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The baud rate is the rate at which information is transferred in a communication channel. Serial ports use two-level (binary) signaling, so the data rate in bits per second is equal to the symbol rate in bauds. Ref: https://en.wikipedia.org/wiki/Serial_port#Speed. ... bit $=12\text{ bits}$. So, number of characters transmitted per second $=\dfrac{9600}{12}=800$. Correct Answer: $B$
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Option is D. $foo$ is printing the lowest digit. But the $printf$ inside it is after the recursive call. This forces the output to be in reverse order $2, 0, 4, 8$ The final value $sum$ printed will be $0$ as $C$ uses pass by value and hence the modified value inside $foo$ won't be visible inside $main$.
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$2(n-1)$ the spanning tree will traverse adjacent edges since they contain the least weight.
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What is feedback on Y indicating...?Do we need to take into account...since it is a combinational circuit
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answer - C There are 2 decision points in this module 1. while condition and 2. if condition hence cyclomatic complexity = number of decision points + 1 = 3
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There is no limit on a large value being considered. f(n)= n^0.0000001 = n^(10^-7) g(n)=lg n(base 2) n= 10^8 f(n)= 1.000001842 , g(n)=26.565 Now, I take n = 10^(10^8) f(n) = n^(10^-7) = (10^(10^8))^(10^-7) = 10^(10^(8-7)) = 10^10 g(n) = 10^8 log10 2 So, f(n) becomes larger. Thus for any x, we can have an n, where nx becomes larger than log n and stays larger from there on wards for any higher n.
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The function terminates for all powers of $2$ (which is infinite), hence (i) is false and (ii) is TRUE. Let $n = 5.$ Now, recursive calls will go like $5 - 14 - 7 - 20 - 10 - 5 -$ And this goes into infinite recursion. And if we multiply $5$ with ... possible, there are infinite recursions possible (even considering this case only). So, (iv) is TRUE and (iii) is false. So, correct answer is (D).
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How query 2 will not produce the correct result? I am getting the desired result from query 2.
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