8.876 to the nearest whole number
8.876 can be rounded to 9 because if we have a number greater than 5 we round to next number i.e is 9 but if we have a number smaller than 5 then we round to the previous number that is 8.
So your answer is 9
Answer:
The % increase is 7.179 but see the remark I made below. What I have given and what it could be is a rounding error. Less than 2 dollars error is not much when you are calculating something for someone just to show them how to do it.
Step-by-step explanation:
242,555 * x/100 = 259,970 Multiply by 100
242,555 * x = 259970 * 100
242,555 * x = 25997000 Divide by 242555
x = 24997000/242444
x = 107.179
That's the total increase.
What you want is 7.179%
If you take 7.179% of 242,555 and add it onto 242555 you should get 259,970
7.1797/100 * 242555 = 17413.02
17413.02 + 242555 = 259968.02
Why isn't it exact. I'm out just about 2 whole dollars. The reason is that the % isn't exact. It's only out to 3 decimal places which apparently is not enough. You can go over my numbers if you want more exactness us use something that is rounded to 7.180 if you like.
Answer:144
Explanation: 25% of the total in other words means that it is 1/4 (a quarter). You could just multiply 36 by 4 to get the total (100%). If u want to verify, just multiply 144 by 0.25 and u should get 36.
Answer:
for the first on its a=24 and for the second its a= 48
Step-by-step explanation:
for the first one you have to find the length which is 8 and then you find the width which is 3 then you multiply that to get 24
for the second one you find the base which is the strait line which is 8 then you find the height which is 6 then you multiply it to get 48 i hopes this helped!
Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
