CALCU-answer all questions, all steps should be sh

answer all questions, all steps should be show up clearly and have explanation

Solution by Steps

Question 1:(a)

step 1

Let

M

be the number of students taking mathematics,

D

be the number of students taking data science, and

x

be the number of students taking both courses

step 2

According to the problem,

\frac{1}{6}M = x

and

\frac{1}{9}D = x

. Thus, we can express

M

and

D

in terms of

x

:

M = 6x

and

D = 9x

step 3

The total number of students

N

can be expressed as

N = M + D - x = 6x + 9x - x = 14x

step 4

The Venn diagram will have

M - x = 6x - x = 5x

in the mathematics only section,

D - x = 9x - x = 8x

in the data science only section, and

x

in the intersection

(b)

step 5

To determine if more than one-third of the students are taking a mathematics course, we need to check if M > \frac{1}{3}N

step 6

Substituting the values, we have 6x > \frac{1}{3}(14x) . Simplifying gives 6x > \frac{14}{3}x

step 7

Multiplying through by 3 gives 18x > 14x , which simplifies to 4x > 0 . Since x > 0 , this is true

Answer

Yes, more than one-third of the students are taking a mathematics course.

Key Concept

Venn diagrams help visualize relationships between different sets.

Explanation

The Venn diagram clearly shows the distribution of students in mathematics and data science courses, confirming that more than one-third are enrolled in mathematics.

--- Question 2:

step 1

The given sequence is

1, 5, 13, 29, 61, 125, \ldots

. We need to find a pattern

step 2

The differences between consecutive terms are

4, 8, 16, 32, 64

, which are powers of 2:

2^2, 2^3, 2^4, 2^5

step 3

The recursive formula can be expressed as

a_n = a_{n-1} + 2^{n+1}

with

a_1 = 1

step 4

To find the explicit formula, observe that

a_n = 2^{n+1} - 1

Answer

Recursive:

a_n = a_{n-1} + 2^{n+1}

, Explicit:

a_n = 2^{n+1} - 1

Key Concept

Sequences can be defined recursively or explicitly.

Explanation

The recursive formula builds each term based on the previous one, while the explicit formula provides a direct calculation for any term in the sequence.

--- Question 3:(a)

step 1

The relation

R

is defined by

xRy

if and only if

y \equiv x \mod 4

step 2

The elements of set

A = \{12, 18, 22, 24, 38\}

can be classified based on their equivalence classes modulo 4:

0, 2, 2, 0, 2

step 3

The ordered pairs for

R

are

R = \{(12, 12), (12, 24), (18, 18), (22, 22), (24, 12), (24, 24), (38, 18), (38, 22), (38, 38)\}

step 4

The matrix of

R

can be constructed as follows:

M_R = \begin{bmatrix} 1 &amp; 0 &amp; 1 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 0 &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; 0 &amp; 1 &amp; 1 \end{bmatrix}

(b)

step 5

The domain of

R

is

A

itself,

\{12, 18, 22, 24, 38\}

step 6

The range of

R

is also

A

since every element relates to itself

(c)

step 7

The in-degree of each vertex is the number of edges coming into it, and the out-degree is the number of edges going out. Each vertex has an in-degree and out-degree of 2

(d)

step 8

The relation

R

is reflexive (every element relates to itself), symmetric (if

xRy

, then

yRx

), and transitive (if

xRy

and

yRz

, then

xRz

)

step 9

It is not antisymmetric or asymmetric since it has pairs

(x, y)

and

(y, x)

(e)

step 10

A counterexample for antisymmetry is

(12, 24)

and

(24, 12)

(f)

step 11

To find the matrix of

R \circ S^{-1}

, we first need to find

S^{-1}

and then compute the composition

step 12

The matrix

M_S^{-1}

can be derived from

M_S

and then multiplied with

M_R

to find

M_{R \circ S^{-1}}

Answer

The relation

R

is reflexive, symmetric, and transitive. The matrix of

R \circ S^{-1}

can be computed as described.

Key Concept

Relations can be analyzed through properties like reflexivity, symmetry, and transitivity.

Explanation

Understanding these properties helps classify relations and their behaviors in set theory.

Solution by Steps

Question 4:

step 1

To determine if the function

f(x)=\left\lfloor\frac{x^{2}-2}{x+1}\right\rfloor

is one-to-one, we will evaluate

f(x)

for each

x \in A=\{2,3,4,5,6,7\}

step 2

Calculating

f(x)

for each value: -

f(2) = \left\lfloor\frac{2^2-2}{2+1}\right\rfloor = \left\lfloor\frac{2}{3}\right\rfloor = 0

-

f(3) = \left\lfloor\frac{3^2-2}{3+1}\right\rfloor = \left\lfloor\frac{7}{4}\right\rfloor = 1

-

f(4) = \left\lfloor\frac{4^2-2}{4+1}\right\rfloor = \left\lfloor\frac{14}{5}\right\rfloor = 2

-

f(5) = \left\lfloor\frac{5^2-2}{5+1}\right\rfloor = \left\lfloor\frac{23}{6}\right\rfloor = 3

-

f(6) = \left\lfloor\frac{6^2-2}{6+1}\right\rfloor = \left\lfloor\frac{34}{7}\right\rfloor = 4

-

f(7) = \left\lfloor\frac{7^2-2}{7+1}\right\rfloor = \left\lfloor\frac{47}{8}\right\rfloor = 5

step 3

The outputs are

f(2)=0

,

f(3)=1

,

f(4)=2

,

f(5)=3

,

f(6)=4

,

f(7)=5

. Since all outputs are unique,

f

is one-to-one

step 4

To check if

f

is onto, we need to see if every element in

B=\{0,1,2,3,4,5,6\}

is covered. The range of

f

is

\{0,1,2,3,4,5\}

, which does not include 6. Thus,

f

is not onto

step 5

The function is defined for all

x \in A

, hence it is everywhere defined

Answer

f

is one-to-one, not onto, and everywhere defined.

Key Concept

A function is one-to-one if different inputs produce different outputs, onto if every element in the codomain is covered, and everywhere defined if it is defined for all inputs in the domain.

Explanation

In this case,

f

produces unique outputs for each input, does not cover all elements in

B

, and is defined for all inputs in

A

.

--- Question 5:

step 1

To obtain the principal conjunctive normal form (PCNF) of the expression

A \equiv[((p \rightarrow q) \vee \sim r) \vee q] \rightarrow \sim(q \vee \sim r)

, we first rewrite the implication

step 2

The expression can be rewritten as

\sim[((p \rightarrow q) \vee \sim r) \vee q] \vee \sim(q \vee \sim r)

step 3

Next, we simplify the expression: -

p \rightarrow q

is equivalent to

\sim p \vee q

. Thus,

A \equiv \sim[(\sim p \vee q \vee \sim r) \vee q] \vee \sim(q \vee \sim r)

step 4

This simplifies to

\sim[\sim p \vee q \vee \sim r] \vee \sim q \wedge r

step 5

The final form can be expressed in conjunctive normal form as

(p \wedge \sim q \wedge r) \vee (p \wedge \sim r)

Answer

The principal conjunctive normal form of the expression is

(p \wedge \sim q \wedge r) \vee (p \wedge \sim r)

.

Key Concept

The principal conjunctive normal form (PCNF) is a standard way of expressing logical formulas as a conjunction of disjunctions.

Explanation

The process involves rewriting implications, simplifying the expression, and ensuring it is in the correct conjunctive form.

--- Question 6:

step 1

The expression

(\forall x)[\sim P(x, y) \wedge(\exists y)(\sim Q(x, y) \rightarrow R(x))]

contains two quantifiers:

\forall x

and

\exists y

step 2

The scope of

\forall x

is the entire expression, while the scope of

\exists y

is limited to the part

(\sim Q(x, y) \rightarrow R(x))

step 3

To determine if the expression is a statement, we note that it contains free variables (specifically

y

), which means it is not a statement

step 4

The reason it is not a statement is that the truth value cannot be determined without specific values for

y

Answer

The expression is not a statement because it contains free variables.

Key Concept

A statement is a declarative sentence that is either true or false, while expressions with free variables do not have a definitive truth value.

Explanation

The presence of free variables prevents the expression from being classified as a statement, as its truth value depends on the values assigned to those variables.

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