Which option enables you to keep the last grammatical change

You are consulting for a trucking company that does a large amount ofbusiness shipping packages between New York and Boston. The

likoan [24]

Answer:

Answer explained with detail below

Explanation:

Consider the solution given by the greedy algorithm as a sequence of packages, here represented by indexes: 1, 2, 3, ... n. Each package i has a weight, w_i, and an assigned truck t_i. { t_i } is a non-decreasing sequence (as the k'th truck is sent out before anything is placed on the k+1'th truck). If t_n = m, that means our solution takes m trucks to send out the n packages.

If the greedy solution is non-optimal, then there exists another solution { t'_i }, with the same constraints, s.t. t'_n = m' < t_n = m.

Consider the optimal solution that matches the greedy solution as long as possible, so \for all i < k, t_i = t'_i, and t_k != t'_k.

t_k != t'_k => Either

1) t_k = 1 + t'_k

i.e. the greedy solution switched trucks before the optimal solution.

But the greedy solution only switches trucks when the current truck is full. Since t_i = t'_i i < k, the contents of the current truck after adding the k - 1'th package are identical for the greedy and the optimal solutions.

So, if the greedy solution switched trucks, that means that the truck couldn't fit the k'th package, so the optimal solution must switch trucks as well.

So this situation cannot arise.

2) t'_k = 1 + t_k

i.e. the optimal solution switches trucks before the greedy solution.

Construct the sequence { t"_i } s.t.

t"_i = t_i, i <= k

t"_k = t'_i, i > k

This is the same as the optimal solution, except package k has been moved from truck t'_k to truck (t'_k - 1). Truck t'_k cannot be overpacked, since it has one less packages than it did in the optimal solution, and truck (t'_k - 1)

cannot be overpacked, since it has no more packages than it did in the greedy solution.

So { t"_i } must be a valid solution. If k = n, then we may have decreased the number of trucks required, which is a contradiction of the optimality of { t'_i }. Otherwise, we did not increase the number of trucks, so we created an optimal solution that matches { t_i } longer than { t'_i } does, which is a contradiction of the definition of { t'_i }.

So the greedy solution must be optimal.

6 0

2 years ago

Which of the following attributes of a website indicates a more reliable source for information?

scoundrel [369]

An attribute of a website that will indicate a more reliable source of information is when the site ends in ".edu". It is a top-level domain for education. It would mean that this particular site is linked to universities, colleges or other educational sites thus it gives more information that is real and factual.

3 0

1 year ago

g Suppose the information content of a packet is the bit pattern 1110 0110 1001 1101 and an even parity scheme is being used. Wh

erma4kov [3.2K]

Answer: see description

Explanation:

first we accommodate the bit pattern in a matrix of 4x4 which is the minimum length checksum field, now with even parity two-dimensional scheme we need to complete this matrix by adding one row and one column by adding at the end of each row a 1 or a 0 to complete pairs of 1's:

we have

$\left[\begin{array}{cccc}1&1&1&0\\0&1&1&0\\1&0&0&1\\1&1&0&1\end{array}\right]$

so we complete with this, adding a row at the end which matches a pair number of 1's

$\left[\begin{array}{ccccc}1&1&1&0&1\\0&1&1&0&0\\1&0&0&1&0\\1&1&0&1&1\\1&1&0&0&0\end{array}\right]$

4 0

2 years ago

When a CPU executes instructions as it converts input into output, it does so with

yuradex [85]

The power of science of course

8 0

2 years ago

A blood bank maintains two tables - DONOR, with information about people who are willing to donate blood and ACCEPTOR, with info

Kipish [7]

Answer:

The sql query is given below.

Since we need to count of males and females for a particular blood group, we put xxx for the blood group.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%") as Male_Donors,

COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%") as Female_Donors

FROM DONOR

WHERE BG = xxx;

Explanation:

The clauses in the query are as follows.

1. SELECT: all the columns required in the output are put in this clause.

2. FROM JOIN ON: the table(s) from which the above columns are taken are put in this clause.

3. WHERE: any condition required to filter the output is put in this clause.

The query is explained below.

1. Find the number of male donors. Number of anything can be found using COUNT() function. A query is required since gender is included in deciding the type of donor.

2. The query is defined to find number of male donors as follows.

COUNT( SELECT DID FROM DONOR WHERE GENDER LIKE "M%"; )

3. In the previous query, LIKE operator is used since it is not defined what value is stored for male donors.

4. Similarly, the query to find the number of female donors is formed.

COUNT( SELECT DID FROM DONOR WHERE GENDER LIKE "F%"; )

5. Next, the final query is formed as follows.

SELECT: both COUNT() functions will come here.

FROM: table name

WHERE: specific blood group will be put here

GROUP BY: this clause is optional and is not used in this query.

HAVING: this clause is optional and is not used in this query.

6. The query after putting all clauses is shown below.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%"),

COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%")

FROM DONOR

WHERE BG = xxx;

7. Alias is used in the above query for each column to get the final query.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%") as Male_Donors, COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%") as Female_Donors

FROM DONOR

WHERE BG = xxx;

7 0

2 years ago