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saul85 [17]
3 years ago
11

function summedValue = SummationWithLoop(userNum) % Summation of all values from 1 to userNum summedValue = 0; i = 1; % Write a

while loop that assigns summedValue with the % sum of all values from 1 to userNum end
Engineering
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

function summedValue = SummationWithLoop(userNum)

% Summation of all values from 1 to userNum

  summedValue = 0;

  i = 0;

  % use a while loop that assigns summedValue with the

  % sum of all values from 1 to userNum

  while(i <= userNum)

      summedValue = summedValue + i;

      i = i + 1;

  end

end

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no

Explanation:

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4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb
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Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

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Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

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Attached below is the remaining answers

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Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra
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Answer:

Explanation:

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struct array{

int arr[10];

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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

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            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

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                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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