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larisa86 [58]
4 years ago
3

Design a circuit that will tell whether a given month has 31 days in it. The month is specified by a 4-bit input, A3:0. For exam

ple, if the inputs are 0001, the month is January, and if the inputs are 1100, the month is December. The circuit output, Y, should be HIGH only when the month specified by the inputs has 31 days in it. However, if the input is not a valid month (such as 0 or 13) then the output is don’t care (can be either 0 or 1).

Engineering
1 answer:
grandymaker [24]4 years ago
7 0

Answer:

  see attachments

Explanation:

A Karnaugh map for the output is shown in the first attachment. The labeled and shaded squares represent the cases where Y = HIGH. The associated logic can be simplified to

  Y = A3 xor A0

when the don't care at 1110 gives an output of HIGH.

__

The second attachment shows a logic diagram using a 4:1 multiplexer to do the decoding. A simple XOR gate would serve as well. If AND-OR-INV logic is required, that would be ...

  Y = Or(And(A3, Inv(A0)), And(A0, Inv(A3)))

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You are given a rectangular piece of cloth with dimensions X by Y, where X and Y are positive integers, and a list of n products
Bond [772]

The proof that recursion is exponential and that dynamic programming is polynomial is given by the formula;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

To prove that the recursion is exponential and that dynamic programming is polynomial. we will do so as follows;

Let us first have the assumption that the cloth is in such a manner that  either way, a product can be oriented. This implies that that after a cut, we will now have two pieces of cloth.

Now, we will make a list of the side lengths of the products that can fit in the piece after which we will consider a vertical cut for each of the side length as well as a horizontal cut for each of the side length, then we apply the same algorithm to each of the two resulting pieces.  

Thus, after the point above, it is likely true that in some instances, there may be a place to cut that is not at a product side length. However, It might be better for us to make a list of lengths composed of one or more pieces side by side as long as the sum is less than the length of the side being considered.

 

Lastly, we would note that this recursive approach is not limited to just two -dimensional problems as It could also be applied to a single or more than two dimensions. A useful proof would be to prove it for one dimension, then assuming it is true for n dimensions, prove it is true for n + 1 dimensions.

Thus;

P(x,y) = max{

P(x,y)

max (1 <= h <= X) { P[h, Y] + P(X - h, Y) }

max (1 <= v <= Y) { P[X, v] + P[X, Y - v] }

}

Read more at; brainly.com/question/11665190

3 0
3 years ago
ممكن الحل ............
Roman55 [17]

Answer:

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Explanation:

4 0
4 years ago
Create a file named students containing the following data in your current directory. Each line in this file represents a studen
Ostrovityanka [42]

Answer:

#!/bin/bash

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# -----------------------------------------------------------------------------

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# init

file="current_line.txt"

let count=0

# this while loop iterates over all lines of the file

while read LINE

do

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   ((count++))

   # write current line to a tmp file with name $file (not needed for counting)

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Explanation:

4 0
4 years ago
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Ad libitum [116K]

Answer:horizontal

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3 0
3 years ago
Read 2 more answers
A fair cubical die is thrown four times. use the binomial probability formula to calculate the probability of at least two 3's.
Marina86 [1]

By using the binomial probability formula, the probability of at least two 3's is equal to 0.1319.

<h3>How to calculate the probability of at least two 3's?</h3>

Since the fair cubical die is thrown four times, the number of times is given by n = 4 and the probability of 3's is P = 1/6.

Mathematically, the binomial probability formula is given by:

P(X \geq x) =\sum^{n}_{r=x} ^nC_r (p)^r (q)^{(n-r)}\\\\P(X \geq 2) = \; ^4C_2 (\frac{1}{6} )^2 (\frac{5}{6} )^{(4-2)} + ^4C_3 (\frac{1}{6} )^3 (\frac{5}{6} )^{(4-3)} + ^4C_4 (\frac{1}{6} )^4 (\frac{5}{6} )^{(4-4)}\\\\P(X \geq 2) = 6 \times \frac{1}{36} \times \frac{25}{36} + 4 \times \frac{1}{216} \times \frac{5}{36} + 1 \times \frac{1}{1296} \times 1\\\\P(X \geq 2) =  \frac{150}{1296} +  \frac{20}{1296}+  \frac{1}{1296}

P(X ≥ 2) = 19/144

P(X ≥ 2) = 0.1319.

Read more on probability here: brainly.com/question/25870256

#SPJ1

6 0
2 years ago
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