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3 years ago

How are graphs used to represent patterns??

2 answers:

5 0

They organize data and can reveal relationships, making it easier to write equations that describe different patterns.

mr Goodwill [35]3 years ago

5 0

A picture is worth a thousand words. It's also worth a thousand data points!

A graph can be defined as :

a diagram showing the relation between variable quantities, typically of two variables, each measured along one of a pair of axes at right angles.

The different uses of graphs are :

Line graphs can be used to compare changes over the same period of time for more than one group. Pie charts are best to use when you are trying to compare parts of a whole. They do not show changes over time. Bar graphs are used to compare things between different groups or to track changes over time.

Other uses of graphs are :

It can show patterns. Like a declining sales line.

It can easily show the breakdown of costs as in a Pie Chart.

It can determine profit and loss in a breakeven analysis.

It can show a comparison of two different processes.

You might be interested in

Which statement comparing these measures is true? A microgram is 1,000 times greater than a milligram.

ololo11 [35]

The statement is false.

The statement below is true.
A milligram is 1000 times greater than a microgram.

3 0

3 years ago

Write 3,200 in scientific notation

spin [16.1K]

Answer:

The answer is 3.2 E 3 or 3.2 x 10^3

You move back 3 spaces which is 3.2, since you moved back 3 spaces, you put 3 as the power/exponet.

To be more satisfied, 3.2 is your base number, it is the big number Below the power/exponent. The power/exponent is 3, you moved 3 spaces and you moved to the left, so it is represented as a positive power/exponent.

Hope this helps.

8 0

3 years ago

0.09X+4=4.9 I need to solve the unknown

saul85 [17]

Answer:

the answer should be x=10

5 0

3 years ago

A football team lost 4 yards on each of 2 plays, gained 14 yards on the third play, and lost 5 yards on the forth play. Write an

yanalaym [24]

2 x -4 + 14 - 5
-8 +14 - 5
6 - 5
1

3 0

3 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

3 years ago