What is 2/3 + 1/2 In fraction form please .
2 answers:
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This load will weigh 1.5 TONS ALTOGETHER.
So, the truck is delivering a
ton of cement blocks, and a 
of bricks. To find how much this load weighs, we'd have to add the two fractions together, which proves to be quite easy, because the denominators are already the same, so we wouldn't have to change anything.
+ = 
And now, we would want to convert that fraction into a whole number, or at least a decimal, to find how many tons the truck is carrying. So we would divide. Once the work is done, here is what would happen:
12 ÷ 8 = 1.5
And there's your answer. The truck is carrying 1.5 TONS.
So, the truck is delivering a
And now, we would want to convert that fraction into a whole number, or at least a decimal, to find how many tons the truck is carrying. So we would divide. Once the work is done, here is what would happen:
12 ÷ 8 = 1.5
And there's your answer. The truck is carrying 1.5 TONS.
Answer:
(a) Number of sets B given that
- A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
- A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)
(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.
Step-by-step explanation:
<h3>(a)</h3>Let denote the 20 elements of set X.
Let denote elements of set X that are also part of set A.
For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain .
For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from .
.
For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus possibilities for set B.
In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves possibilities.
Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.
Start by considering the case when set A and set B are indeed disjoint.
.
Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus possibilities for a set B that is disjoint with set A.
There are 20 elements in X so that's possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only
possibilities left.