJECT</p>
<table align="center" border="2">
<tr>
<td rowspan="2">YEAR</td>
<td colspan="5">Vocational course</td>
</tr>
<tr>
<td rowspan="1">IT</td>
<td rowspan="1">CS</td>
<td rowspan="1">EL</td>
</tr>
<tr>
<td colspan="1">2016</td>
<td colspan="1">66</td>
<td colspan="1">68</td>
<td colspan="1">60</td>
</tr>
<tr>
<td colspan="1">2017</td>
<td colspan="1">77</td>
<td colspan="1">78</td>
<td colspan="1">80</td>
</tr>
<tr>
<td colspan="1">2018</td>
<td colspan="1">60</td>
<td colspan="1">67</td>
<td colspan="1">70</td>
</tr>
</table>
</body>
</html>
Answer:
Answer is provided in the explanation section
Explanation:
Given data:
Bandwidth of link = 10* 106 bps
Length of packet = 12* 103 bits
Distance of link = 40 * 103m
Transmission Speed = 3 * 108 meters per second
Formulas:
Transmission Delay = data size / bandwidth = (L /B) second
Propagation Delay = distance/transmission speed = d/s
Solution:
Transmission Delay = (12* 103 bits) / (10* 106 bps) = 0.0012 s = 1.2 millisecond
Propagation Delay = (40 * 103 meters)/ (3 * 108mps) = 0.000133 = 0.13 millisecond
Answer: The answer would be Bethlehem!
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