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sveta [45]
2 years ago
15

A city planner is using simulation software to study crowd flow out of a large arena after an event has ended. The arena is loca

ted in an urban city. Which of the following best describes a limitation of using a simulation for this purpose?
a) The model used by the simulation software cannot be modified once the simulation has been used.
b) The model used by the simulation software often omits details so that it is easier to implement.
c) Running a simulation requires more time to generate data from trials than observing the crowd exiting the arena at various events.
d) Running a simulation requires a large number of observations to be collected before it can be used to explore a problem.
Computers and Technology
1 answer:
Damm [24]2 years ago
6 0

Answer:

d)

Explanation:

The main limitation of simulations is that running a simulation requires a large number of observations to be collected before it can be used to explore a problem. In a real life situation there are thousands of variables to take into consideration which can drastically affect the way that the situation unfolds at any given time. Therefore, in order to replicate/simulate such a scenario all of these variables need to be taken into consideration. This data can take a large amount of time to observe and collect in order to implement into the simulation so that it provides an accurate depiction of the problem.

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Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

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2 years ago
Lucas put a lot of thought into the design for his company's new white paper. He made sure to include repeating design elements
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C. provide consistency
8 0
2 years ago
Read 2 more answers
Which type of word processing programs enables us to include illustrations within the program?
bonufazy [111]

Answer:

C.  full featured

Explanation:

You need to make use of the full-featured version of the word processing programs to use the illustrations that are part of the program. And simple versions might not have that. By fully featured it means you need to have the license included or else you will be able to use it for a few days or a maximum of 3 months, and that is certainly not a good idea.

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2 years ago
Add the following 2's complement binary numbers. Also express the answer in decimal. a. 01+ 1011b. 11+ 01010101c. 0101+ 110d. 01
kenny6666 [7]

Answer:

(a) 01 + 1011

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 1011

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  0 1  1 </u>      ------------------> - 5 (in decimal)

<u>1  1  0 0 </u>    -------------------> - 4 (in decimal)

Explanation for (a)

<em>01 or 0001</em>

Since the most significant bit is 0 (leftmost bit), it shows that it is a positive number. Therefore, it can be directly converted to its decimal representation as follows:

0001 = 0 x 2^{3} + 0 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 1

0001 is + 1 in decimal.

<em>1011</em>

Its most significant bit(MSB) is 1 showing that it is negative.

Flipping all its bits and adding 1 to the result, will convert it to it's positive counterpart.

i.e 1011 => 0100 + 1 = 0101.

Converting the result (0101) to decimal gives

0101 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 5

1011 is -5 in decimal

<em>01 + 1011 or 0001+ 1011 = 1100</em>

The result (1100), of the addition of the two numbers, is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1100 => 0011 + 1 = 0100.

Converting the result (0100) to decimal gives

0100 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 0 x 2^{0} = 4

Therefore, the result, 1100 is -4 in decimal.

===================================================

(b) 11 + 01010101

Convert the two numbers into their 8-bit representation since one of them is in 8-bit.

PS: Taking 11, its MSB is 1 showing that it is a negative number. Also, it's 8-bit representation will mean pre-padding it with ones(1s) rather than zeros as follows. In other words, pre-padding a 2's complement  number is done using its sign bit.

=> 11 = 11111111

=> 01010101

Addition of the two numbers (in their 2's complement form) gives the following:

  1  1  1  1  1  1  1  1       ------------------> - 1 (in decimal)

<u>   0 1  0 1  0  1 0 1 </u>      ------------------> + 85 (in decimal)

<u>1 0 1  0 1  0  1 0 0</u>    ------------------->  + 84 (in decimal)

Discard the carry-out bit (1) making the result 01010100

Explanation for (b)

<em>11 or 11111111</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

11111111 => 00000000 + 1 = 00000001

Converting the result (00000001) to decimal gives 1

11 or 11111111 is  - 1 in decimal.

<em>01010101</em>

Its MSB is 0 showing that it is also positive. Now, convert to decimal

01010101 = 85

01010101 is + 85 in decimal.

<em>11111111 + 01010101 = 01010100</em>

The result (01010100), is also positive as the MSB is 0. Now, convert to decimal.

01010100 = 84

Therefore the result (01010100) is + 84 in decimal.

===================================================

(c) 0101 + 110

Convert the two numbers into their 4-bit representations.

=> 0101 = 0101

=> 110 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

     0 1 0 1       ------------------> + 5 (in decimal)

<u>      1  1 1  0 </u>      -----------------> - 2 (in decimal)

<u>   1 0 0 1  1 </u>    -------------------> + 3 (in decimal)

Dicard the carry-out bit (leftmost bit) to get 0011 as result.

Explanation for (c)

<em>0101</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now, convert to decimal.

0101 =  5

0101 is +5 in decimal.

<em>110 or 1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 =  2

1110 is -2 in decimal

<em>0101 + 110 = 0101+ 1110 = 0011</em>

The result (0011), is a positive number as the MSB is 0.

Converting the result (0011) to decimal gives

0011 = 3

Therefore, the result, 0011 is 3 in decimal.

===================================================

(d) 01 + 10

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 10 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  1  1  0 </u>      ------------------> - 2 (in decimal)

<u>1  1  1  1 </u>    -------------------> - 1 (in decimal)

Explanation for (d)

<em>01 or 0001</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now convert to decimal.

0001 = 1

0001 is + 1 in decimal.

<em>1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010  = 2

1110 is -2 in decimal

<em>01 + 1011 or 0001+ 1110 = 1111</em>

The result (1111), is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1111 => 0000 + 1 = 0001.

Converting the result (0001) to decimal gives

0001 = 1

Therefore, the result, 1111 is -1 in decimal.

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The correct answer is

A.  the incorporation of technology into objects we use regularly

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