A quiz has 8 multiple choice questions and 4 options for each. How many ways are there to answer the questions

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:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ 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 $x_1, x_2, \cdots, x_{10}$ .

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 $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

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 $2^{10} = 1024$ 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 $2^{10} -2 = 1024 - 2 = 1022$ 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.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

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 $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

For f(x) = 3x - 1 and g(x) = x+2, find (f – g)(x).

kobusy [5.1K]

Well, I bet you want your answer right away! So here it is.

Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(gf)(x)=g(x)f(x)= \small{\dfrac{3x+2}{4-5x}}=4−5x3x+2

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}(gf)(x)=4−5x3x+2

Hope I helped! :) If I did not help that's okay.

-Duolingo

7 0

3 years ago

A car travelling at a constant speed travels 90km in 50 min .How far travelling at the same constant speed will the car travel i

vovikov84 [41]

Answer:

261 km

Step-by-step explanation:

given,s=constant

d1/T1=d2/T2

d1=90km,t1=50 min

d2=? ,t2=(120+25)min

=145min

therefore,d2=(90×145)/50

=261 km

6 0

3 years ago