If possible, find AB, BA, and A. (If not possible, enter IMPOSSIBLE.) 8-9 0 1 0 0 0 1 0 0 -6 В 3 A 0 0 1 -1 4 15 (а) АВ АВ 3 (b) ВА ВА %3 (c) A2 A2

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A: It is easy we can solve this problem by using the concept of union and intersection of a set.

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A: Given : For first class  class = 50-52  frequency = 5

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Q: AB N I 1 15°19'27" / OBC -DE OEA OCD AB = 87° 50'17" OBC= 78°53'09" OCD 120° 47'17" DE-69°32'42.5"…

A: Solution is given below:

Q: 14 H0: μ1 ≥ μ2  H1: μ1 < μ2   Z  = 2.789 Z0 = 2.879 Thus, we ................    Select one: a.…

A: Given information: H0 : μ1≥μ2 HA : μ1&lt;μ2 z  = 2.789z0 = 2.879

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A: Answer

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Q: A = BAB=C A=C AnB 0= AOB #0 A = BAB C A=C AUB 0 A#0v B#0

A: a) Let x∈A ⇒x∈B (∵A=B)⇒x∈C (∵B=C)⇒A⊂C Similarly C⊂A. Here A⊂C and C⊂A⇔A=C

Q: a A 1. A = 23.5°, b = 10 2. A = 41.5°, b = 3. A = 52.6°, c = 54 4. A = 54.8°, c = R - 23 8° h = %3D

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Q: _1. V3 • V1Z _2. V27 • V8 3. VZ • V7 _6. V8÷ vZ 7. V24 + V2 _8. V10 + V7 3 2 4. 3 3 9. 25 9. 5. V7…

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Q: 13. Evaluate. ,zE (-x,0] 24 h(z) = zE (0, 8] (2– 1)(z+3) , z€ (8, x) h(-3) = [ %3D A. 27 B. 20 C.…

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A: We are given some statements to verify De Morgan's theorem. We can do this in the following way:

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A: Solution: The given claim is that the average review time of student per subject is 45 mins.

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Q: 5 7 -5 A 0 4 -1 2 8 -3 matrisini köşegen hale getiren P matrisini elde ediniz.

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Q: 12. H0: μ1 ≥ μ2 H1: μ1 < μ2 Z = 2.789 Z0 = 2.879 Thus, we ................ Select one: a. Don’t…

A: Given information: H0 : μ1≥μ2H1 : μ1&lt;μ2 z = 2.789 z0 = 2.879

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