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Vikentia [17]
2 years ago
7

State which of the following sets of ordered pairs represent a function.

Mathematics
1 answer:
zalisa [80]2 years ago
7 0

Answer:

Option (D) is correct.

Step-by-step explanation:

A function can be termed as a process from a set of x values to a set of possible y-values where each x-value relates to exactly one y-value.

In other words, a relation can only be a function there is no repetition of x-values. It means a relation can not have duplicated inputs.

Given the sets

Set A: (5,2), (4, 3), (3, 4), (2,5)

Set B: (-1,-6), (0, 2), (1, 2), (3, 6)

Set C: (2, 1), (4, 2), (2, 3), (8,4)

It is clear that sets A and B represent the function because, in these sets, each x-value relates to exactly one y-value. In other words, there are no duplicated inputs.

But, set C has duplicated inputs i.e. x = 2 is repeated twice, and a function can not have a repeated input.

Therefore, we conclude that set A and set B represent a function.

Thus, option (D) is correct.

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During the 2000 season, the home team won 138 out of 240 regular season National Football League games. (15 points) a) Construct
jeka94

Answer:

Step-by-step explanation:

A) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 240

x = 138

p = 138/240 = 0.58

q = 1 - 0.58 = 0.42

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, the z score for a confidence level of 95% is 1.96

Therefore, the 95% confidence interval is

0.58 ± 1.96√(0.58)(0.42)/240

Confidence interval is

0.58 ± 0.062

B) winning more than halve of the games would be winning 120 games and above.

p = 120/240 = 0.5

We would set up the hypothesis test.

For the null hypothesis,

P ≥ 0.5

For the alternative hypothesis,

P < 0.5

Considering the population proportion, probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 138

n = number of samples = 240

P = 138/240 = 0.58

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.58 - 0.5)/√(0.5 × 0.5)/240 = 2.48

Recall that this is a left tailed test. We would determine the probability value of the area to the right of the z score from the normal distribution table.

P value = 1 - 0.9934 = 0.0066

Since alpha, 0.01 > the p value, 0.0066, then we would reject the null hypothesis.

Therefore, At the 0.01 significance level, there is no strong evidence of a home field advantage (they win more than half of the games) in professional football.

3 0
2 years ago
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