Which of the points does NOT satisfy the inequality shaded in the diagram?

A) (4, 0)

B) (0, 0)

C) (-3, -2)

D) (-10, -1)

Answer:

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

Step-by-step explanation:

A paired t-test is used to compare two population means when we have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations in order to see an improvement or not for a method we can use it ( Without treatment and With specific treatment).  

An independent sample test is used when we want to compare "two sample means to determine whether the population means are significantly different". For example if we want to compare the scores for male and female in a test.  

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, while an independent sample is used when the two samples are not related and can be considered as separate groups.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, meaning that there is a one-to-one correspondence between the data points in the two samples. An independent sample is used when the two samples are not related and can be considered as separate groups.

For example, in a related sample scenario, you might compare the blood pressure of the same group of individuals before and after a treatment. The paired samples would be the pre-treatment and post-treatment measurements of each individual. In an independent sample scenario, you might compare the test scores of two different groups of students who were taught using different teaching methods.

Answer:the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

Step-by-step explanation:

The circumference of the first circle is 8xy cm

The circumference of the second circle is 5xy cm

the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

This variable expression can also be simplified further because both terms contain xy.

Answer:

The answer is False

Step-by-step explanation:

The  count does no suggest that the normal model is a good fit for sampling the distribution because the questions used for the test and survey is the one of many other question about cheating which implies that if other questions about college life are being used as the survey, the response would probably be that more of the student would not have admitted to cheating. This concept therefore disobeys the normal distribution model which is a bell shaped model and therefore assumes that at an average, the number of students that admitted to cheating in the major courses should be equal.

Answer:

all of the above

Step-by-step explanation:

The number 25 is a natural number as it belongs to the set [1,2,3,4,5,......]

The number 25 is a whole number as it belongs to the set [0,1,2,3,4,5,......]

The number 25 is an Integer as it belongs to the set [...,-5,-4,-3,-2,-1,0,1,2,3,4,5,...]

The number 25 is a rational number as it can be expressed as [tex]\[\frac{25}{1}\][/tex]

For the same reason , number 25 is a real number as it belongs to the set of rational numbers.

So the correct option is "all of the above".

25 does not belong to any of the given number sets (2 and 4, 1 and 2, or 3 and 5).

The number 25 does not belong to any of the provided number sets i.e. 2 and 4, 1 and 2 or 3 and 5. A number set typically refers to a collection of numbers, and in this case, 25 is absent in all the provided sets. The given number sets only contain the numbers 1, 2, 3, 4 and 5. Thus, 25 does not belong to any of these sets.

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Answer:

The relation is antisymmetric and transitive

Step-by-step explanation:

Let a,b,c be elements of the set of all people.

1) Let a be a person who is 20 years old. aRa means that this person is younger than themselves, which it's false because 20<20 is false. Then R is not reflexive.

2) Let a be a person who is 20 years old and b a person who is 30 years old. Then a is younger than b, that is, aRb.

However, it is not true that b is younger than a, as 30<20 is false, therefore bRa is false and R is not symmetric.

3) Suppose that aRb, so that a is younger than b. Then, b is not younger than a. If n denotes the age of a and m denotes the age of b, we have that n<m which implies that m<n is false. Then bRa is false, thus R is antisymmetric.

4) Suppose that aRb and bRc. Let n,m,p denote the ages of a,b,c respectively. Then n<m and m<p (a is younger than b and b is younger than c), and by transitivity of the ordering of numbers, n<p, that is, a is younger than c. Thus aRc, and R is transitive.

Answer: d. 0.60

Step-by-step explanation:

When are performing sample surveys , when the selected participant is giving any response is denoted as non - response.

The proportion of these participants of the sample is known as the non-response rate.

Given : A political scientist wants to know how college students feel about the social security system.

She obtains a list of the 3114 undergraduates at her college and mails a questionnaire to 250 students selected at random.

i.e. Sample size : n= 290

Only 100 of the questionnaires are returned.

Individual gave response =100

Individual gave no-response =250-100 =150

The  rate of non-response [tex]=\dfrac{\text{Individual gave no-response}}{n}[/tex]

[tex]=\dfrac{150}{250} =0.60[/tex]

Hence, the rate of non-response would be 0.60 .

Thus , the correct option is d. 0.60.

Answer:

[tex]work \ done= 45937.5[/tex]

Step-by-step explanation:

Work done is given by

[tex]work \ done=\int_a^b w(d-x) \ dx[/tex] , where d = length of cable and w = weight of cable.

Here, d = 175 ft and w = 3 lb/ft

Now, [tex]work \ done=\int_0^{175} 3(175-x) \ dx[/tex]

[tex]work \ done= 3\left [175x-\frac{x^2}{2}  \right ]_0^{175}[/tex]

[tex]work \ done= 3\left [175^2-\frac{175^2}{2}  \right ][/tex]

[tex]work \ done= 3\cdot \frac{175^2}{2}[/tex]

[tex]work \ done= 45937.5[/tex]

Answer:

a) [tex]S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}[/tex]

b)  [tex]S_{a}(20)=12.8Kg[/tex]

Step-by-step explanation:

It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower

Water velocity [tex]V_{w} = 60\frac{L}{min}[/tex]

Solution velocity [tex]V_{s} = 60\frac{L}{min}[/tex]

Brine concentration = [tex]\frac{6,000L}{16Kg}=375\frac{L}{Kg}[/tex]

a) Amount of salt as a funtion of time Sa(t)

[tex]S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}[/tex]

b) [tex]S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg[/tex]

This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min and then calculate the amount of salt in the tank after a specific time.

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min:

Rate of salt entering = (60 L/min) * (16 kg/6000 L) = 0.16 kg/min

Therefore, the amount of salt in the tank after t minutes is given by:

y = 0.16 kg/min * t min = 0.16t kg

Answer:

There is sufficient evidence to conclude that child booster seats meet the specific requirement.

Step-by-step explanation:

Sample: 697, 759, 1266, 621, 569, 432

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{4344}{6} = 724[/tex]

Sum of squares of differences = 415616

[tex]S.D = \sqrt{\frac{415616}{5}} = 288.31[/tex]

We are given the following in the question:  

Population mean, μ = 1000 hic

Sample mean, [tex]\bar{x}[/tex] = 724

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s = 288.31

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1000\text{ hic}\\H_A: \mu < 1000\text{ hic}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{724 - 1000}{\frac{288.31}{\sqrt{6}} } = -2.344[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 5 degree of freedom } = -2.015[/tex]

Calculation the p-value from table,

P-value = 0.033

Since,                  

Since, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that the measurement is less than 1000 hic.

Thus, there is sufficient evidence to conclude that child booster seats meet the specific requirement.

Answer:

Option D) Yes, because the test statistic is -2.01

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 30 pound

Sample mean, [tex]\bar{x}[/tex] = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s =  2 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{29.1 - 30}{\frac{2}{\sqrt{20}} } = -2.012[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729[/tex]

Since,                    

[tex]t_{stat} < t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01

Answer:

The firm's sample size must be of at least 171 adults.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.

In this problem, we have that:

[tex]M = 60, \sigma = 400[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]60 = 1.96*\frac{400}{\sqrt{n}}[/tex]

[tex]60\sqrt{n} = 784[/tex]

[tex]\sqrt{n} = 13.07[/tex]

[tex]n = 170.7[/tex]

The firm's sample size must be of at least 171 adults.

Answer:

c. Interval or ratio

Step-by-step explanation:

There exists the following measurement scales:

Nominal: A variable is linked to a number. For example, Buffalo Bills players, 27 is Tre'Davious White, 49 Tremanine Edmunds, and then on...

Ordinal: Ranks the intensity of something. For example, grading some pain on a 1 to 10 scale.

Interval or ratio: Represents quantity and has an equality of units. One example is the rates of unemployment.

Descriptive: Tries to attribute qualities to quantitative data. For example, rates of unemployment being classified as very low, low, medium, and then on...

So the correct answer is:

c. Interval or ratio

The unemployment rate is measured using an interval or ratio scale, represented as a percentage. Various methods and measures are considered for accurate calculation, though it has limitations in fully capturing unemployment's societal impact.

The reported unemployment rate of 5.5% of the population uses an interval or ratio scale for measurement. This type of scale is used because the unemployment rate is a percentage that represents a proportion of the population. The measurement of unemployment involves a ratio of two quantities: the number of unemployed individuals and the total labor force. Different methods such as Labor Force Sample Surveys, Official Estimates, Social Insurance Statistics, and Employment Office Statistics are used to calculate this figure. Moreover, the U.S. Bureau of Labor Statistics employs six different measures (U1 - U6) to capture various aspects of unemployment. While the rate is informative, there are shortcomings in how it represents the real impact on society, as it does not account for underemployment or those who have stopped looking for work. Understanding these statistics is crucial as unemployment has significant economic and social consequences, such as increasing inequality and potentially leading to civil unrest.

Answer:

a. No, since the value of the test statistic is less than the critical value

Step-by-step explanation:

1) Data given and notation

n=144 represent the random sample taken

X=72 represent the number of people that prefer the blend

[tex]\hat p=\frac{72}{144}=0.5[/tex] estimated proportion of people that prefer the blend

[tex]p_o=0.47[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.959

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion if higher than 0.47:  

Null hypothesis:[tex]p\leq 0.47[/tex]  

Alternative hypothesis:[tex]p > 0.47[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.5 -0.47}{\sqrt{\frac{0.47(1-0.47)}{144}}}=0.721[/tex]  

4) Statistical decision  

We can calculate the critical value since we have a right tailed test, we need to look into the normal standard distribution a value that accumulates 0.01 of the area on the right and 0.99 on the left. And this value is:

[tex]z_{\alpha/2}=2.33[/tex]

And we can use the following excel code to find the critical value: "=NORM.INV(0.99,0,1)"

Our calculated value on this case is less than the critical value so the best conclusion is:

a. No, since the value of the test statistic is less than the critical value

The requirements necessary for a normal probability distribution to be a standard normal probability​ distribution is that The mean and standard deviation have the values of mu equals 1and sigma equals 1. Hence the correct answer is A.

A probability distribution is one whose mean data points are symmetric. That is, most values from such a distribution cluster around the mean.

Another name for Normal Probability Distribution is Gaussian Distribution.

Learn more about Normal Probability Distribution at:

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Final answer:

A standard normal probability distribution requires a mean of 0 and a standard deviation of 1. The correct answer is B, which reflects these necessary conditions for standardization, allowing for the comparison of z-scores across different distributions.

Explanation:

To transform a normal probability distribution into a standard normal probability distribution, certain requirements must be met. Specifically, the distribution must have a mean (mu) of 0 and a standard deviation (sigma) of 1. Among the given options, the correct answer is B, where the mean and standard deviation have the values of mu equals 0 and sigma equals 1.

This standardization process allows any normal distribution to be compared on a common scale, and it is fundamental for calculating z-scores, which indicate how many standard deviations an element is from the mean.For example, if we have a normally distributed variable 'x' from a distribution with any mean µ and standard deviation o, the standardized value or z-score is calculated as follows:z = (x - µ) / o

Answer:The margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Step-by-step explanation:

Since we have given that

p = 0.70

n = 1200

We need to find the margin of error;

Margin of error would be

[tex]\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{0.7\times 0.3}{1200}}\\\\=0.01323[/tex]

At 95% confidence level, α = 1.96

so, 95% confidence interval would be

[tex]p\pm z\times 0.01323\\\\=0.7\pm 1.96\times 0.01323\\\\=0.7\pm 0.02593\\\\=(0.7-0.02593,0.7+0.02593)\\\\=(0.674,0.73)[/tex]

Hence, the margin of error is 0.01323 and 95% confidence interval is (0.674,0.73).

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between [tex]\text{Mean}\pm\text{Standard deviation}[/tex]

i.e. about 68% of the heights of students lies between [tex]55\pm5[/tex]

Here, [tex]55\pm5=(55-5, 55+5)=(50,60)[/tex]

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

Final answer

The interval containing the middle 68 of a typically distributed class height is one standard divagation from the mean, which is( 50, 60) elevation for the given mean of 55 elevation and standard divagation of 5 elevation.

Explanation

The Empirical Rule countries that for a typically distributed set of data, roughly 68 of data values will fall within one standard divagation of the mean, 95 within two standard diversions, and99.7 within three standard diversions. In this case, the mean height is 55 elevation and the standard divagation is 5 elevation. thus, to find the interval that contains the middle 68 of the heights, we add and abate one standard divagation from the mean.

55 elevation 5 elevation = 60 elevation( Mean height plus one standard divagation)

55 elevation- 5 elevation = 50 elevation( Mean height minus one standard divagation)

This means the interval that contains the middle 68 of the heights is( 50, 60) elevation. Hence, the correct answer is option( c).

Answer: 525

Step-by-step explanation: As I read the question I’m getting the idea of division. The community program chooses 16 students every year. The 8,400 dollars from last year was the amount of money the students receive all together. Therefore 8,400 divided by 16 is 525

Final answer:

Each of the 16 fifth grade students received $525 from the community program to attend music or art camp, calculated by dividing the total funds of $8,400 by 16 students.

Explanation:

The question asked is about calculating the amount of money awarded to each of the 16 fifth grade students by a community program for attending music or art camp. Since the program awarded a total of $8,400 last year and 16 students were chosen, we need to perform a simple division to find out how much money each student received. To do this, we divide the total amount of money ($8,400) by the number of students (16).

Step-by-step Calculation:

Divide the total amount of money by the number of students: $8,400 \/ 16.

Calculate the result to determine the amount per student.

Therefore, each student received $525 to attend the music or art camp.

Answer:

Step-by-step explanation:

If you pay cash, the total amount that you will pay for the television is $349

If you pay $75 down, the balance would be paid in 18 monthly payments of 22.50 which is the installment price of the TV. Total amount paid in 18 months would be

22.5 × 18 = $405

Total cost of the TV when you pay in installments would be

405 + 75 = $480

Difference between the installment price and the cash price would be

480 - 349 = $131

The percent by which the installment price would be greater than the cash price is

131/349 × 100 = 37.5%

Answer:

Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]

Or in the alternative way would be:

 Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

On this case the claim that they want to test is: "The means for the two groups (American Idol and 60 Minutes) is the same". So we want to check if we have significant differences between the two means, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.

Null hypothesis:[tex]\mu_1 -\mu_2 = 0[/tex]

This null hypothesis can be expressed like this:

Null hypothesis:[tex]\mu_1 = \mu_2 [/tex]

Alternative hypothesis:[tex]\mu_1 -\mu_2 \neq 0[/tex]

Or in the alternative way would be:

Alternative hypothesis:[tex]\mu_1 \neq \mu_2 [/tex]

Final answer:

Lucy Baker's alternate hypothesis for her demographic analysis of American Idol and 60 Minutes would suggest a difference in the mean ages of the two audiences, represented as either (Ha: µ₁ ≠ µ₂), (Ha: µ₁ < µ₂), or (Ha: µ₁ > µ₂), depending on the direction of the difference she anticipates.

Explanation:

Lucy Baker's hypothesis test within her demographic analysis involves comparing the mean ages of audiences of two television programs: American Idol and 60 Minutes. This is a hypothesis test for two independent sample means, assuming that population standard deviations are unknown and that the samples are random.

Given that the null hypothesis (
H) posits no difference in the mean ages of the two audiences (
µ₁ = µ₂), the alternate hypothesis (
Ha) Lucy should consider would suggest that there is a difference. Her alternate hypothesis could be that the mean age of one audience is either higher or lower than the other, which can be denoted as either (
Ha: µ₁ ≠ µ₂). However, if she has a specific direction in mind (e.g., assuming one program's audience is younger than the other), she might opt for (
Ha: µ₁ < µ₂) or (
Ha: µ₁ > µ₂), accordingly.

It's crucial to select an appropriate alternate hypothesis as it signifies the anticipated outcome that stands opposed to the assumption of the null hypothesis. In this case, given that the previous study indicates no difference, Lucy's alternative hypothesis should represent the possibility that a difference indeed exists.

Answer:

The indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

Step-by-step explanation:

Let the drying times of type A be the first population and the drying times of type B be the second population. Then

We have small sample sizes [tex]n_{1} = 11[/tex] and [tex]n_{2} = 9[/tex], besides [tex]\bar{x}_{1} = 71.5[/tex], [tex]s_{1} = 3.4[/tex] , [tex]\bar{x}_{2} = 68.5[/tex] and [tex]s_{2} = 3.6[/tex]. Therefore, the pooled

estimate is given by  

[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(11-1)(3.4)^{2}+(9-1)(3.6)^{2}}{11+9-2} = 12.1822[/tex]

The 99% confidence interval for the true mean difference between the mean drying time of type A and the mean drying time of type B is given by

[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex], i.e.,

[tex](71.5-68.5)\pm t_{0.005}(3.4903)\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex]

where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (11+9-2) = 18 degrees of freedom. So

[tex]3\pm(-2.8784)(3.4903)(0.4495)[/tex], i.e.,

the indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

Answer: The function is constant from x = -2 to x=1

Final answer:

The function is decreasing from x = 1 to x = 8, as the segment in that interval shows a decrease in y values as x increases.

Explanation:

To determine where the function is decreasing, we need to look at the segments provided. A function is decreasing if, as x increases, the y value of the function decreases. From the description of the segments, we can analyze behavior in each interval:

Therefore, the function is decreasing from x = 1 to x = 8.

Answer:

The x axis in the function represents, the number of hours after 7:00 A.M. , the person reaches her car. The person reaches the car at 1:00 P.M.

Step-by-step explanation:

The x axis denotes the no. of hours and and the y axis denotes the distance from the car.

X Intercept is a point where the line intersects the X axis, we can easily notice the fact that at that point, y=0 ie. The person has reached his/her respective car.

The line intersects x at 6.

Therefore, a total of 6 hours are taken from the beginning of the hike.

Thus, the person reaches the car at 1:00 P.M.

Answer:

The person will take 6 hours to get back to her car.

Step-by-step explanation:

Answer:

a)[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Quality management        Excellent      Good     Fair    Total

Excellent                                40                35         25       100

Good                                      25                35         10         70

Fair                                         5                   10          15        30

Total                                       70                 80         50       200

Part a

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

The level of significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*100}{200}=35[/tex]

[tex]E_{2} =\frac{80*100}{200}=40[/tex]

[tex]E_{3} =\frac{50*100}{200}=25[/tex]

[tex]E_{4} =\frac{70*70}{200}=24.5[/tex]

[tex]E_{5} =\frac{80*70}{200}=28[/tex]

[tex]E_{6} =\frac{50*70}{200}=17.5[/tex]

[tex]E_{7} =\frac{70*30}{200}=10.5[/tex]

[tex]E_{8} =\frac{80*30}{200}=12[/tex]

[tex]E_{9} =\frac{50*30}{200}=7.5[/tex]

And the expected values are given by:

Quality management        Excellent      Good     Fair       Total

Excellent                                35              40          25         100

Good                                      24.5           28          17.5        85

Fair                                         10.5            12           7.5         30

Total                                       70                 80         65        215

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(3-1)(3-1)=4[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent     Ex=Reputation of company excellent

G= Quality Management good     Gx=Reputation of company good

F= Quality Management fait     Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Answer:

a) [tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

b) [tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

c) [tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X[/tex] represent the sample mean

[tex]s[/tex] represent the standard deviation for the sample

[tex]n[/tex] sample size      

[tex]\mu_o =20[/tex] represent the value that we want to test    

[tex]\alpha[/tex] represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 20[/tex]      

Alternative hypothesis:[tex]\mu > 20[/tex]      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

 First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

The P-value helps decide whether to reject the null hypothesis in a t-test. If the P-value is less than the significance level α, the null hypothesis is rejected. For each given scenario, the P-value is found from the t-distribution considering the provided t-statistic and degrees of freedom (n-1).

The problem is about conducting a t-test to check if the reflectometer reading (μ) for a new type of road paint is greater than a specified value (20). The P-value of the t-test will tell us if we should reject the null hypothesis H0: μ = 20 in favor of the alternative hypothesis Ha: μ > 20. The P-value is the probability of observing a t-score as extreme as the one calculated from the sample data (or more extreme), given that the null hypothesis is true.

(a) For n = 16, t = 3.3, and α = 0.05, we can use the t-distribution table or a statistical software to find the P-value. Since t is positive and we are dealing with a one-tailed test (because Ha: μ > 20), the P-value is the area to the right of the t-score (3.3) under the t-distribution. If the calculated P-value is less than α (0.05), we reject the null hypothesis.
(b) The same process applies for n = 8, t = 1.8, and α = 0.01. However, due to the smaller α level, we would need a larger t statistic (and thus a smaller P-value) to reject the null hypothesis.
(c) For n = 26, t = -0.6, if the calculated P-value is greater than the chosen α value, we do not reject the null hypothesis believing that the true mean reflectometer reading is 20.

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Answer:

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

[tex]p_v =P(Z>1.5617)=0.059[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}=23[/tex] represent the mean for the sample A

[tex]\bar X_{B}=18[/tex] represent the mean for the sample B

[tex]\sigma_{A}=4[/tex] represent the population standard deviation for the sample A

[tex]\sigma_{B}=5[/tex] represent the population standard deviation for the sample B

[tex]n_{A}=4[/tex] sample size selected A

[tex]n_{B}=4[/tex] sample size selected B

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for A is higher than the mean for B, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{A}-\mu_{B}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{A}-\mu_{B}>0[/tex]

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{A}-\bar X_{B})-0}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

P-value

Since is a one right tailed test the p value would be:

[tex]p_v =P(Z>1.5617)=0.059[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

To determine the appropriate conclusion at the 5% significance level, conduct a hypothesis test for the difference in means between the two fertilizer distributors.

To determine the appropriate conclusion at the 5% significance level, we need to conduct a hypothesis test for the difference in means between the two fertilizer distributors. The test statistic calculated in Excel is 1.5617. We compare this test statistic to the critical value of the t-distribution at the desired significance level of 5% with 6 degrees of freedom (8 samples - 2). If the test statistic is greater than the critical value, we reject the null hypothesis that the means are equal and conclude that there is evidence to suggest that distributor A's fertilizer contains more nitrogen than distributor B's.

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Answer:

Brownian Motion- The usual model for the time-evolution of an asset price S(t) is given by the geometric Brownian motion.

Now the geometric Brownian motion is represented by the following stochastic differential equation:

To solve the problem now we have the been Data provided:

μ= 0.12,

σ=0.24,

Step-by-step explanation:

Step A:

we have, the variables of Black Scholes Model, by putting the values of variables available, we get:

Next is, "r" the risk free rate,

Step B:

We now need to calculate the parameter d₂ of the Black Scholes Model. .

Step C:

As step C is done on excel for further calculations so, do use it if you are solving it on computer.

Final answer:

To find the probability that a call option will be exercised and the risk-neutral arbitrage-free valuation of the call option, we can use the Black-Scholes-Merton model and the risk-neutral pricing framework respectively.

Explanation:

To find the probability that a call option will be exercised, we can use the Black-Scholes-Merton model. In this case, we have:

The current price of the security (S) = $40

The strike price of the option (K) = $42

The time to expiration (T) = 4 months (or 0.33 years)

The risk-free interest rate (r) = 8% (or 0.08)

The volatility of the security (σ) = 0.24

Using these values, we can plug them into the Black-Scholes-Merton formula to calculate the probability of the call option being exercised.

For part (b), we can use the risk-neutral pricing framework to calculate the arbitrage-free valuation of the call option. This involves discounting the expected future payoff of the option at the risk-free interest rate.

To calculate the risk-neutral valuation, we use the same values as in part (a) and discount the expected payoff to the present value using the risk-free interest rate.

Answer:

Step-by-step explanation:

Given

span of bridge [tex]L=1270\ ft[/tex]

height of span [tex]h=157\ ft[/tex]

Equation of Parabola

[tex]y=0.00039x^2[/tex]

[tex]|x|<635[/tex] i.e.

[tex]-635<x<635[/tex]

[tex]\frac{dy}{dx}=2\times 0.00039[/tex]

length of Arc[tex]=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times (660.08)[/tex]

[tex]=1320.16\ ft[/tex]

The approximate length of the cables is approximately 4534.24 feet.

To approximate the length of the cables that stretch between the tops of the two towers of the suspension bridge, we can use the integral calculus to find the length of the curve defined by the equation [tex]\(y = 0.00039x^2\).[/tex]

The formula for finding the length of a curve between two points [tex]\([a, b]\)[/tex] is given by the integral:

[tex]\[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\][/tex]

Where:

- L is the length of the curve.

- a and b are the x-coordinates of the two points between which we want to find the length.

- f(x) is the function representing the curve.

- f'(x) is the derivative of the function.

In this case, we want to find the length of the cables between the two towers, which corresponds to the x-values from -635 to 635 since the width of the bridge is 1270 feet. The curve is defined by [tex]\(y = 0.00039x^2\)[/tex], so:

- a = -635

- b = 635

- [tex]\(f(x) = 0.00039x^2\)[/tex]

- [tex]\(f'(x) = 2 \cdot 0.00039x\)[/tex]

Now, let's calculate the length:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + (2 \cdot 0.00039x)^2} \, dx\][/tex]

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx\][/tex]

Now, we can evaluate this integral:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx \approx 4534.24\][/tex]

So, the approximate length of the cables that stretch between the tops of the two towers of the suspension bridge is approximately 4534.24 feet.

Learn more about Length of suspension bridge cables here:

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Answer:

[tex]E[R][/tex] = 99 Ω

[tex]\sigma_R[/tex] = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, [tex]E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2}[/tex] = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

[tex]V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333[/tex]

Hence,

The standard deviation of resistance, [tex]\sigma_R = \sqrt{V(R)} = \sqrt5.333[/tex] = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

[tex]z_1 = \frac{102-99}{2.3094} = 1.299[/tex]

[tex]z_2 = \frac{98-99}{2.3094} = -0.433[/tex]

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

Answer:

[tex]P(X<0.1)= 5.5x10^{-5}[/tex]

Step-by-step explanation:

Previous concepts

Beta distribution is defined as "a continuous density function defined on the interval [0, 1] and present two parameters positive, denoted by α and β, both parameters control the shape. "

The probability function for the beta distribution is given by:

[tex] P(X)= \frac{x^{\alpha-1} (1-x)^{\beta -1}}{B(\alpha,\beta)}[/tex]

Where B represent the beta function defined as:

[tex]B(\alpha,\beta)= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}[/tex]

Solution to the problem

For our case our random variable is given by:

[tex] X \sim \beta (\alpha=5, \beta =2)[/tex]

We can use the following R code to plot the distribution for this case:

> x=seq(0,1,0.01)

> plot(x,dbeta(x,5,2),main = "Beta distribution a=5, b=2",ylab = "Probability")

And we got as the result the figure attached.

And for this case we want this probability, since we want the probability that she has at most 10% or 0.1 change of winning:

[tex]P(X<0.1)[/tex]

And we can find this probability with the following R code:

> pbeta(0.1,5,2)

[1] 5.5e-05

And we got then this : [tex]P(X<0.1)= 5.5x10^{-5}[/tex]

This equation is separable, as

[tex]\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx[/tex]

Integrate both sides; on the left, expand the fraction as

[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]

Then

[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C[/tex]

[tex]\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C[/tex]

Since [tex]y(0)=1[/tex], we get

[tex]\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1[/tex]

so that the particular solution is

[tex]\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}[/tex]