Answer:
4
Step-by-step explanation:
Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:
CW = UCL - LCL
Where,
CW= Class width
UCL= Upper class limit
LCL= Lower class limit
From the table above:
For class 1, CW = 64 - 60 = 4
For class 2, CW = 69 - 65 = 4
For class 3, CW = 74 - 70 = 4
For class 4, CW = 79 - 75 = 4
For class 5, CW = 84 - 80 = 4
Therefore, the class width of the grouped data = 4
Answer:
The value of the proposition is FALSE
Step-by-step explanation:
~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]
Let's start with the smallest part: ~X. The symbol ~ is negation when X is true with the negation is false and vice-versa. In this case, ~X is true (T)
~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ T) v (B ⊃ X)]
Now the parts inside parenthesis: (A ⊃ Y),(X ⊃ B),(A ≡ T) and (B ⊃ X). The symbol ⊃ is the conditional and A ⊃ Y is false when Y is false and A is true, in any other case is true. The symbol ≡ is the biconditional and A ≡ Y is true when both A and Y are true or when both are false.
(A ⊃ Y) is False (F)
(X ⊃ B) is True (T)
(A ≡ T) is True (T)
(B ⊃ X) is False (F)
~[(F) v ~(T)] ⋅ [~(T) v (F)]
The two negations inside the brackets must be taken into account:
~[(F) v F] ⋅ [F v (F)]
The symbol left inside the brackets v is the disjunction, and A v Y is false only with both are false. F v (F) is False.
~[F] ⋅ [F]
Again considerating the negation:
T⋅ [F]
Finally, the symbol ⋅ is the conjunction, and A v Y is true only with both are true.
T⋅ [F] is False.
Step-by-step explanation:
F -------> 1/d²
12 -------> 1/3²
12 ×(3/6)² ----->1/6²
3----->1/6²
Force = 3N if d=6
Answer:
1 2 3 6 7 14 21 42Step-by-step explanation:
all of them are divisible by 42
