Final Exam S11 Solutions P 1 A B C P 2 A B P 3 P 4 P 5 P 6 P 7 P 8 P

Final Exam S11 Solutions P 1 A B C P 2 A B P 3 P 4 P 5 P 6 P 7 P 8 P P 1 a b c p 2 a b. p 3. p 4 p 5 p 6 p 7 p 8 p 9. p 10 a b. total. the circuit in figure 1 is the current amplifier with gain il ii = 10. (a) find the required value for r (b) find the input and output resistances r i and ro (c) if rl = 1 k and vo is clipped at 12v, what range of il is possible?. Ecse 2050 intro to electronics final exam name: 1 p 1 a b c p 2 a b p 3 p 4 p 5 p 6 p 7 p 8 p 9 p 10 a b total 1. the circuit in figure 1 is the current amplifier with gain l i i i = 10.

Solution Final Exam Version Complete Solutions Class Version Studypool If x ~ n (2, 9) ⇒ μ = 2. σ 2 = 9 ⇒ σ = 3. a) find p (x ≥ 2) p (x ≥ 2) = p (x − μ σ ≥ 2 − 2 3) = p (z ≥ 0) = 1 − p (z <0) = 1 − 0.5 p (x ≥ 2) = 0.5. explanation: in z table, the area left to the z score of 0 is 0.5. subtract 0.5 from 1 to get the required p view the full answer step 2. A sample space contains six sample points and events a, b, and c as shown in the venn diagram. the probablities of the sample points are p(1)=0.2, p(2)=0.25, p(3)=0.35, p(4)=0.05, p(5)=0.1, p(6)=0.05. use the venn diagram and the probabilities of the sample points to find: p(c)= 0.15. 8. 1 7 cp. study with quizlet and memorize flashcards containing terms like consider the following premises of a natural deduction proof in propositional logic. 1. t ≡ (~s • ~s) 2. t ⊃ (s • r) 3. ~s • ~r, consider the following premises of a natural deduction proof in propositional logic. 1. ~ (e ∨ i) 2. The variance (σ2) of a discrete random variable x is the number. σ2 = ∑(x − μ)2p(x) which by algebra is equivalent to the formula. σ2 = [∑x2p(x)] − μ2. the standard deviation, σ, of a discrete random variable x is the square root of its variance, hence is given by the formulas. σ = √∑(x − μ)2p(x) = √[∑x2p(x)] − μ2.