Waiting in line. A quality - control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determinate if a new loading/unloading procedure if efective in reducing wait time in line, he measured the amount of time (in minutes) people are waiting in line for 7 days. After implementing the new procedure, he again measures the amount of time in minutes and people are waiting in line 7 days and obtains the following data.

Wait time before new procedure
Day
Mon Tues Wed Thurs Fri Sat Sat Sun Sun
11.6 25.9 20.0 38.2 57.3 32.1 81.8 57.1 62.8

Wait time after new procedure
10.7 28.3 19.2 35.9 59.2 31.8 75.3 54.9 62.0

test the claim that the new loading/unloading procedure is effective in reducing wait time (H0: µd=0 and H1: µd<0)at α=.05 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers (use the classical approach and the p-value approach).

Answer:

There is a 55.97% that a male can find a full time job in his chosen profession.

Step-by-step explanation:

We have these following probabilities:

A 60% probability that a college graduates cannot find a full time job in their chosen profession.

A 40% probability that a college graduates can find a full time job in their chosen profession.

57% of the college graduates who cannot find a job are female

43% of the college graduates who cannot find a job are male

18% of the college graduates who can find a job are female

82% of the college who can find a job are male.

Given a male college graduate, find the probability he can find a full time job in his chosen profession?

The total males are 43% of 60%(Those who cannot find a job) and 82% of 40%(Those who can find a job). So the percentage of males is [tex]P(M) = 0.43*0.60 + 0.82*0.40 = 0.586[/tex]

Those who are males and find a job in their chosen profession are 82% of 40%. So [tex]P(M \cap J) = 0.82*0.40 = 0.328[/tex]

[tex]P = \frac{P(M \cap J)}{P(M)} = \frac{0.328}{0.586} = 0.5597[/tex]

There is a 55.97% that a male can find a full time job in his chosen profession.

Answer:

The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

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

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Step-by-step explanation:

Previous concepts

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:

                      F          B         S      Bs     Total

home wins    39      156      25      83      303

Visitor wins   31       98        19      75       223

Total              70      254     44      158      526

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

H0: The number of home team and visiting team losses is independent of the sport.

H1: The number of home team and visiting team losses is dependent of the sport.

The level of significance assumed for this case is [tex]\alpha=0.01[/tex]

The statistic to check the hypothesis is given by:

[tex]\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*303}{526}=40.32[/tex]

[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]

[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]

[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]

[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]

[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]

[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]

[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]

And the expected values are given by:

                        F           B            S          Bs     Total

home wins    40.32   146.32    25.35    91.02    303

Visitor wins   29.68   107.68    18.65    66.98    223

Total                70         254         44        158      526

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]

The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]

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

"=1-CHISQ.DIST(3.29,3,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.

Answer:

-2; Inferior good

Step-by-step explanation:

Given that,

Initial Quantity = 10 boxes

New Quantity = 8 boxes

Percentage increase in Sally's income = 10%

Change in consumption:

= 8 boxes - 10 boxes

= - 2 boxes

Percentage change in quantity demanded:

= (Change in quantity demanded ÷ Initial quantity) × 100

= (-2 ÷ 10) × 100

= - 20%

Therefore,

Income elasticity of demand:

= percentage change in quantity demanded ÷ Percentage change in income

= - 20% ÷ 10

= -2

Inferior goods are generally have a negative income elasticity of demand which means that an increase in the income of the consumer will lead to reduce the quantity demanded for inferior good and vice versa.

Hence, the good is a inferior type of good.

Answer:

Step-by-step explanation:

let A represent the amount of the 10% fungicide solution

let B represent the amount of the 50% fungicide solution

B milliliters = 50% of B = (50/100)A = 0.5B

Total milliliters = 26% of 200 milliliters = 0.26 * 200 = 52

A milliliters + B milliliters = 200 milliliters

0.1A + 0.5B = 52

using substitution method to solve A and B

0.1(200-B) + 0.5B = 52

20 - 0.1B + 0.5B = 52

20 + 0.4B = 52

0.4B = 52 -20 = 32

0.4B = 32

B = 32/0.4 = 80

since B = 80

A = 200 -B = 200 - 80 = 120

Answer: he must use 120 milliliters of the 10% solution and 80 milliliters of the 50% solution.

Step-by-step explanation:

Let x represent the amount of 10% fungicide solution that Ngoc must use.

Let y represent the amount of 50% fungicide solution that Ngoc must use.

The total volume of the fungicide solution that he wants to create is 200 milliliters. It means that

x + y = 200

Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. This means that

(10/100 × x) + (50/100 × y) = (26/100 × 200)

0.1x + 0.5y = 52 - - - - - - - - - -1

Substituting x = 200 - y into equation 1, it becomes

0.1(200 - y) + 0.5y = 52

20 - 0.1y + 0.5y = 52

- 0.1y + 0.5y = 52 - 20

0.4y = 32

y = 32/0.4 = 80 milliliters

x = 200 - y = 200 - 80

x = 120 milliliters

Answer:

15.65

Step-by-step explanation:

Suppose we have two points:

[tex]A = (x_{1}, y_{1})[/tex]

[tex]B = (x_{2}, y_{2})[/tex]

The distance between these points is:

[tex]D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]

So, for points (-5, -10) and (2, 4)

[tex]D = \sqrt{(2 - (-5))^{2} + (4 - (-10))^{2}}[/tex]

[tex]D = \sqrt{7^{2} + 14^{2}[/tex]

[tex]D = 15.65[/tex]

So the correct answer is:

15.65

yes, because 6 miles equals 1 hour, and they are asking the hours for 2 and 18 miles. So all you need to do is  6 x2+ 18 miles

Answer:

(a) Ratio of A to B = 17566 : 5827

(b) Ratio of B to A = 5827 : 17566

(c) A is 301.46% percent of B.

Step-by-step explanation:

We are given A = 52,698 number of deaths due to a deadly disease in the United States in 2005   and   B = 17,481 number of deaths due to the same disease in the United States in 2009.

(a) Ratio of A to B = [tex]\frac{A}{B}[/tex] = [tex]\frac{52,698}{17,481}[/tex] = 17566 : 5827

(b) Ratio of B to A = [tex]\frac{B}{A}[/tex] = [tex]\frac{17,481}{52,698}[/tex] = 5827 : 17566

(c) Let A is x% of B so the equation formed will be;

    A = x% of B

     52,698 = x% of 17481

  Therefore x = [tex]\frac{52,698}{17,481}*100[/tex] = 301.46%

Hence, A is 301.46% of B.

The ratio of A to B is 3:1, the ratio of B to A is 1:3, and A is approximately 301.45 percent of B.

a. Find the ratio of A to B:

We can find the ratio of A to B by dividing A by B, which gives us 52,698/17,481. Evaluating this division gives a ratio of approximately 3:1, meaning that for every 3 deaths in 2005, there was 1 death in 2009.

b. Find the ratio of B to A:

To find the ratio of B to A, we divide B by A, which gives us 17,481/52,698. Simplifying this division gives a ratio of approximately 1:3, which is the inverse of the previous ratio.

c. Complete the sentence: A is ____ percent of B:

To find the percentage of A relative to B, we divide A by B, then multiply by 100. Evaluating this division gives us (52,698/17,481) * 100, which is approximately 301.45%. Therefore, we can complete the sentence by saying 'A is approximately 301.45 percent of B'.

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Answer:a) observational study

b) Randomization

c)Observational study

Step-by-step explanation:

a) The participants in (a) are monitored closely and data collected are reviewed.

b) Participants are randomly assigned to work and study.

a) The research method described is an observational study. b) The research method described is a randomized experiment. c) The research method described is an observational study.

a) The research method described is an observational study. The students are simply keeping a log of the hours they spend studying in a group and reporting it after the course is completed. The researcher is not manipulating any variables or assigning students to different groups.

b) The research method described is a randomized experiment. The students are randomly assigned to study groups and the teacher tells each group how often to meet. This allows for the manipulation of the independent variable (number of hours studying in a group) and the measurement of its effect on the dependent variable (final course grade).

c) The research method described is an observational study. The students voluntarily join groups based on how often the groups will meet. The researcher is not manipulating any variables or assigning students to different groups. The students' choice of groups is a naturally occurring phenomenon that is being observed.

Answer:

c. neither significantly low nor significantly high.

Step-by-step explanation:

We have been given that 1700 births are randomly selected and 857 of the births are girls. We are asked to choose the correct option about number of girls using subjective judgement.

We could expect that there would be 850 boys and 850 girls on average among 1700 births.

In our given scenario there are 857 girls, so there would be 843 boys. 857 girls compared to 850 girls are close.

Now probability of choosing 857 girls or more from 1700 births is approximately 0.5041 or 50%, therefore, the outcome is neither significantly low nor significantly high.

The number of girls in the study, which is about 50.4% of the total births, does not significantly deviate from the expected natural ratio of 100 girls to 105 boys, and therefore is described as neither significantly low nor significantly high.

When we talk about the natural ratio of births, we refer to the expected proportion of girls to boys in a population, under natural circumstances without human intervention. According to Newsweek, the natural ratio is 100 girls to 105 boys. If we look at the provided study data where 1700 births were randomly selected and 857 of the births are girls, we need to determine if this proportion of girls is unusually high or low compared to the natural ratio.

The expected number of girls, according to the natural ratio, would be about 50% of the population, given there is no gender preference or other factors affecting the gender birth rate. In this case, the proportion of girls is 857/1700, or approximately 50.4% which is very close to the expected 50%. Therefore, using subjective judgment, we would describe the number of girls in this study as c. neither significantly low nor significantly high.

It is important to remember that small deviations from the expected ratio can occur due to natural variability, and the number observed in this study does not seem to deviate significantly from the natural ratio.

Answer:

Option C and D are false

Step-by-step explanation:

All the mentioned option are correct in the given scenario except option C and D.

The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.

Now consider all options

A) Gender is categorical ; dose is ordinal

This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.

B) Gender is discrete; weight is continuous

This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.

Option C and D are already discussed an option E is discussed in option B.

The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.

In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.

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Answer:

[tex] P(X>9) = 0.3593[/tex]

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=25, p=0.3089)[/tex]  

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

For this case we want this probability:

[tex] P(X >9)[/tex]

And we can use the complement rule like this:

[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:

[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]  

[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]  

[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]  

[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]  

[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]  

[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]  

[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]

[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]

[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"  

And we got:

[tex] P(X>9) = 0.3593[/tex]

Answer:

Yes, one of the solutions of this differential equation is a constant solution equation.

Step-by-step explanation:

9xy' + 5y = 10.

If y' = 0, 9xy' = 0

5y = 10

y = 2 = c.

So, for all real values of x such that -∞ < x < ∞, 9xy' will be 0 and one of the solutions of the differential equation will be y = 2.

Hope this helps!

Answer:

[tex]$ \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]

Step-by-step explanation:

The [tex]$ n^{th} $[/tex] term of an arithmetic sequence is given by:

                       [tex]$ \textbf{a}_{\textbf{n}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{a} \hspace{1mm} \textbf{+} \hspace{1mm} \textbf{(n - 1)d} $[/tex]

where a is the first term of the sequence

and     d is the common difference.

We are given the [tex]$ 3^{rd} $[/tex] and the [tex]$ 13^{th} $[/tex] term of the sequence.

We are asked to find the [tex]$ 17^{th} $[/tex] term.

From the formula, we can write

[tex]$ a_3 = a + (3 - 1)d $[/tex]

[tex]$ \implies 13 = a + 2d \hspace{6mm} \hdots (1) $[/tex]  

Also, [tex]$ a_{13} = a + (13 - 1)d $[/tex]

[tex]$ \implies 43 = a + 12d \hspace{6mm} \hdots (2) $[/tex]

Now, we solve Equation (1) and (2) for a and d.

Solving we get:

a = 7; d = 3

Therefore, [tex]$ 17^{th} $[/tex] term, [tex]$ a_{17} $[/tex] can now be calculated.

[tex]$ a_{17} = a + (17 - 1)d $[/tex]

[tex]$ \implies a_{17} = 7 + 16(3) $[/tex]

[tex]$ \implies \textbf{a}_{\textbf{17}} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{55} $[/tex]

Therefore, the [tex]$ 17^{th} $[/tex] term of the sequence is 55.

Hence, the answer.

Answer:

a. 4)

b. 4)

Step-by-step explanation:

Hello!

An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information.

An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).

(a) The U.S. Census Bureau tracks population age. In 1900, the percentage of the population that was 19 years old or younger was 44.4%. In 1930, the percentage was 38.8%; in 1970, the percentage was 37.9%; and in 2000, the percentage in the age group was down to 28.5% (The First Measured Century, T. Caplow, L. Hicks, B. J. Wattenberg).

In this item, the researcher merely obtained the records of the population that were 19 or younger over several years and compared the obtained percentages. This is a clear example of an observational study, the researcher did not manipulate any factor or variable, he just looked up the information and documented it.

(b) After receiving the same lessons, a class of 100 students was randomly divided into two groups of 50 each. One group was given a multiple-choice exam covering the material in the lessons. The other group was given an essay exam. The average test scores for the two groups were then compared.

In this item, the students were randomly assigned to one of two groups and each group was given a test, group one: multiple choice and group two: essay. This is an example of an experimental study, after the class, the researcher controlled the variable "type of test" giving one type to each group and the subjects were randomly selected fro the groups assuring that they will not receive biased results. And at the end of the experiment, the response variables "test scores" where compared.

I hope it helps!

Final answer:

The studies presented about U.S. Census Bureau population tracking and exam type comparison in a classroom represent an observational study and an experiment, respectively. The former is so because no treatment was applied, while the latter involved purposefully assigning different tests to observe outcomes.

Explanation:

When classifying the given studies as either an observational study or an experiment, it's important to understand the key characteristics of each. An observational study involves monitoring subjects without any intervention, while an experiment involves the deliberate imposition of a treatment to observe potential changes.

(a) The U.S. Census Bureau tracking population age is an example of an observational study because the data on the percentage of the population that was 19 years old or younger in various years was collected without any manipulation or treatment being applied to the subjects.

(b) Dividing a class into two groups and administering different types of exams constitutes an experiment. This is because a treatment (the type of exam) was deliberately imposed on the students to observe the difference in test scores, which we interpret as the outcome of interest.

Answer:

ahsbbsnssisiskak sjsnsnaan

The sample space consists of all sets of 5 numbers that can be picked from 40, and each outcome has an equal chance of occurring. The probability that exactly three out of five numbers are even is calculated as the combination of choosing three even numbers and two odd numbers, divided by the total number of outcomes.

(a) In this case, the sample space consists of all possible outcomes of the experiment. There are 40 distinct numbers, and we're picking 5. The total number of distinct sets of 5 that can be chosen from 40 is given by the combination formula C(n, k) = n! / [k!(n - k)!]. So, in this case, C(40, 5), which would give the total number of outcomes. The probability measure P assigns each outcome a probability. Since we're choosing the numbers randomly and uniformly, each outcome has an equal probability of occurring, which is 1 / C(40, 5).

(b) To find the probability that exactly three out of the five numbers are even, we first note that there are 20 even numbers and 20 odd numbers in the set 1, 2, ..., 40. The number of ways to choose three even numbers is C(20, 3) and the number of ways to choose two odd numbers is C(20, 2). The probability of this happening is thus [C(20, 3) * C(20, 2)] / C(40, 5).

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Answer:

B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation =  [tex]\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}[/tex]

Suppose,

     X             X - Xbar       [tex](X - Xbar)^{2}[/tex]

     3              3 - 6 = -3         -3 * -3 = 9

     5              5 - 6 = -1          -1 * -1 = 1

     10            10 - 6 = 4          4 * 4 = 16

Mean of above data, Xbar = [tex]\frac{3+ 5+10}{3}[/tex] = 6

Standard Deviation of data = [tex]\sqrt{\frac{26}{3-1} }[/tex] = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

     X                X - Xbar           [tex](X - Xbar)^{2}[/tex]

 3*2 = 6        6 - 12 = -6         -6 * -6 = 36

 5*2 = 10       10 - 12 = -2       -2 * -2 = 4

10*2 =20      20 - 12 = 8         8 * 8 = 64

Mean of new data, Xbar = [tex]\frac{6+ 10+20}{3}[/tex] = 12

Standard Deviation of new data = [tex]\sqrt{\frac{104}{3-1} }[/tex] = 7.2111

Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as

  3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.

Answer:

a) [tex] W=0,2[/tex]

b) [tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]

Step-by-step explanation:

Part a

For this case we have the following differential equation:

[tex] W \sqrt{4-2W}=0[/tex]

If we square both sides we got:

[tex] W^2 (4-2W) =0[/tex]

And we have two possible solutions for this system [tex] W=0, W=2[/tex]

So then that represent the constant solutions for the differential equation.

So then the solution for this case is :

[tex] W=0,2[/tex]

Part b: Solve the differential equation in part (a)

For this case we can rewrite the differential equation like this:

[tex] \frac{dW}{dx} =W \sqrt{4-2W}[/tex]

And reordering we have this:

[tex] \frac{dW}{W \sqrt{4-2W}} = dx[/tex]

Integrating both sides we got:

[tex] \int \frac{dW}{W \sqrt{4-2W}} = \int dx[/tex]

Using CAS for the left part we got:

[tex] -tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c[/tex]

We can multiply both sides by -1 we got:

[tex] tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c[/tex]

And we can apply tanh in both sides and we got:

[tex] \frac{1}{2} \sqrt{4-2W} = tanh(-x-c)[/tex]

By properties of tanh we can rewrite the last expression like this:

[tex]\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)[/tex]

We can square both sides and we got:

[tex] \frac{1}{4} (4-2W) = tanh^2 (x+c) [/tex]

[tex] 1-\frac{1}{2}W = tanh^2 (x+c)[/tex]

And solving for W we got:

[tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]

And that would be our solution for the differential equation

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]

Where:

[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]

a. Let's find h(3) and h(4) using the data provided by the problem:

[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]

The average velocity over the interval [3, 4] is :

[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]

b. Let's find h(3.5) using the data provided by the problem:

[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]

The average velocity over the interval [3, 3.5] is :

[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]

c. Let's find h(3.1) using the data provided by the problem:

[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]

The average velocity over the interval [3, 3.1] is :

[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]

d. Let's find h(3.01) using the data provided by the problem:

[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]

The average velocity over the interval [3, 3.01] is :

[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]

e. Let's find h(3.001) using the data provided by the problem:

[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]

[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]

The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.

Given :

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].

a) [3,4]

At time t = 3 and 4, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]

b) [3,3.5]

At time t = 3 and 3.5, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]

c) [3,3.1]

At time t = 3 and 3.1, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]

d) [3,3.01]

At time t = 3 and 3.01, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]

e) [3,3.001]

At time t = 3 and 3.001, the value of h(t) is given below:

[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]

[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]

The average velocity is given by:

[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]

For more information, refer to the link given below:

https://brainly.com/question/18153640

Answer:

The probability that the good exam belongs to student X is 0.8571.

Step-by-step explanation:

It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,

Compute the probability that exactly one student does well in the exam as follows:

[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]

Then the probability that X is the one who did well in the exam is:

[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]

Thus, the probability that the good exam belongs to student X is 0.8571.

Answer:

[tex]T(p) = \frac{1}{4}p[/tex]

The new price is $6.75

Step-by-step explanation:

The new price of the toy after the discount (T) is given by the original price (p) subtracted by the discounted amount (3/4 of p):

[tex]T(p) = p-\frac{3}{4}p \\T(p) = (1-\frac{3}{4})p\\T(p) = \frac{1}{4}p[/tex]

If the original price was $26.99, the new price is:

[tex]T= \frac{1}{4}*\$26.99\\T=\$6.75[/tex]

Answer:

y = - x + 5

Step-by-step explanation:

L(5.0), M(0,5)

y = mx + b

m = (5 - 0) / (0 - 5) = 5 / -5 = - 1

b = y - mx = 5 - ((-1) x 0) = 5

y = - x + 5

Answer:

a) Lower inner fence = 77.6168 = 77.62 to 2 d.p

Lower outer fence = 48.9044 = 48.90 to 2 d.p

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159

Step-by-step explanation:

Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Q₁ = 25th percentile = lower quartile

IQR = Inter quartile Range = Q₃ - Q₁

Q₃ = 75th percentile = upper quartile

To calculate Q₁ for a normal distribution with only mean and standard deviation known,

We need the standardized score whose probability is 0.25 P(z) = 0.25

From the normal distribution table

z = (± 0.674)

z = (x - xbar)/σ

x = the value in the data we're interested in,

xbar = mean = 115.9

σ = standard deviation = 14.2

Lower quartile corresponds to (z = - 0.674)

- 0.674 = (x - 115.9)/14.2

Q₁ = X = 106.3292

The upper quartile, Q₃ corresponds to z = (+0.674)

Q₃ = 125.4708

IQR = 125.4708 - 106.3292 = 19.1416

Lower inner fence = Q₁ – (1.5 × IQR)

Lower outer fence = Q₁ – (3 × IQR)

Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168

Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044

b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)

We standardize 74 by obtaining its z-score

z = (x - xbar)/σ

z = (74 - 115.9)/14.2 = - 2.95

P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)

Final answer:

The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.

Explanation:

To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.

The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:

Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5

Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.

Answer:

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Step-by-step explanation:

We have that  Mohammed can knit 5 sweaters or bake 240 cookies.

In one week Aisha can knit 15 sweaters or bake 480 cookies.  

We calculate how much is Mohammed's opportunity cost knitting one sweater. We get  

\frac{240}{5}= 48.

We conclude,  Mohammed's opportunity cost knitting one sweater is 48 cookies.

Answer:

Type I error: The correct option is (C).

Type II error: The correct option is (D).

Step-by-step explanation:

The type-I-error is the probability of rejecting the null hypothesis when the null hypothesis is true.

The type-II-error is the probability of filing to reject the null hypothesis when in fact it is false.

The hypothesis in this problem can be defined as follows:

Null hypothesis (H₀): The percentage of adults who have a job is equal to 88%.

Alternate Hypothesis (Hₐ): The percentage of adults who have a job is different from 88%.

The type-I-error in this case will be committed when we conclude that the percentage of adults who have a job is different from 88% when in fact it is equal to 88%.

The type-II-error in this case will be committed when we conclude that the percentage of adults who have a job is equal to 88% when in fact it is different than 88%.

Answer:

(a) Yes, f(x) is a valid probability mass function.

(b) The probability that you will walk at least two dogs this week = 0.74.

(c)  The expected number of dogs you will walk this week = 3 dogs.

(d) The expected value of X2 = 11.29

(e) Var[x] = E(X2) – (E[X])2 = 3.2811

Step-by-step explanation:

We are given with the probability mass function (pmf),f(x), o'is defined as follows:

Firstly let X = Number of dogs you walk this week

    X                 P(X = x)

    0                      0.14

    1                       0.12

    2                      0.15

    3                      0.23

    4                      0.18

    5                      0.09

    6                      0.08

    7                       0.01

(a) Now f(x) to be a valid probability mass function, two conditions should be    met :

So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1

Hence both the conditions are satisfied so f(x) is a valid probability mass function.

(b) Probability that we will walk at least two dogs this week = P(X>=2)

      = 1 - P(X = 0) - P(X = 1) = 1 -  0.14 - 0.12 = 0.74

(c) To Compute the expected number of dogs you will walk this week we will use expectation formula which says:

           E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]

                                           = 2.83 or 3 after rounding off.

Therefore the expected number of dogs you will walk this week are 3 dogs.

(d) The expected value of X2 [E(X2)] =   [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]

      = [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29

(e) Var[x] = E(X2) – (E[X])2

     We have E(X2) = 11.29 and E(X) = 2.83

      Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.

The random variable X represents the number of dogs walked in a week for a dog walking business. By verifying that the sum of the probabilities equals 1, we've confirmed the pmf is valid. The probability of walking at least two dogs is 0.74, and calculations for the expected number and variance involve using the given probabilities and applying the respective formulas.

Understanding Probability Mass Functions

In the context of a dog walking business, let us define the random variable X as the number of dogs walked in a week. The values that X can take on are 0, 1, 2, 3, 4, 5, 6, and 7. The probabilities of these events are given by P(X = x), where x are the respective values X can take.

To verify if the given probability mass function (pmf) is valid, the sum of all probabilities P(X = x) should be equal to 1. We calculate the sum as follows:

Adding them gives 1 which confirms it is a valid pmf.

To find the probability of walking at least two dogs this week, which is P(X ≥ 2), we need to add up the probabilities for all events where X is 2 or more:

P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)
= 0.15 + 0.23 + 0.18 + 0.09 + 0.08 + 0.01
= 0.74

To compute the expected number of dogs walked in a week, or E[X], we use the formula:

E[X] = Σ x * P(X = x)

For E[X2], we calculate the expectation of X-squared by multiplying each value of x by itself and then by its probability, before summing these products.

Finally, the variance Var[X] of the random variable X is given by E(X2) - (E[X])2. This involves first calculating E[X2] as explained, then squaring the previously found E[X] value, and subtracting that square from E[X2].

You can't deduce the length of a side from the perimeters of a rectangle.

Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.

So, we know that

[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]

But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.

The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.

To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.

However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.

Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.

https://brainly.com/question/29595517

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Answer:

variance = (27*24656384-20182^2)/(27*26) =368104.3

standard devaition SD= sqrt(368104.3) =606.716

maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD

= (20182/27)+(2*606.716)

= 1960.913

=$1960913

Answer:

A. Control group  

Step-by-step explanation:

A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.

An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.

Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.          

Answer:

1 hr 52 minutes

Step-by-step explanation:

As per Newton law of cooling we have

[tex]T(t) = T_s +(T_0-T_s)e^{-kt}[/tex]

where T0 is the initial temperature of the body

Ts = temperature of surrounding

t = time lapsed

k = constant

Using this we find that T0 = 98.6 : Ts= 65

Let x hours be lapsed before the body was found.

Then we have

[tex]T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917[/tex]

Next after 1 hour temperature was 80

[tex]T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464[/tex]

Dividing we get

[tex]e^k = 1.325408\\k = 0.2817[/tex]

Substitute this in

[tex]e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863[/tex]

approximately 1 hour 52 minutes have lapsed.