You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not appropriate to use the results of the Central Limit Theorem (CLT) to calculate the confidence intervals for the mean.

Answer:

95% confidence interval for the mean efficiency is [84.483 , 85.517].

Step-by-step explanation:

We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;

          P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\mu[/tex] = sample average efficiency = 85

            [tex]\sigma[/tex] = sample standard deviation = 2

            n = sample of motors = 60

            [tex]\mu[/tex] = population mean efficiency

So, 95% confidence interval for the mean efficiency, [tex]\mu[/tex] is ;

P(-2.0009 < [tex]t_5_9[/tex] < 2.0009) = 0.95

P(-2.0009  < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.0009 ) = 0.95

P( [tex]-2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95

P( [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ]

                                                 = [ [tex]85 -2.0009 \times {\frac{2}{\sqrt{60} }[/tex] , [tex]85 +2.0009 \times {\frac{2}{\sqrt{60} }[/tex] ]

                                                 = [84.483 , 85.517]

Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].

Final answer:

The 95% confidence interval for the mean efficiency of electric motors, given a sample mean of 85, a standard deviation of 2, and a sample size of 60, is approximately (84.494, 85.506) when rounded to three decimal places.

Explanation:

To calculate the 95% confidence interval for the mean efficiency of electric motors, we use the sample mean, standard deviation, and the sample size along with the z-score for the 95% confidence level. Since the sample size is large (n > 30), we can use the z-distribution to approximate the sampling distribution of the sample mean.

The formula for a confidence interval is:

Confidence Interval = sample mean "+/-" (z-score * (standard deviation / sqrt(n)))

Given that the sample mean is 85, the standard deviation is 2, and the sample size (n) is 60, and using the z-score of approximately 1.96 for 95% confidence, we can compute the confidence interval.

Confidence Interval = 85 "+/-" (1.96 * (2 / √(60)))

After the calculation, we get the two ends of the interval:

Lower Limit = 85 - (1.96 * (2 / √(60))) = 84.494

Upper Limit = 85 + (1.96 * (2 / √(60))) = 85.506

Rounding these to three decimal places, the 95% confidence interval for the mean efficiency of electric motors is approximately (84.494, 85.506). This means we can be 95% confident that the true average efficiency of all electric motors is between 84.494% and 85.506%.

Answer:

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find 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 size of the sample.

[tex]M = 2.575*\frac{62}{\sqrt{38}} = 25.90[/tex]

The lower end of the interval is the mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.

The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

Answer:

Option A

Step-by-step explanation:

The 99% confidence interval is (472.69, 527.31). We are 99% confident that the true population mean of sodium loss for tennis players will be between 472.69 milligrams per pound and 527.31 milligrams per pound.

Answer:

Perimeter = 72

Step-by-step explanation:

Mathematics Puzzle

The area of a rectangle is the product of the base and the height

A=b.h

We know the base and the height are two consecutive prime numbers, not much of useful information so far.

We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.

We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.

The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.

The perimeter of that rectangle is 2*17+2*19= 72

Answer:

72 units

Step-by-step explanation:

72 units

Answer:

A) He should take the route with 4 crossings

B) The probability that he took the 4 crossing route is 0.2158

Step-by-step explanation:

Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.

Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1

The probability that the Professor is on time on route 1 is equal to

P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477

On the other hand, the probability that the professor is on time on route 2 is

P(X|B = 0) = 0.9² = 0.81

Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.

B) Lets call L the event 'The professor is late'. We know that

P(L|A) = 1-0.9477 = 0.0524

P(L|B) = 1-0.81 = 0.19

Also

P(A) = P(B) = 1/2 (this only depends on the result of the coin.

For the Bayes theorem we know, therefore that

[tex]P(A|L) = \frac{P(L|A) * P(A)}{P(L|A)*P(A) + P(L|B)*P(B)} = \frac{0.0523*0.5}{0.0523*0.5 + 0.19*0.5 } = 0.2158[/tex]

Hence, the probability that he took the 4 crossing route is 0.2158.

Answer:

x = 6 units.

Step-by-step explanation:

By Geometric mean property:

[tex]x = \sqrt{3 \times 12} = \sqrt{36} = 6 \\ \hspace{20 pt} \huge \orange{ \boxed{ \therefore \: x = 6}}[/tex]

Hence, x = 6 units.

Answer:

7.51% probability that none of the 9 employees will say their company is loyal to them.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

25% of employees said their company is loyal to them.

This means that [tex]p = 0.25[/tex]

9 employees are selected randomly

This means that [tex]n = 9[/tex]

What is the probability that none of the 9 employees will say their company is loyal to them?

This is P(X = 0).

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

[tex]P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751[/tex]

7.51% probability that none of the 9 employees will say their company is loyal to them.

Answer:

For this case we can define the following two events A and B.

In order to classify A and B as independent we needd to satisfy this condition:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

None of the above.

True, because none of the options were correct.

See explanation below

Step-by-step explanation:

For this case we can define the following two events A and B.

In order to classify A and B as independent we need to satisfy this condition:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

So let's analyze one by one the possible options:

the sum of their probabilities must be equal to one.

False, the sum of the probabilities can be <1 so this statement is not true

they must be mutually exclusive.

False when we talk about mutually exclusive events we are saying that:

[tex] P(A \cap B) =0[/tex]

But independence not always means that we have mutually exclusive events

their intersection must be zero.

False the intersection of the probabilities is 0 just if we have mutually exclusive events, not independent events

None of the above.

True, because none of the options were correct.

Independent events refer to the concept in which the outcome of one event doesn't influence the outcome of another. If two events are independent, the probability of both events taking place is the product of their individual probabilities.

In mathematics, when we say that events are independent, it means that the outcome of one event does not affect the outcome of the other. The correct statement is: 'If two events are independent, the probability of both events happening is the product of the probabilities of each event.' When the sum of their probabilities is equal to one, they are mutually exclusive, not independent. And their intersection being zero also refers to mutually exclusive events not independent ones.

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Answer:

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

We can estimate the mean, variance, and standard deviation by using class midpoints and the number of observations. First, find the midpoint of each age group. Then, to find the mean, multiply each midpoint by the number of shoplifters in that age group, sum those products, and divide by the total number of observations. To find the variance, subtract the mean from each midpoint, square the result, multiply by the number of shoplifters in that group, sum those products, and divide by the total number of observations minus one. The standard deviation is the square root of the variance.

To estimate the mean age, sample variance, and sample standard deviation for the shoplifters from a random sample of 895 incidents we need to follow a few steps. Here's how to do it:

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Answer:

1/3

Step-by-step explanation:

60-90 is 30 numbers, right? So it is 30/90, or 1/3

Answer:

Probability that a smoker has lung disease = 0.2132

Step-by-step explanation:

Let L = event that % of population having lung disease, P(L) = 0.07

So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93

S = event that person is smoker

% of population that are smokers given they are having lung disease, P(S/L) = 0.90

% of population that are smokers given they are not having lung disease, P(S/L') = 0.25

We know that, conditional probability formula is given by;

                        P(S/L) = [tex]\frac{P(S\bigcap L)}{P(L)}[/tex]  

                        [tex]P(S\bigcap L)[/tex] = P(S/L) * P(L)

                                      = 0.90 * 0.07 = 0.063

So,  [tex]P(S\bigcap L)[/tex] = 0.063 .

Now, probability that a smoker has lung disease is given by = P(L/S)

      P(L/S) = [tex]\frac{P(S\bigcap L)}{P(S)}[/tex]

P(S) = P(S/L) * P(L) + P(S/L') * P(L')

       = 0.90 * 0.07 + 0.25 * 0.93 = 0.2955

Therefore, P(L/S) = [tex]\frac{0.063}{0.2955}[/tex] = 0.2132

Hence, probability that a smoker has lung disease is 0.2132 .

Answer:

A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.

B) slope: (5.6-1.8)/(2-0) = 1.9

y = 1.9x + 1.8

(Line of best fit)

x = 13,

y = 1.9(13) + 1.8 = 26.5

Predicted no. of games won after 13 months of practice is 26.5

Final answer:

The y-intercept, representing initial games won and the equation for line of best fit predicting future games won, are determined from the graph data.

Explanation:

Part A:

Part B:

The equation for the line of best fit in slope-intercept form is y = 1.4x + 1.8.

To predict the number of games won after 13 months of practice, substitute x = 13 into the equation:

y = 1.4(13) + 1.8 = 19.5

So, the predicted number of games that could be won after 13 months of practice is 19.5.

Answer:

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Step-by-step explanation:

Total plants = 11

Domestic plants = 7

Outside the US plants = 4

Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure

Here we have n=4, p=4/11 and q=7/11

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰

          = 1 - 0.16399

P(X≥1) = 0.836

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Answer:

Step-by-step explanation:

18 a)Since line DE is parallel to line BC, it means that triangle ADE is similar to triangle ABC. Therefore,

AB/AD = AC/AE = BC/DE

AC = AE + CE = 18 + 9

AC = 27

AB = AD + DB

AB = AD + 5

Therefore,

27/18 = (AD + 5)/AD

Cross multiplying, it becomes

27 × AD = 18(AD + 5)

27AD = 18AD + 90

27AD - 18AD = 90

9AD = 90

AD = 90/9

AD = 10

b) AC = AE + EC = 13 + 3

AC = 16

To find AD,

16/13 = 24/AD

16 × AD = 13 × 24

16AD = 312

AD = 312/16

AD = 19.5

DB = 24 - 19.5

DB = 4.5

Answer:

a-True

b-True

c-True

d- False

e-True

f-True

g-True

h- False

i-True

j- False

Step-by-step explanation:

See the attached pictures.

Answer:

(a) The probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.

(b) The probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.

Step-by-step explanation:

Denote the events as follows:

X = a mower is manufactured at the Portland factory

Y = a mower is manufactured at the Dallas factory

Z= a mower is manufactured at the Buffalo factory

A = a mower passes the quality check.

The information provided is:

[tex]P(X)=0.30\\P(A|X)=0.80\\P(Y)=0.50\\P(A|Y)=0.70\\P(Z)=0.20\\P(A)=0.72[/tex]

(a)

The probability that a mower manufactured at the Buffalo factory will pass the quality control check is:

P (A|Z)

Compute the value of P (A|Z) as follows:

[tex]P(A)=P(A\cap X)+P(A\cap Y) + P (A\cap Z)\\0.72=(0.80\times0.30)+(0.70\times0.50)+(0.20\times P(A|Z))\\0.20\times P(A|Z)=0.72-0.24-0.35\\P(A|Z)=\frac{0.13}{0.20}\\=0.65[/tex]

Thus, the probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.

(b)

Compute the value of P (Y|A) as follows:

[tex]P(Y|A)=\frac{P(A|Y)P(Y)}{P(A)}=\frac{0.70\times0.50}{0.72}=0.4861[/tex]

Thus, the probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.

The probability that a mower from the Buffalo factory will pass the quality control check is 13%, and if a customer receives a mower that has passed the check, there is a 49% probability that it was manufactured in Dallas.

To find the probability that a robot lawn mower made at the Buffalo factory will pass the quality control check, we first understand that the total probability of a mower passing the check is a sum of all the probabilities from the three factories. This is given as 0.72.

The Portland factory which makes 30% of the mowers has a 0.8 probability of passing the check. Therefore, the contribution of Portland to the total probability is 0.3*0.8 = 0.24. Similarly, the Dallas factory makes 50% of the mowers and each has a passing probability of 0.7, so the Dallas contribution is 0.5*0.7 = 0.35.  

Knowing this, we can subtract the total contributions of Portland and Dallas from the overall probability of 0.72 to get Buffalo's contribution, which is the Buffalo passing probability. Therefore, the Buffalo passing probability becomes 0.72 - 0.24 - 0.35 = 0.13 or 13%.

In the second part, if the customer receives a mower that has passed the quality control check, the probability that it was manufactured in Dallas is the contribution of the Dallas factory to the passing mowers i.e., 0.35 ÷ 0.72 = 0.486 or approximately 49%.

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Answer:

The​ P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is 4.4 which is LOW so there IS sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

Step-by-step explanation:

Hello!

Remember:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Let's say that the significance level of this correlation test is α: 0.05

If the p-value is the probability of obtaining

you can express it as a percentage: 4.4%Is a very low probability. The decision rule using the p-value is:

p-value < α ⇒ Reject the null hypothesis

p-value ≥ α ⇒ Do not reject the null hypothesis.

The p-value is less than the significance level, the decision is to reject the null hypothesis.

In a linear correlation analysis the statement "there is no linear correlation between the two variables" is always in the null hypothesis, so if you reject it, you can conclude that there is a linear correlation between the variables.

I hope it helps!

In statistical analysis, a P-value of 0.044 indicates there is a 4.4% chance of obtaining a linear correlation as extreme as the observed correlation coefficient. A P-value under 0.05 provides enough evidence to reject the null hypothesis of no correlation, therefore suggesting a significant correlation. Hence, there is sufficient evidence of a linear correlation between car weight and highway fuel consumption.

The P-value of 0.044 in this context represents the likelihood of obtaining a linear correlation coefficient for the data points in your dataset that is as extreme as, or more extreme than, the one you calculated, assuming there is no linear relationship between the two variables (weight and highway fuel consumption in automobiles). This probability is 4.4% - rather low. A common threshold for significance in many fields is 0.05, or 5%. If your P-value is below this threshold, we reject the null hypothesis that there is no correlation and conclude there may be a correlation. Therefore, as the P-value is 0.044, which is below the 0.05 threshold, there is sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.

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Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

[tex] E(\theta) = \mu[/tex]

And we can find the expected value of each estimator like this:

[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]

So then we conclude that [tex] \theta_1 [/tex] is unbiased.

For the second estimator we have this:

[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]

And then we conclude that [tex]\theta_2[/tex] is unbiaed too.

b) For this case first we need to find the variance of each estimator:

[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]

And for the second estimator we have this:

[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]

And the relative efficiency is given by:

[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]

Step-by-step explanation:

For this case we assume that we have a random sample given by: [tex] X_1, X_2,....,X_7[/tex] and each [tex] X_i \sim N (\mu, \sigma)[/tex]

Part a

In order to check if an estimator is unbiased we need to check this condition:

[tex] E(\theta) = \mu[/tex]

And we can find the expected value of each estimator like this:

[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]

So then we conclude that [tex] \theta_1 [/tex] is unbiased.

For the second estimator we have this:

[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]

And then we conclude that [tex]\theta_2[/tex] is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]

And for the second estimator we have this:

[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]

And the relative efficiency is given by:

[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]

Both θ1 and θ2 are unbiased estimators of the population mean μ. However, θ1 is more efficient due to having a lower variance compared to θ2. The relative efficiency of θ1 compared to θ2 is 10.5.

To determine if the estimators [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased and which one is best, we follow these steps:

A. Evaluating Unbiasedness

B. Evaluating Efficiency

The relative efficiency of [tex](\theta_1)[/tex] to [tex](\theta_2)[/tex] is:

Both [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased estimators of [tex](\mu)[/tex], but [tex](\theta_1)[/tex] is the best in terms of efficiency since it has a lower variance.

Answer:

The code in R for the time series is given as below.

Step-by-step explanation:

As the complete question is not presented thus a sample code for the variables is given as below

# Libraries

library(ggplot2)

library(dplyr)

library(plotly)

library(hrbrthemes)

# Load dataset

filepath=" Your dataset path here in csv format"

data <- read.table("filepath", header=T)

data$date <- as.Date(data$date)

# plot

data %>%

 ggplot( aes(x=date, y1=value1,x=date, y2=value2)) +

   geom_line(color="#69b3a2") +

   ylim(0,22000) +

   theme_ipsum()

In R, stock prices and return series can be visualized using the ts.plot() or plot() functions in a line graph. The time series graph will then help in identifying and understanding trends and patterns in the data over time. Comparing the plots of the stock prices and the return series can provide insights into market behavior.

To plot both stock prices and return series using R, one would use either the ts.plot or plot function. Firstly, you need to import or generate the stock prices and the return series data. The horizontal axis of the plot corresponds to the date or time increments, while the vertical axis corresponds to the values of the variable (i.e., stock prices or return series).

Time series graphs facilitate spotting trends and are used to graphically represent the same variable values recorded over an extended period of time. In the context of stock prices and return series, these patterns over time can provide valuable insights into market behavior and investment opportunities.

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In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.

The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.

To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:

1.9 = 1/(sqrt(1 - (γ/2)^2))

Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.

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The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.

The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.

The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.

The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.

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Answer:

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continous

Step-by-step explanation:

Continuous:

Real numbers, can be integer, decimal, etc.

Discrete:

Only integer(countable values). So can be 0,1,2...

(a) number of traffic fatalities per year in the state of Florida

You cannot have half of a traffic fatality, for example. So this is discrete

(b) distance a golf ball travels after being hit with a driver

The ball can travel 10.25m, for example, which is a decimal number. So this is continuous.

(c) time required to drive from home to college on any given day

You can take 10.5 minutes, for example, which is a decimal value. So this is continuous.

(d) number of ships in Pearl Harbor on any given day

There is no half ship, for example. So this is discrete.

(e) your weight before breakfast each morning continuous discrete

You can weigh 80.4kg, for example, which is a decimal number. So this is continuous.

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continuous

Discrete data is the numerical type of data which includes whole, concrete numbers that has specific and fixed data values. therefore, they can be determined by counting.

example: Number of students in the class, ect.,

Continuous type of data includes complex numbers or the varying data values that are measured over a specific time interval.

example: Height of students in a school, etc.,

(a) Number of traffic fatalities per year in the state of Florida.

This is a Discrete data, as it is a countable data and will be always a whole be number. Number of traffic fatalities per year will be 75, 150, 200, etc.

(b) distance a golf ball travels after being hit with a driver.

This is a Discrete data, as the data can be measures and will be not always be a whole number. distance a golf ball travels after being hit can be 1.25 meters,  10.9 meters, 21.1 meters, etc.

(c) time required to drive from home to college on any given day.

This is a Discrete data, as the data can be measures and will be not always be a whole number. time required to drive from home to college can be 5minutes 30 seconds, 15 minutes 26 seconds.

(d) number of ships in Pearl Harbor on any given day.

This is a Discrete data, as it is a countable data and will be always a whole number. number of ships in Pearl Harbor on any given day will be always 1, 10, 8, etc.

(e) your weight before breakfast each morning.

This is a Discrete data, as the data can be measures and will be not always be a whole number. weight can be 44.56 kgs., 52.3 kgs, etc.

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Answer:

( 5 x 10 ) cm

Step-by-step explanation:

Given:

- The dimensions of the bar are:

                                  ( 2 x 5 x 10 ) cm

Find:

Which two faces with dimensions ( _x _ ) should be connected to get smallest possible resistance.

Solution:

- The electrical resistance R of any material with density ρ and corresponding dimensions is expressed as:

                                R = ρ*L / A

- Where, A: cross sectional Area

              L: The length in between the two faces.

- We need to minimize the electrical resistance of the bar. For that the Area must be maximized and Length should be minimized.

-                               A_max & L_min ---- > R_min

                               (5*10) & ( 2) ------> R_min

Hence, the electrical resistance is minimized by connecting the face with following dimensions ( 5 x 10 ) cm

Answer: the third option is correct

Step-by-step explanation:

The first matrix is a 2 × 3 matrix while the second matrix is a 3 × 2 matrix. To get the product of both matrices, we would multiply each term in each row by the terms in the corresponding column and add.

1) row 1, column 1

1×2 + 3×3 + 1×4 = 2 + 9 + 4 = 15

2) row 1, column 2

1×-2 + 3×5 + 1×1 = - 2 + 15 + 1 = 14

3) row 2, column 1

-2×-2 + 1×3 + 0×4 = - 4 + 3 + 0 = - 1

4) row 2, column 2

- 2×-2 + 1×5 + 0×1 = 4 + 5 = 9

The solution becomes

15 14

- 1 9

Answer: 0.06 ft long

Step-by-step explanation:

Porcelli's board = 2% feet long; to convert 2% into fraction, divide by 100= 2/100 = 0.02 ft

Smith's board, from the question is 3 times porcelli's board = 3 x 0.02= 0.06 ft

I hope this helps.

Use simple unitary method:

∵ 1.5 cm on map = 25 miles in reality

∴ x cm on map = 190 miles in reality

[tex]\frac{1.5}{x} = \frac{25}{190}\\ x = 11.4 cm[/tex]

Thus, the two cities are 11.4 cm apart on the map

Final answer:

To determine the map distance between two cities that are 190 miles apart using the scale 1.5 cm = 25 miles, divide the actual distance by the scale ratio (16.67 miles/cm), resulting in 11.4 cm.

Explanation:

For the map with scale 1.5 cm = 25 miles, we first calculate how many miles one centimeter represents by dividing the miles by the centimeters in the scale:

25 miles / 1.5 cm = 16.67 miles/cm

Next, we find the map distance for 190 miles by dividing the actual distance by the distance one centimeter represents:

190 miles / 16.67 miles/cm = 11.4 cm

So, the two cities are 11.4 cm apart on the map.

Answer:

(a) No, the coffee industry's claim is not supported by this sample.

(b) Null hypothesis: The average adult drinks 1.7 cups of coffee per day.

Alternate hypothesis: The average adult drinks more than 1.7 cups of coffee per day.

Step-by-step explanation:

(a) Test statistic (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 1.9 cups per day

population mean = 1.7 cups per day

sd = 0.6 cups per day

n = 40

z = (1.9 - 1.7) ÷ 0.6/√40 = 0.2 ÷ 0.095 = 2.11

The test is a one-tailed test. Using alpha (significance level) = 0.1, the critical value is 2.326.

Conclusion:

Reject the null hypothesis because the test statistic 2.11 falls within the rejection region of the critical value 2.326.

The coffee industry's claim is contained in the null hypothesis, hence it is not supported by the sample because the null hypothesis is rejected.

(b) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Multiply the cost per hour by number of hours x and for option 1 you need to add the monthly fee:

A) option 1: C= 4x +35

Option2: C = 6.50x

B)

4x+35 < 6.50x

Subtract 4x from both sides:

35 < 2.50x

Divide both sides by 2.50 :

X < 14

You would need to rent more than 14 hours for option 1 to be cheaper.

Answer:

The probability that the next customer will arrive in 3 minutes or less is 0.45.

Step-by-step explanation:

Let N (t) be a Poisson process with arrival rate λ. If X is the time of the next arrival then,

[tex]P(X>t)=e^{-\lambda t}[/tex]

Given:

[tex]\lambda=\frac{12}{60}[/tex]

t = 3 minutes

Compute the probability that the next customer will arrive in 3 minutes or less as follows:

P (X ≤ 3) = 1 - P (X > 3)

              [tex]=1-e^{-\frac{12}{60}\times3}\\=1-e^{-0.6}\\=1-0.55\\=0.45[/tex]

Thus, the probability that the next customer will arrive in 3 minutes or less is 0.45.

The probability that the next customer will arrive in 3 minutes or less is; 0.4512

This is a Poisson distribution problem with the formula;

P(X > t) = e^(-λt)

Where;

λ is arrival rate

t is arrival time

We are given;

λ = 12/60

t = 3 minutes

P (X ≤ 3) = 1 - (P(X > 3))

Thus;

P (X ≤ 3) = 1 - e^((12/60) × 3)

P (X ≤ 3) = 1 - 0.5488

P (X ≤ 3) = 0.4512

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Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

The confidence intervals for the two experiments have overlapping intervals, suggesting that the results are not very different.

The first experiment resulted in a 95% confidence interval of 3.070-3.164 g for the population mean weight of newborn girls. The second experiment had a 95% confidence interval of 3.035-3.127 g. Although the two confidence intervals are not identical, the mean for each experiment is within the confidence interval of the other experiment. This suggests that the results are not very different and there is an appreciable overlap between the two intervals. Therefore, the answer is C. No, because each confidence interval contains the mean of the other confidence interval.

Answer:

P ( X > 5) = 0.0374

Step-by-step explanation:

Given:

n = 12

p = 0.23

Using Binomial distribution formula,

X ~ Binomial ( n = 12, p = 0.23)

[tex]=\frac{n!}{(n-x)! x!}. p^{x} q^{n-x}[/tex]

Substitute for n = 12, p = 0.23, q = 1-0.23 for  x = 6,7,8,9,10,11 and 12

P (X > 5)  = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

P ( X > 5 ) = 0.0285 + 0.007299 + 0.00136 + 0.000181 + 0.0000162 + 1E-6 + 1E-6

P ( X > 5) = 0.0374