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Question 12 5 pts Which are ways which we could describe the relationship between a language L1 and a Finite state machine M1: L1 = {0, 10, 110, 000, 1110, 111000, ...} M1: 1 0 $s_0$ $s_1$ 0 1 L1 is a subset of M1 L1 = L(M1) L1 recognizes the alphabet (Sigma) M1 recognizes L1 The language of M1 is L1 All of the above Question 13 5 pts Choose the regular expression which describes the language that is accepted by the following FSA: Hint: Think of strings the Regular Expression could generate but are NOT accepted by the FSA? and visa versa $s_1$ a $s_0$ b a $s_2$ a $s_3$ a,b a,b $s_2$ a*b*aba(a|b)(a|b)* a*bb+a*(a|b)(a|b)* a*b+aa(a|b)(a|b)* a*b*a*b None of the above 

          Question 12
5 pts
Which are ways which we could describe the relationship between a language L1 and a Finite state
machine M1:
L1 = {0, 10, 110, 000, 1110, 111000, ...}
M1:
1
0
$s_0$
$s_1$
0
1
L1 is a subset of M1
L1 = L(M1)
L1 recognizes the alphabet (Sigma)
M1 recognizes L1
The language of M1 is L1
All of the above
Question 13
5 pts
Choose the regular expression which describes the language that is accepted by the following FSA:
Hint: Think of strings the Regular Expression could generate but are NOT accepted by the FSA? and visa
versa
$s_1$
a
$s_0$
b
a
$s_2$
a
$s_3$
a,b
a,b
$s_2$
a*b*aba(a|b)(a|b)*
a*bb+a*(a|b)(a|b)*
a*b+aa(a|b)(a|b)*
a*b*a*b
None of the above
Show more…
Question 12 5 pts Which are ways which we could describe the relationship between a language L1 and a Finite state machine M1: L1 = 0, 10, 110, 000, 1110, 111000, ... M1: 1 0 s0 s1 0 1 L1 is a subset of M1 L1 = L(M1) L1 recognizes the alphabet (Sigma) M1 recognizes L1 The language of M1 is L1 All of the above Question 13 5 pts Choose the regular expression which describes the language that is accepted by the following FSA: Hint: Think of strings the Regular Expression could generate but are NOT accepted by the FSA? and visa versa s1 a s0 b a s2 a s3 a,b a,b s2 a*b*aba(a|b)(a|b)* a*bb+a*(a|b)(a|b)* a*b+aa(a|b)(a|b)* a*b*a*b None of the above

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Trishna Knowledge Systems 2018 Edition
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Question 12 5 pts Which are ways which we could describe the relationship between a language L1 and a Finite state machine M1: L1={0,10,110.000,1110.111000,...} M1: L1 is a subset of M1 L1 = L(M1) L1 recognizes the alphabet (Sigma) M1 recognizes L1 The language of M1 is L1 All of the above Question 13 5pts Choose the regular expression which describes the language that is accepted by the following FSA Hint: Think of strings the Regular Expression could generate but are NOT accepted by the FSA? and vice versa a^(x)b^(x)aba(a|b)(a|b)^(x) a^(2)bb+a^(2)(a|b)(a|b)^(2) a^(2)b+aa(a|b)(a|b)^(2) a^(x)b^(x)a^(x)b None of the above
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2. (a) Define and draw a finite State Machine (FSA) M1 for the following patterns. Mark your Final States clearly. all strings that begin OR end with an a. Note that the alphabet A = {a, b} (in other words contains letters a and b only [Hint: first write the regular expression to make it easy for you]. (6 points) Show that the following strings will be accepted by the FSA M1 (by providing the sequence of states and the type of the end state): (5 points) • abb • bba • abba • a • aa Show that the following strings will not be accepted by the FSA M1 (by providing the sequence of states and the type of the end state): (5 points) • baab • bb • bbaab • b • λ (b) Define and draw a finite State Machine (FSA) M2 for the following patterns. Mark your Start State and Final States clearly. all strings that have even number of 0's followed immediately by even number if 1s. The string may start with anything other than 0 or 1 from the alphabet {0, 1, 2, 3} (6 points) Show that the following strings will be accepted by M2: (by providing the sequence of states and the type of the end state): (3 points) • 2001111 • 333322211 • 323200 Show that the following strings will not be accepted by M2: (by providing the sequence of states and the type of the end state): (3 points) • 0011 • 20111 • 33320101

Supreeta N.
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Q#1 Write the Regular Expression of the language L that will accept any string consisting entirely of b's and will also accept any string in which the number of a's is divisible by 3. Also, make its Finite Automata and remember that the machine does not accept null, and the start and ending state cannot be the same. There is a clear representation of state numbers along with their signs and clearly represents the character that's transition is taking place. Also, label dead-end states properly. Q#2 Consider a Language L that accepts all strings that have an odd number of occurrences of the substring abc over the alphabet = {ab.c}. Build a Finite Automata. There is a clear representation of state numbers along with their signs and clearly represents the character that's transition is taking place. Also, label dead-end states properly. Also, justify your Finite Automata that it accepts the given Language and rejects those strings which are not included in the above Language. Hint: The final state can be more than one.

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Consider the Deterministic Finite Automata, A = (QA, ΣA, δA, q0A, FA) and B = (QB, ΣB, δB, q0B, FB) with QA ∩ QB = ∅, ΣA ∩ ΣB = ∅ where ∅ stands for the null set. Let LA ⊆ ΣA* and LB ⊆ ΣB* be the languages accepted by A and B respectively and define the interleaved language: LA || LB := {s ∈ (ΣA ∪ ΣB)* | s ↑ A ∈ LA and s ↑ B ∈ LB} where s ↑ A and s ↑ B stand for the projection of s on ΣA and ΣB obtained by erasing all the symbols of s in ΣB and ΣA respectively. (a) Define the interleaving product A || B of A and B as a DFA that accepts the language LA || LB (b) Compute a DFA that accepts the language L = (0.1)* || (a.b)*

Sri K.
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Transcript

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00:01 Hello student in this problem the mu that is population mean is given 70 rupees 70 dollars and standard deviation of a population is given 11 dollars.
00:10 So let's define x be the customer charge by the value of the credit card customer charge, okay, so in first we have to probability that the customer charge is vary between 70 to 85 in between 83 dollars so this find this probability we are going to use the rate score formula that is z is equal to x minus mu upon sigma let's find the z1 value for both limits that is 70 and 83 so z1 be the value for 70 70 on a 70 and we have mu as a 11 this compares 0 and z2 for 83 minus 70 upon sigma 11 and this value come up as 0 .37 so the required probability is equal to the probability that z is lie between 0 to 1 point 1 .182 and this equal to probability of z is less than 1 .182 minus probability of z less than 0 so this calculation is 0 .879 minus z less than 0 is nothing but due to symmetry the half portion is 0 .5 this value come up as 3 .79 similarly for part b.
01:48 We have to find the probability that the applied charge is lie between 55 and 82 so z1 for 55 minus mu value of 70 upon sigma that is 11 compares minus 1 .36 and z2 value for 82 come up as 1 .09 therefore the required property is equal to the z value lie between 1 .36 minus 1 .36 and minus 1 .09 from calculating standard normal distribution table this answer come up as 0 .86214 minus 0 .06891 that is is equal to 0 .77523 that's answer of part 3.
02:46 That's a probability of part 2 and part 3 we have to find the credit card charges are less than 52...
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