Question 4. 348 +668 = 1228
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Q: Consider the value of a=40 and value of b=52. a=a+b b=a+b c=a+b d=a+b+c print(c) print(d) What is…
A: Answer: a = a+b , a=40 , b= 52 so , a = 40 + 52 = 92 , a = 92.b = a+b , b = 92 + 52 =…
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A: answers c=132 d=264
Q: Consider the value of a=38 and value of b=28. a=a+b b=a+b C=a+b d=a+b+c print(c) print(d) What is…
A: answers: 1) 160 2) 320 Explanation: a=38 b=28 a=a+b so a will become 66 b=a+b here a is 66 + 28…
Q: Consider the value of a=35 and value of b=56. a=a+b b=a+b C=a+b d=a+b+c print(c) print(d) What is…
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Q: The question is in the picture below.
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Q: 4) Convert the following binary fractions to decimal: 1)100001.111 2)111111.10011 3)1001100.1011
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A: Answer is 1966904.2
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A: Answer Print(c) will give: 262 print (d) will give: 524
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A: I Have answered this question in step 2.
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A: given
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Q: hi can u send me BMI calculation on python by using if, else
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Q: Consider the value of a=38 and value of b=60. a=a+b b=a+b c=a+b d=a+b+c print(c) print(d) What is…
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- Make a truth table for the given statement: p and not q A. TTFF B. FTFF C. FFFF D. NONE OF THESE ANSWERS6: Prove the identity of each of the following Boolean equations, usingalgebraic manipulation:(a) XY + XY + XY = X + Y(b) AB + BC + AB + BC = 1(c) Y + XZ + XY = X + Y + Z(d) XY + YZ + XZ + XY + YZ = XY + XZ + YZQuestion 4the complement of the following logical expression xy'+yw
- Computer Science: Use appended Table to show that the following Boolean identities are satisfied. Boolean variable and/or Boolean expressions postfixed with ’ symbol are considered negated. For instance, X’ is equivalent to NOT X.a. X + ( X’ Y + X Y )’ = X + Y’b. (( X Y ) Z ) + ( Y Z ) = Y Zc. X’ Y = X’ Y + X’ Y Zd. (X + Y) = XYZ’ + XY’Z’ + X’YZ’2. Simplify the following Boolean expressions, using three variable K-maps: (a) F(x,y,z) = xy+x′y′z′+x′yz′ (b) F(x,y,z) = x′y′+yz+x′yz′ (c) F(x,y,z) = x′y+yz′+y′z′1) a) simplify the following Boolean expression(show all steps): W'X'YZ'+WX'YZ'+WXYZ+W'X'Y'Z'T+W'X'YZ'+WXY'Z+WX'YZ'+WX'Y'Z+W'XYZ+W'X'Y'Z'T' the answer should be X'YZ'+W'X'Z'+WY'Z+XYZ b) prove the equality of these two expressions using a truth table. show all intermediate steps c) show that the simplifcation process in (a) has significantly reduced the number of logic gates required to implement the expression by: * drawing the two networks using only 2-input AND and OR gates. assume that you can use complemented variables as inputs. do not simplify * how many gates have been saved by this simplification ?
- Simplifies the following Boolean algebraic expressions. ABC (ABC’ + AB’C + A’BC) XY (X’YZ’ + XY’Z’ + X’Y’Z’) XY + XYZ’ + XYZ’ + XYZ4. Simplify the following Boolean expressions, using four-variable K-maps: (a) A′B′C′D′+AC′D′+B′CD′+A′BCD+BC′D (b) w′z+xz+x′y+wx′zSimplify the following Boolean expressions to a minimum number of literals indicating theorems used for each step . 1. ABC + A 'B + ABC' 2. x'yz + xz 3. (x + y)'(x' + y') 4. xy + x(wz + wz') 5. (BC' + A' D)(AB' + CD ')
- i) Carefully explain why p ↔ q can be written as “p is necessary and sufficient for q”. ii) Verify the second part of De Morgan’s laws, ∼ (p∨q) ≡ (∼ p) ∧(∼ q), in two ways: (a) Explain it using conversational English, using your own particular examples of statements for p and q. (b) Use a truth table.Can you explain me step by step, without using any calculator Calculate Gray code of the numbers:101010110110000111101010 without using any calculator without using any calculator without using any calculator Calculate the BCD code of the numbers:567898563 solve all or will dislike the answerhow this boolean expression works in what terms of the laws?: x'y'z' + x'y'z + x'yz + xy'z = x'y'(z+z') + z(x'y + xy') = x'y' + x'z + y'z I understood z+z' becomes 1 by the the inverse law, but how z(x'y + xy') goes x'z + y'z?