Answer:
<em>estimated sales on Wednesday is 19000 pounds.</em>
<em></em>
Step-by-step explanation:
On Monday, he sold 25196 pounds. Estimated to the nearest thousand that is 25000 pounds.
On Tuesday, he sold 18023 pounds. Estimated to the nearest thousand, that is 18000 pounds
Wednesday's sales is unknown. We designate as x
All in all he sold 62409. Estimated to the nearest thousand, that is 62000
The sales on Monday, plus sales on Tuesday, plus sales on Wednesday, must all sum up to the total sales.
25000 + 18000 + x = 62000
43000 + x = 62000
x = 62000 - 43000 = 19000
therefore <em>estimated sales on Wednesday is 19000 pounds.</em>
There you good luck with everything
Answer: 0.24 +/- 0.088 = (0.152, 0.328)
Step-by-step explanation:
The point estimate p is given by;
p= 24/ 100 = 0.24
Z value for 96% confidence interval is 2.05
The solution for the given confidence interval is derived using the equation
p +/- z√(pq/n)
Where p = 0.24 q= 1-p = 0.76, n=100 z= 2.05
= 0.24 +/- 2.05√(0.24×0.76/100)
= 0.24 +/- 2.05(0.0427)
=0.24 +/- 0.088
= ( 0.152, 0.328)
Answer:
To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x+3)(x+4) ( x + 3 ) ( x + 4 ) and (x+4)(x+5) ( x + 4 ) ( x + 5 ) , then the LCD would be (x+3)(x+4)(x+5) ( x + 3 ) ( x + 4 ) ( x + 5 ) .
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.
