MATHEMATICS COLLEGE

Which table represents a function?

Answers

Answer 1

Answer:

Answer:

The bottom left table

Step-by-step explanation:

the same x value cannot have different y values

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The filling variance for boxes of cereal is designed to be .02 or less. A sample of 41 boxes of cereal shows a sample standard deviation of .16 ounces. Use α = .05 to determine whether the variance in the cereal box fillings is exceeding the design specification.

Answers

Answer:

$\chi^2 =(41-1)/(0.02) 0.0256 =51.2$

$p_v =P(\chi^2_(40) >51.2)=0.1104$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(51.2,40,TRUE)"

If we compare the p value and the significance level provided 0.05 we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 0.02, so there is no a violation of the specifications.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=41$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.16^2 =0.0256$ represent the sample variance obtained

$\sigma^2_0 =0.02$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is higher than the standard of 0.02, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 0.02$

Alternative hypothesis: $\sigma^2 >0.02$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =(n-1)/(\sigma^2_0) s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =(41-1)/(0.02) 0.0256 =51.2$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case $df = 41-1=40$ . And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2_(40) >51.2)=0.1104$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(51.2,40,TRUE)"

Conclusion

Right triangle ABC is shown. Which of these is equal to cos(A)?A) cos(B)
B) cos(C)
C) sin(B)
D) sin(C)

Answers

Answer: C

cosA=AC/AB

sinB=AC/AB

hence cosA=sinB

Answer:

C).

Step-by-step explanation:

Since angles A and B are complementary, their cofunctions are equal. So, cos(A) = sin(B).

1.25 = (d/32) + (d/48)

Answers

Answer:

d =24

Step-by-step explanation:

1.25=3d/96+2d/96

1.25=3d+2d/96

1.25=5d/96

96•1.25=96•5d/96

96•1.25=5d

120=5d

120/5=5d/5

d =24

Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 19.2pounds with a standard deviation of 4.4 pounds. You want to construct a 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds

(b) Construct the 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
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(c) Are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 18 pounds and why?

No, because 18 is above the lower limit of the confidence interval.

Yes, because 18 is below the lower limit of the confidence interval.

No, because 18 is below the lower limit of the confidence interval.

Yes, because 18 is above the lower limit of the confidence interval.

(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?

Because the sample size is greater than 10.

Because we do not know the distribution of the parent population.

Because the parent population is assumed to be normally distributed.

Because the sample size is less than 100.

Answers

Answer:

a) $\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ b) The 90% confidence interval is given by (17.3;21.1)

c) No, because 18 is above the lower limit of the confidence interval.

d) Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Step-by-step explanation:

1) Notation and definitions

n=17 represent the sample size

Part a

$\bar X= 19.2$ represent the sample mean. And that represent the best estimator for the population mean since $\hat \mu =\bar X=19.2$ Part b

$s=4.4$ represent the sample standard deviation

m represent the margin of error

Confidence =90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The degrees of freedom are given by:

$df=n-1=17-1=16$

We can find the critical values in excel using the following formulas:

"=T.INV(0.05,16)" for $t_(\alpha/2)=-1.75$

"=T.INV(1-0.05,16)" for $t_(1-\alpha/2)=1.75$

The critical value $tc=\pm 1.75$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c (s)/(√(n))$

$m=1.75 (4.4)/(√(17))=1.868$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_(c) (s)/(√(n))$

And calculating the limits we got:

$19.2 - 1.75 (4.4)/(√(17))=17.332$

$19.2 + 1.75 (4.4)/(√(17))=21.068$

The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)

Part c

No, because 18 is above the lower limit of the confidence interval.

Part d

Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

A typical combine harvester sells for $500,000. If the value of the combine depreciates 5.52% each year, how many years will it take to lose half of it's value? Round your answer to the nearest whole number of years.

Answers

To model the given problem, we use the following exponential function:

$V(t)=500,000(1-0.0552)^t.$

Now, we set the above equation to

$V(t)=(500,000)/(2)=250,000=500,000(1-0.0552)^t.$

Solving for t, we get:

$\begin{gathered} (250,000)/(500,000)=0.9448^(t,) \n tln(0.9448)=ln((1)/(2)), \n t=(ln((1)/(2)))/(ln(0.9448)). \end{gathered}$

Finally, we get:

$t\approx12\text{ years.}$

Answer:

$12\text{ years.}$

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best reate with 80 % of its flights arriving on time. A test is conducted by randomly selecting 10 Southwest flights and observing whether they arrive on time. (a) Find the probability that at least 3 flights arrive late.

Answers

Answer:

There is a 32.22% probability that at least 3 flights arrive late.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And $\pi$ is the probability of X happening.

In this problem, we have that:

There are 10 flights, so $n = 10$ .

A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that $\pi = 0.2$ .

(a) Find the probability that at least 3 flights arrive late.

Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

$P(X = 0) = C_(10,0).(0.2)^(0).(0.8)^(10) = 0.1074$

$P(X = 1) = C_(10,1).(0.2)^(1).(0.8)^(9) = 0.2684$

$P(X = 2) = C_(10,2).(0.2)^(2).(0.8)^(8) = 0.3020$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6778 = 0.3222$

There is a 32.22% probability that at least 3 flights arrive late.

Final answer:

The problem is solved by calculating the probability of the complementary event (0,1,2 flights arriving late) using the binomial distribution, then subtracting this from 1 to find the probability of at least 3 flights arriving late.

Explanation:

This problem is typically solved by using a binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'.
Here, our 'success' is a flight arriving late. The probability of success, denoted as p, is thus 20% or 0.2 (since 80% arrive on time, then 100%-80% = 20% arrive late). The number of trials, denoted as n, is 10 (the number of randomly selected flights).
We want to find the probability that at least 3 flights arrive late, in other words, 3,4,...,10 flights arrive late. The problem can be solved easier by considering the complementary event: 0,1,2 flights arrive late. Then subtract the sum of these probabilities from 1.

The binomial probability of exactly k successes in n trials is given by:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where C(n, k) is the binomial coefficient, meaning choosing k successes from n trials.
We calculate like so:
P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10
P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9
P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8
Sum these up and subtract from 1 to get the probability that at least 3 flights arrive late. This gives the solution to the question.

Learn more about binomial probability here:

brainly.com/question/34083389

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Simplify x