A spinner is divided into four equal sections labeled 1, 2, 3, and 4. Another spinner is divided into three equal sections label
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For this case, the first thing we must do is define a variable.
We have then:
x: number of weeks
y: amount of cereal that Melanie eats
We now write the function that models the problem:
Then, for 5 weeks we have:
Answer:
A model that best represents the number of pounds of cereal she eats is:
For 5 weeks we have:
We have then:
x: number of weeks
y: amount of cereal that Melanie eats
We now write the function that models the problem:
Then, for 5 weeks we have:
Answer:
A model that best represents the number of pounds of cereal she eats is:
For 5 weeks we have:
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 .
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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.