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3 years ago

What is the purpose of filters in a data base

1 answer:

8 0

Filters let you filter out certain things or search for them. For example if i'm looking through a database with all the students grades. I could filter out everything that doesnt math a students name. This would only show results for what I searched for.

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Credible sites contain __________ information,

Reptile [31]

Answer:

All of the above. Explanation:Credible sites include the date of any information, cite the source of the information presented, are well designed and professional.

6 0

3 years ago

Read 2 more answers

You are responsible for performing all routine maintenance on your company's laser printers. Which of the following maintenance

Luden [163]

Answer:

B. installing the maintenance kit

Explanation:

page count keeps track of how many number of pages that have been printed since the installation of the last maintenance kit. When a new maintenance kit is installed on most Laser printers, the page count is reset because it does not automatically clears most times, hence, there's the need to clear the count manually.

4 0

3 years ago

Weak Induction

slega [8]

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

$\to 3^7$

when n=k for the true condition:

$\to 3^k$

when n=k+1 it tests the value:

$\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to$

since k>6 hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

$\log(1!)$

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

$\log(k!)$

when n=k+1

$\log(k+1)!=\log(k!)+\log(k+1)\\$

$\because k \log k$ $[\therefore KH>K \Rightarrow \log(KH>\loK)]$

In option 3:

when n=1:

$A_1 \cup B=A_1 \cup B\\\\$

when n=k

$\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\$

hence n=k+1 is true.

7 0

3 years ago

What is a perfect hashing function?

antiseptic1488 [7]

Answer:

First we understand what is hash function.A hash function is mostly used in Hashmaps. It maps different keys to a set of values.There may occur a case when we have same key but different values.This case is called collision.So we have to use different collision handling techniques that are open addressing and separate chaining.

A perfect hash function maps key-value pair such that there are no collisions.

3 0

3 years ago

Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th

Evgen [1.6K]

Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)

=> P(f(A), A, g(A)), P(f(A), A, g(A))

Thus , mgu = {x/y, z/y , y/A }.

Explanation:

Unification: Any substitution that makes two expressions equal is called a unifier.

a) P(B,A,B), P(x,y,z)

Use { x/B}

=> P(B,A,B) , P(B,y,z)

Now use {y/A}

=> P(B,A,B) , P(B,A,z)

Now, use {z/B}

=> P(B,A,B) , P(B,A,B}

Hence, most general unifier = {x/B , y/A , z/B }

b. P(x,x), Q(A,A)

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q

c. Older(Father(y),y), Older(Father(x),John)

Use {y/x}

=> Older(Father(x),x), Older(Father(x),John)

Now use { x/John }

=> Older(Father(John), John), Older(Father(John), John)

Thus , mgu ={ y/x , x/John }

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

Use { y/x }

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))

This is not unifiable as x cannot be bound for both A and B

e) P(f(x), x, g(x)), P(f(y), A, z)

Use {x/y}

=> P(f(y), y, g(y)), P(f(y), A, z)

Now use {z/g(y)}

P(f(y), y, g(y)), P(f(y), A, g(y))

Now use {y/A}

=> P(f(A), A, g(A)), P(f(A), A, g(A))

Thus , mgu = {x/y, z/y , y/A }.

7 0

3 years ago