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3 years ago

When is it better to use a table When is it better to use a scatter plot

1 answer:

34kurt3 years ago

6 0

Management information system requires tables, Tables are used only if a person is trying to analyse data and understand its business by summarizing.It is the 1st step of data analysis collecting data. Scatter plot is used when there is a question of association of two or three variable.

Step-by-step explanation:

Table is the matrix to stack up the data in there respective columns.
It is the most important setup to analysis.
It's better used collecting data for any projects or business.
Tables are even used for Bio statistics and Analytics.
Databases uses tables for one query table to another query table.
Scatter plot is dotted plot which analyse two variables.
Example if a project is eating chocolates in gram in x axis.
Gaining weight in y axis two variables are used to predict and outcome.
If a person eats this amount of chocolate it will gain this wait.
Association rules, contingency statistics and rest.

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John drove 450 miles in 8 hours. he averaged 50 miles per hour on the highway and 60 miles per hour on the turnpike. how many ho

Alexeev081 [22]

Total distance John drove=450 miles
total time=8 hours
let the number of hours he drove on highway be x, number of hours he drove on turnpike will be (8-x) hours
since
distance=speed*time
distance traveled in highway=50x
distance traveled in turnpike=60(8-x)
thus the total distance will be:
50x+60(8-x)=450
simplifying the above
50x+480-60x=450
-10x=450-480
-10x=-30
thus
x=3 hours
Thus he drove on highway for 3 hours.

3 0

3 years ago

At the start of the football game there were 640 fans in the stadium. It begun to rain during halftime, so 10% of the fans went

choli [55]

10% of 640 is 64. So 640-64= 576 fans. Then after the five touchdowns 26% of 576 is 144 . So 576-144= 432 fans left

4 0

3 years ago

Can someone help me with these five problems please i will mark as brainliest if the answers are correct

kondaur [170]

Answer:

1. C

2. B

3. C

4. A

5. D

Step-by-step explanation:

4 0

3 years ago

Arrange the following polynomial into descending order for x, then interpret the degree of the 3rd term. 2xy4 + 2x2y − 3y2 + 10x

Darina [25.2K]

The arrangement should be based on the decreasing degree of the terms which may be determined by the variable x as it should come first in alphabetical order. The polynomial is then,
10x³ + 2x²y + 2xy^4 - 3y²
The degree is equal to the sum of the exponents. Thus, the answer is 5 and it is letter D.

8 0

3 years ago

Read 2 more answers

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e

Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let $\%R_{A}$ be the quantity of nonzero elements in $R_{A}.$

Let $C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right)$ .

Parity demands that $\%R_{A}$ and $\%R_{B}$ must equal 2 or 4.

Case 1: $\%R_{A}$ =4 and $\%R_B$ =4. There are $\left(\begin{array}{l}3\\ 2\end{array}\right)$ =3 ways to put 2-1's in $R_A$ , so there are 3 ways.

Case 2: $\%R_{A}$ =2 and $\%R_B$ =4. There are 3 ways to position the -1 in $R_A$ , 2 ways to put the remaining -1 in $R_B$ (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of $\%R_{A}=4,\%R_{B}=2$ for a complete of 24 ways.

Case 3: $\%R_A=\%R_B=2$ . There are 3 ways to put the -1 in $R_{A}$ . Now, there are two cases on what happens next.

The 1 in $R_B$ goes directly under the -1 in $R_A$ . There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in $R_C$ and $R_D$ . (Either the 1 comes first in $R_C$ or the 1 comes first in $R_D$ .)
The 1 in $R_B$ doesn't go directly under the -1 in $R_A$ . There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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3 0

2 years ago

Read 2 more answers