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ZOO507 GDB 1 Solution and Discussion

ZOO507 - Principles of Animal Ecology

zareen Cyberian's Gold last edited by

Total Marks 5
Starting Date Monday, January 20, 2020
Closing Date Monday, January 27, 2020
Status Open
Question Title Ecology
Question Description

Differentiate between mimicry and camouflage with one example?
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zareen Cyberian's Gold last edited by

please share idea
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CS611 Assignment 3 Solution and Discussion
CS611 - Software Quality Engineering • cs611 assignment 3 solution discussion spring 2020 • • zaasmi

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Solution Ideas Please!
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SOLVED PHY301 Assignment 1 Solution and Discussion
PHY301 - Circuit Theory • phy301 assignment 1 solution discussion fall 2019 • • zareen

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Q. 1 Solution:
Starting from lower side of circuit network, we can see that 2Ω and 2Ω are in series, so their combined effect is
2+2=4Ω
97b5d636-b0d8-40d1-8f33-9266b3779585-image.png

This 4Ω is in parallel of 4Ω to lower side, so their equivalent is

1b08526d-0c95-43b2-9f73-3927652469d8-image.png

0888b8c5-aeba-4acd-9184-9f1a8f70b1ef-image.png

This 2Ω is in series of 5Ω and 1Ω, so the sum of series resistances is

2Ω+ 5Ω+1Ω=8Ω
f1ffb19b-69a3-4eb0-8d8d-4eb1316289f4-image.png

Now at upper side, 4Ω and 2Ω are in series
4Ω+ 2Ω=6Ω
b3a9072f-b02e-4881-aaf3-9b389aab13e0-image.png

This 6Ω is in parallel of 6Ω, so their equivalent is
df9ccb62-3a78-4fec-a9d8-9044a372be75-image.png

cac80a33-5165-4e48-b124-810aaad24d2e-image.png

This 3Ω is in series of 3Ω and 1Ω, so the sum of series is

3Ω+ 3Ω+1Ω=7Ω
051e272e-1e7b-48b8-9359-4810dcb187d5-image.png

Now we can see 7Ω becomes in series of 8Ω, so equivalent resistance is
Req =7Ω + 8Ω
Req=15Ω

Q.2 Solution:
1)
To calculate the source current Is, Firstly, we calculate the total resistance
RT= R1+R2+R3
=6Ω
V=IR
IS=VS/RT
= 12/6
=2A

2)
Since all resistances are in series, same 2A current pass through each resistance.
V1= R1IS
=12
=2V
V2=R2Is
=22
=4V
V3=R3IS
= 32
=6V

3)
P1 =I2R1
= (2)2 *1
=4W
P2 =I2R2
= (2)2 *2
= 8W
P3 =I2R3
=(2)2 *3
= 12W
= 12W

4)-
Ps = VsIs
P =122
=24 W
PT=P1+P2+P3 or PT=I2RT
=4+8+12
=24 W
SOLVED CS507 Assignment 2 Solution and Discussion
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CS507 2 SOLUTION.docx
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HR Digital Transformation Mobile App
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SOLVED CS614 Assignment 2 Solution and Discussion
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@zareen said in CS614 Assignment 2 Solution and Discussion:

Calculate the total space reserved in memory using normalization

Q. 1. Calculate the total space reserved in memory using normalization1. Calculate the total space reserved in memory using normalization

Q. 1 Solution:
Total space used
= 10 (Million records) x 40 (Bytes header) + 110 (Million records) x 70 (Bytes header)
= (1000000040) + (11000000070)
=8,100,000,000
= 8.1 GB

Q. 2. Calculate the total space reserved in memory after de-normalization using pre-joining technique

Q. 2 Solution:
Total space used
= (70 + 40 – 10) (Bytes header) x 110 (Million records)
=100*110000000
=11,000,000,000
= 11 GB
M

SOLVED STA631 GDB 1 Solution and Discussion
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@moaaz said in STA631 GDB 1 Solution and Discussion:

Part b):
What are the types of sampling design? Explain each type.

Types of sampling design in Research Methodology

There are different types of sample designs based on two factors viz., the representation basis and the element selection technique. On the representation basis, the sample may be probability sampling or it may be non-probability sampling. Probability sampling is based on the concept of random selection, whereas non-probability sampling is ‘non-random’ sampling. On element selection basis, the sample may be either unrestricted or restricted. When each sample element is drawn individually from the population at large, then the sample so drawn is known as ‘unrestricted sample’, whereas all other forms of sampling are covered under the term ‘restricted sampling’. The following chart exhibits the sample designs as explained above.

Thus, sample designs are basically of two types viz.,
Non-probability sampling Probability sampling.
We take up these two designs separately.
abacf2b1-11a7-4f87-9833-a3d3d02c6c65-image.png
CHART SHOWING BASIC SAMPLING DESIGNS
Reff
MCM101 Assignment 2 Solution and discussion
MCM101 - Introduction to Mass Communication • mcm101 assignment 2 solution discussion fall 2019 • • zareen

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@Ãyêshä-Dõgãr ok let me try plz wait

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ZOO507 GDB 1 Solution and Discussion

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