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pychu [463]
3 years ago
14

Technician A says that a common type of spontaneous combustion occurs when rags or papers that are soaked in solvents end up in

a garbage can. Technician B says you must clearly mark smoking and nonsmoking areas with easily recognizable symbols. Who is correct? A. Technician A only B. Technician B only C. Neither Technician A nor B D. Both Technicians A and B
Computers and Technology
2 answers:
Ne4ueva [31]3 years ago
4 0

Answer:

B

Explanation:

took the pf test

lakkis [162]3 years ago
3 0
Hello!

I believe the correct answer would be B. Technician B, because a common type of spontaneous combustion occurs only when something gets really hot and then ignites itself, so Technician A is obviously incorrect.

I really hope my answer helped you! :)
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Only one person can receive the same email at the same time true or false
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When you sustain program implementation by staying true to the original design, it is termed A. Goals and objectives B. Program
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Determine and prove whether an argument in English is valid or invalid. About Prove whether each argument is valid or invalid. F
yawa3891 [41]

Answer:

Each understudy on the respect roll got an A.  

No understudy who got a confinement got an A.  

No understudy who got a confinement is on the respect roll.  

No understudy who got an A missed class.  

No understudy who got a confinement got an A.  

No understudy who got a confinement missed class  

Explanation:

M(x): x missed class  

An (x): x got an A.  

D(x): x got a confinement.  

¬∃x (A(x) ∧ M(x))  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ M(x))  

The conflict isn't considerable. Consider a class that includes a lone understudy named Frank. If M(Frank) = D(Frank) = T and A(Frank) = F, by then the hypotheses are overall evident and the end is counterfeit. Toward the day's end, Frank got a control, missed class, and didn't get an A.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope got a confinement.  

Penelope missed class.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a confinement.  

Each understudy who missed class got a confinement.  

Penelope is an understudy in the class.  

Penelope didn't miss class.  

Penelope didn't get imprisonment.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got imprisonment.  

Each understudy who missed class or got imprisonment didn't get an A.  

Penelope is an understudy in the class.  

Penelope got an A.  

Penelope didn't get repression.  

M(x): x missed class  

S(x): x is an understudy in the class.  

D(x): x got a repression.  

An (ax): x got an A.  

H(x): x is on the regard roll  

An (x): x got an A.  

D(x): x got a repression.  

∀x (H(x) → A(x)) a  

¬∃x (D(x) ∧ A(x))  

∴ ¬∃x (D(x) ∧ H(x))  

Real.  

1. ∀x (H(x) → A(x)) Hypothesis  

2. c is a self-self-assured element Element definition  

3. H(c) → A(c) Universal dispatch, 1, 2  

4. ¬∃x (D(x) ∧ A(x)) Hypothesis  

5. ∀x ¬(D(x) ∧ A(x)) De Morgan's law, 4  

6. ¬(D(c) ∧ A(c)) Universal dispatch, 2, 5  

7. ¬D(c) ∨ ¬A(c) De Morgan's law, 6  

8. ¬A(c) ∨ ¬D(c) Commutative law, 7  

9. ¬H(c) ∨ A(c) Conditional character, 3  

10. A(c) ∨ ¬H(c) Commutative law, 9  

11. ¬D(c) ∨ ¬H(c) Resolution, 8, 10  

12. ¬(D(c) ∧ H(c)) De Morgan's law, 11  

13. ∀x ¬(D(x) ∧ H(x)) Universal hypothesis, 2, 12  

14. ¬∃x (D(x) ∧ H(x)) De Morgan's law, 13  

4 0
3 years ago
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KengaRu [80]

Answer:

Explanation:

pop(): Remove an item from the end of an array  

push(): Add items to the end of an array  

shift(): Remove an item from the beginning of an array  

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