A building inspector believes that the percentage of new construction with serious code violations may be even greater than the previously claimed 7%. She conducts a hypothesis test on 200 new homes and finds 23 with serious code violations. Is this strong evidence against the .07 claim?

John Cacioppo's study uses independent samples while the car cleanliness study uses matched subjects because the samples are dependent as couples are paired.

In two given scenarios, the research designs differ in whether they use related samples or not. Related samples refer to a study design where the same subjects are measured more than once, or their measures are compared to those of individuals with whom they share a distinct relationship (e.g., married couples).

In John Cacioppo's study, the research design uses independent samples. Participants were randomly assigned to either the group of lonely people or nonlonely people and the similarities or differences in sleep quality were observed. As there is no mention of matching based on particular characteristics, and individuals from one group are not directly linked to those in the other, these samples are not related.

In contrast, the car cleanliness study deals with matched subjects, as the samples are paired together based on marriage. The husbands' ratings are compared directly with their respective wives' ratings, giving rise to related, paired measurements within each couple.

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The correct answer for the first scenario is that the design uses a related sample with matched subjects. For the second scenario, the design uses a related sample with repeated measures.

Explanation for the first scenario:

In the first scenario, John Cacioppo is comparing the sleep quality of lonely people to nonlonely people. Since the researcher is comparing two different groups of people based on their loneliness status, this is an example of a matched subjects design. The subjects are matched on the characteristic of loneliness, meaning that each lonely individual is paired with a nonlonely individual who is similar in other relevant characteristics that could affect sleep quality, such as age, gender, or health status. This allows the researcher to control for these extraneous variables and focus on the effect of loneliness on sleep quality.

Explanation for the second scenario:

In the second scenario, the researcher is interested in the difference in car cleanliness importance ratings between husbands and wives. By collecting data from both partners within each of the 100 married couples, the researcher is using a repeated measures design. Each couple serves as their own control, with the husband and wife providing two sets of measurements (ratings) on the same variable (importance of car cleanliness). This design allows for the examination of within-subject differences, controlling for the influence of other variables that might differ between different families, such as socioeconomic status or cultural background.

In summary, the first scenario is an example of a related sample with matched subjects, while the second scenario is an example of a related sample with repeated measures.

Answer:

Step-by-step explanation:

The detailed steps and analysis is as shown in the attached file.

To write the system of equations in matrix form, use p = [R, F] and p' = [1.5R - 0.2F + 100]. To find the vector of rabbits and foxes after time V, integrate p' twice. To find the vector of rabbits and foxes after time a, integrate p' once. To find the vector of rabbits and foxes after 2 years, substitute n=2 and use summation notation.

(a) To write the system of equations in matrix form, we have:

R' = 1.5R - 0.2F + 100

We can represent the variables R and F as a vector p, so that p = [R, F]. The derivative of p with respect to time (p') is then given by:

p' = [1.5R - 0.2F + 100]

(b) To find the vector of rabbits and foxes after time V, we can integrate the equation given in part (a) with respect to time twice:

p'' = ∫[1.5R - 0.2F + 100]dt

(c) To find the vector of rabbits and foxes after time a, we can integrate the equation given in part (a) with respect to time:

p(a) = ∫[1.5R - 0.2F + 100]dt

(d) To find the vector of rabbits and foxes after 2 years, we can substitute n=2 into the equation given in part (c) and use summation notation:

p(2) = ∫[1.5R - 0.2F + 100]dt + ∑[∫[1.5R - 0.2F + 100]dt]^n

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Answer:

a) 75%

b) 84%

c) 95%                                                    

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 6.7 hours

Standard Deviation, σ = 1.8 hours

Chebyshev's Theorem:

Empirical Formula:

a) minimum percentage of individuals who sleep between 3.1 and 10.3 hours

[tex]10.3 = 6.7 + 2(3.1) = \mu + 2(\sigma)\\3.1 = 6.7 - 2(3.1) = \mu - 2(\sigma)[/tex]

Minimum percentage:

[tex]1-\dfrac{1}{4} = 75%[/tex]

Thus, minimum 75% of individuals who sleep between 3.1 and 10.3 hours.

b) minimum percentage of individuals who sleep between 2.2 and 11.2 hours.

[tex]11.2 = 6.7 + 2.5(3.1) = \mu + 2.5(\sigma)\\2.2 = 6.7 - 2.5(3.1) = \mu - 2.5(\sigma)[/tex]

Minimum percentage:

[tex]1-\dfrac{1}{(2.5)^2} = 84\%[/tex]

Thus, minimum 84% of individuals who sleep between 2.2 and 11.2 hours.

c) percentage of individuals who sleep between 3.1 and 10.3 hours per day

According to Empirical rule about 95% of the data lies within 2 standard deviations from the mean.

Thus, 95% of individuals who sleep between 3.1 and 10.3 hours.

d) Comparison with Chebyshev's theorem

This is greater than the results obtained from the Chebyshev's theorem in part (a)

According to Chebyshev's theorem, the minimum percentage of individuals who sleep between 3.1 and 10.3 hours is approximately 93.75%. This minimum percentage also applies to individuals who sleep between 2.2 and 11.2 hours. However, using the empirical rule, approximately 95% of individuals are expected to sleep between 3.1 and 10.3 hours per day. The result from the empirical rule suggests that the distribution of sleep hours is closer to a normal distribution than what Chebyshev's theorem predicts.

(a) According to Chebyshev's theorem, at least 75% of the individuals will sleep between 3.1 and 10.3 hours. The formula for Chebyshev's theorem is P(X - μ < kσ), where μ is the mean, σ is the standard deviation, and k is the number of standard deviations from the mean.

(b) Using the same steps as in part (a), the minimum percentage of individuals who sleep between 2.2 and 11.2 hours is also approximately 93.75%.

(c) According to the empirical rule, for a normal distribution, approximately 68% of the individuals will sleep between μ - σ and μ + σ, and approximately 95% will sleep between μ - 2σ and μ + 2σ. Since the interval 3.1 and 10.3 hours falls within 2 standard deviations from the mean, we can estimate that around 95% of the individuals will sleep between 3.1 and 10.3 hours per day.

The result obtained using the empirical rule in part (c) is slightly higher than the value obtained using Chebyshev's theorem in part (a), indicating that the distribution of sleep hours is closer to a normal distribution than a general distribution predicted by Chebyshev's theorem.

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Answer: The disadvantage of using a related sample is that "A study that uses related samples to compare two drugs (specifically, one sample of participants with repeated measures) can have a carryover and/or order effect such that the efects of the drug taken before the first measurement may not wear off before the second measurement".

Step-by-step explanation: Related sample is when a particular sample or two sample that are the same is used to study an effect.

The carryover effect is one of the major disadvantages of using a related sample. Because when a first treatment is applied, their is a tendency of it not being fully consumed or it's effect not being fully neutralized before the second treatment is applied. This will increase error in the result, because in a drugs intake for instance, the drug taken at a particular time may not be the one that cure a sickness, but the drug that was previously taken. But the study will assume the drug taken a that particular moment to be the cure of that sickness.

Answer:

Related samples (especially, one sample of Participants with repeated measures) can have a carryover effect such that Participants can leam from their first measurement and therefore do better on their second Measurement.

Explanation:

Related samples/groups (i.e., dependent measurements) The subjects in each specimen, or organization, are identical. This indicates that the subjects in the first group are also in the second group.

Various Disadvantages of using a related sample versus using two independent samples:

Thus we can say using a related sample versus using two independent samples has various disadvanages.

To learn more about related sample, refer:

Answer:

D) No, because either np or n(1−p) are less than 15.

Step-by-step explanation:

Percentage of students who had tattoos = 36%

Percentage of students who were smokers = 4%

Sample size = n = 100

The condition to use the Normal distribution as an approximation to construct the confidence interval for population proportion is:

Both np and n(1-p) must be equal to or greater than 15.

Since, we are interested in smokers only, so p = 4% = 0.04

np = 100  x 0.04 = 4

n (1 - p) = 100 x 0.96 = 96

Since, np < 15, we cannot use the Normal distribution as an approximation here.

Therefore, the correct answer is:

No, because either np or n(1−p) are less than 15.

Answer:

  g(n) = 0.05·2^n

Step-by-step explanation:

The paper with no folds is 0.05 mm thick. Each fold multiplies the thickness by 2, so the function is ...

  g(n) = 0.05·2^n

_____

Comment on paper folding

In practice, where the paper must bend around the fold, it is impossible to fold an ordinary piece of paper 9 times. You may be able to fold a very large, very thin piece of paper that many times.

The thickness of a standard piece of paper, after it has been folded n times, can be expressed as g(n) = 0.05 * 2^n. This formula takes into account that each fold doubles the thickness of the paper.

In order to calculate the thickness of a standard piece of paper after it has been folded a certain number of times, we may use the concept of exponential growth. In this case, we use a power of 2 corresponding to the number of folds made, because with each fold, the thickness doubles. This gives us the function g(n) = 0.05 * 2^n, where n corresponds to the number of folds.

For example, if we fold the paper once (n=1), we get a thickness of 0.05 * 2^1 = 0.10 mm. If we fold the paper twice (n=2), we get a thickness of 0.05 * 2^2 = 0.20 mm, and so on, with the thickness doubling each time we make a new fold.

This mathematical model assumes perfect folding and does not account for physical limitations we would encounter when attempting to fold paper in reality.

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Answer:

B

Step-by-step explanation:

The data doesn't follow any linearity. Infact, the given data set is in a range of values. It is distributed around some mean value. So it is plausible that it is normally distributed

The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible.

The question asks whether it is plausible that the compressive strength for this type of concrete is normally distributed. In order to answer this question, we can examine the normal probability plot. If the plot is acceptably linear, it suggests that a normal population distribution is plausible. On the other hand, if the plot is not acceptably linear, it suggests that a normal population distribution is not plausible.

According to the options provided, the correct answer would be C. The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible. This implies that the compressive strength for this type of concrete is not normally distributed.

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Final answer:

Relation R defines a specific type of relationship within set theory, where subsets of set A are related if one is a proper subset of the other within the power set of A.

Explanation:

The student's question pertains to the concept of a relation in set theory, particularly concerning the power set of a given finite non-empty set A. In this context, the domain for the relation R is the power set of A, which includes all subsets of A. The relation R is defined such that if X and Y are subsets of A, then X is related to Y if and only if X is a proper subset of Y, denoted as X ⊂ Y. This means that every element in subset X is also contained in subset Y, and Y contains at least one additional element that is not in X.

To illustrate this, consider a simple set A = {1, 2}. Its power set, which is the domain of R, will include { }, {1}, {2}, and {1, 2}. In this case, the set {1} is a proper subset of {1, 2}, since it contains all elements of {1} and A contains an additional element, which is 2. Hence, the ordered pair ({1}, {1, 2}) is part of the relation R.

Answer: his investment will be worth $1606.5 after 3 years

Step-by-step explanation:

We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as

S = R[{(1 + r)^n - 1)}/r][1 + r]

Where

S represents the future value of the investment.

R represents the regular payments made(could be weekly, monthly)

r = represents interest rate/number of interval payments.

n represents the total number of payments made.

From the information given,

R = $250

r = 0.04/2 = 0.002

n = 2 × 3 = 6

Therefore,

S = 250[{(1 + 0.02)^6 - 1)}/0.02][1 + 0.02]

S = 250[{(1.02)^6 - 1)}/0.02][1.02]

S = 250[{(1.126 - 1)}/0.02][1.02]

S = 250[{0.126}/0.02][1.02]

S = 250[6.3][1.02]

S = 250 × 6.426

S = $1606.5

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 79% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 100 found that 89% of the readers owned a personal computer.

Let Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 0.79  {means that the percentage is actually less than or equal to the reported percentage}

Alternate Hypothesis, [tex]H_1[/tex] : p > 0.79  {means that the percentage is actually more than the reported percentage}

The test statics that will be used here is One-sample proportions test;

         T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = % of the readers who owned a personal computer in a sample of 100 = 89%

           n = sample size = 100

So, test statistics = [tex]\frac{0.89 -0.79}{\sqrt{\frac{0.89(1-0.89)}{100} } }[/tex]

                             = 3.196

Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that percentage is actually more than the reported percentage which means we have sufficient evidence at the 0.02 level to support the executive's claim.

Answer:

0.8151 is the probability that the weight will be less than 4293 grams.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4004 grams

Variance = 103,684

[tex]\sigma = \sqrt{103684} = 322[/tex]

We are given that the distribution of weight of newborn baby is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(weight will be less than 4293 grams)

P(x < 4293)

[tex]P( x < 4293) \\\\= P( z < \displaystyle\frac{4293 - 4004}{322}) \\\\= P(z < 0.8975)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 4293) =0.8151 = 81.51\%[/tex]

0.8151 is the probability that the weight will be less than 4293 grams.

Answer:

Probability that the weight will be less than 4293 grams is 0.8133.

Step-by-step explanation:

We are given that weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684.

Let X = weight of newborn baby boys

So, X ~ N([tex]\mu =4004,\sigma^{2}=322^{2}[/tex])

The z score probability distribution is given by;

               Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)  

where, [tex]\mu[/tex] = population mean

           [tex]\sigma[/tex] = population standard deviation

(a) Probability that weight will be less than 4293 grams is given by = P(X < 4293 grams)

    P(X < 4293) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{4293-4004}{322 }[/tex] ) = P(Z < 0.89) = 0.8133

Therefore, if a newborn baby boy born at the local hospital is randomly selected, probability that the weight will be less than 4293 grams is 0.8133.

Answer:

49 cars

Step-by-step explanation:

Probability of cars to be sold in a week,a = 0.2

Probabiity of cars not sold in a week, b = 0.8

Number of cars estimated to be sold in a week = 20 and 60 cars in 3 weeks

Using, P(x) = nCx *(a)∧x * (b)∧n - x, where n = 3 weeks, x = 1 week

P(x=1) = 3C1 * (0.2) * (0.8)² = 3 X 0.2 X 0.64 X 60 cars = 23 cars

P(x=2) = 3C2 * (0.2)² * (0.8) = 3 X 0.04 X 0.8 X 60 cars =  6 cars

Number of cars three weeks from now: 20 + 23 + 6 = 49 cars

Answer:

The forecast of the number of cars 3 weeks from now is 52 cars.

Step-by-step explanation:

As per the trent the number of the cars per week is 5 cars

The current level of cars is 40 cars per week

Number of cars sold in current week=20 cars

Forecast of the cars sold 3 weeks from now is given as

[tex]L_t=\alpha Y_t+(1-\alpha)(L_{t-1}+T_{t-1})\\[/tex]

From the data

Y_t=20 cars

L_t-1=40 cars

T_t-1=5 cars

α=β=0.2

So the equation becomes

[tex]L_t=\alpha Y_t+(1-\alpha)(L_{t-1}+T_{t-1})\\L_t=0.2*20+(1-0.2)(40+5)\\L_t=40[/tex]

Now the trend is calculated as

[tex]T_t=\beta(L_{t}-L_{t-1})+(1-\beta)T_{t-1}[/tex]

By putting the values the equation becomes

[tex]T_t=\beta(L_{t}-L_{t-1})+(1-\beta)T_{t-1}\\T_t=0.2(40-40)+(1-0.2)5\\T_t=0+0.8*5\\T_t=4[/tex]

Now the forecast of the cars sale 3 weeks from now is given as

[tex]L_{t+k}=L_t+kT_t[/tex]

where k is 3 so

[tex]L_{t+k}=L_t+kT_t\\L_{t+3}=40+3*4\\L_{t+3}=40+12\\L_{t+3}=52\\[/tex]

So the forecast of the number of cars 3 weeks from now is 52 cars.

Answer:

y=1/2x+2

Step-by-step explanation:

Answer: y = 1/2x + 2

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the graph,

y2 = 4

y1 = 2

x2 = 4

x1 = 0

Slope,m = (4 - 2)/(4 - 0) = 2/4 = 1/2

To determine the y intercept, we would substitute x = 4, y = 4 and

m = 1/2 into y = mx + c. It becomes

4 = 1/2 × 4 + c

4 = 2 + c

c = 4 - 2 = 2

The equation becomes

y = x/2 + 2

Answer:

Test statistic = 3.1587

Step-by-step explanation:

We are given that the Toy-lot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A).

Also, a sample of eleven motors was tested, and it was found that the mean current was x = 1.20 A, with a sample standard deviation of s = 0.42 A.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 0.8 { claim of 0.8 A is not low}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 0.8 { claim of 0.8 A is too low}

Now, the test statistics used here will be;

           T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, X bar = sample mean = 1.20 A

                 s = sample standard deviation = 0.42 A

                 n = sample size = 11 motors

So, Test statistics = [tex]\frac{1.20 - 0.8}{\frac{0.42}{\sqrt{11} } }[/tex] ~ [tex]t_1_0[/tex]

                             = 3.1587

At 1% level of significance, t table gives a critical value of 2.764 at 10 degree of freedom. Since our test statistics is higher than the critical value so we have sufficient evidence to reject null hypothesis .

Therefore, we conclude that Toy-lot claim of 0.8 A is too low.

To determine if the Toylot company claim of 0.8 A is too low, a one-sample t-test can be used. The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, divided by the sample standard deviation divided by the square root of the sample size. This statistic can then be compared to the critical value from the t-distribution at the specified level of significance to determine if the claim is too low.

To determine if the Toylot company claim of 0.8 A is too low, we can conduct a hypothesis test using the sample data. We will use a one-sample t-test with a null hypothesis that the mean current is equal to 0.8 A. The alternative hypothesis will be that the mean current is greater than 0.8 A.

The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, and dividing by the sample standard deviation divided by the square root of the sample size. In this case, the sample test statistic is (1.20 - 0.8) / (0.42 / sqrt(11)) = 5.11.

To determine if this test statistic is statistically significant, we compare it to the critical value from the t-distribution with 10 degrees of freedom at a 1% level of significance. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the Toylot claim of 0.8 A is too low.

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Answer: increase the sample size

Step-by-step explanation: the margin of error for any confidence interval is given by the formulae below.

Margin of error = critical value × standard deviation/√n

From what we can see the critical value and standard deviation are constants, the only variables here are the margin of error and sample size which are inversely proportional to each other, that is the margin of error is inversely proportional to the square of the sample size.

Hence, reducing the sample size will increase the margin of error while increasing the sample size will reduce the margin of error.

Answer:

a) v ∈ ker(L) if only if [tex][V]_{E}[/tex] ∈ N(A)

b) w ∈ L(v) if and only if [tex][W]_{F}[/tex] is in the column space of A

See attached

Step-by-step explanation:

See attached the proof Considering the vector spaces V and W with other bases E and F respectively.

Let L be the Linear transformation form V and W and A is the matrix representing L relative to E and F

Answer:

a) 16.7% probability that both dice show the same number

b) 83.3% probability that both dice show different numbers

c) 41.67% probability that the second die lands on a lower value than does the first.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have these possible outcomes:

Format(Dice A, Dice B)

(1,1), (1,2), (1,3), (1,4), (1,5),(1,6)

(2,1), (2,2), (2,3), (2,4), (2,5),(2,6)

(3,1), (3,2), (3,3), (3,4), (3,5),(3,6)

(4,1), (4,2), (4,3), (4,4), (4,5),(4,6)

(5,1), (5,2), (5,3), (5,4), (5,5),(5,6)

(6,1), (6,2), (6,3), (6,4), (6,5),(6,6)

There are 36 possible outcomes.

(a) What is the probability that both dice show the same number?

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

6 outcomes in which both dice show the same number.

6/36 = 0.167

16.7% probability that both dice show the same number

(b) What is the probability that both dice show different numbers?

The other 30 outcomes

30/36 = 0.833

83.3% probability that both dice show different numbers

(c) What is the probability that the second die lands on a lower value than does the first?

(2,1)

(3,1), (3,2)

(4,1), (4,2), (4,3)

(5,1), (5,2), (5,3), (5,4)

(6,1), (6,2), (6,3), (6,4), (6,5)

15 outcomes in which the second die lands on a lower value than does the first.

15/36 = 0.4167

41.67% probability that the second die lands on a lower value than does the first.

Answer:

Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.

Step-by-step explanation:

We use the central limit theorem to solve this question.

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a sample size larger than 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

So

Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.

Answer: FIRST OPTION.

Step-by-step explanation:

First, it is important to remember that the Slope-Intercept form of the equation of a line is the shown below:

[tex]y=mx+b[/tex]

Where "m" is the slope of the line and "b" is the y-intercept.

By definiton, given a System of Linear equations, if they are exactly the same line, then the System of equations have Infinely many solutions.

In this case you have the following System of Linear equations given in the exercise:

[tex]\left \{ {{y=-2x+5} \atop {y=ax+b}} \right.[/tex]

So, since you need the system has Infinite solutions, you know that the slope and the y-intercept of both lines must be equal.

Therefore, you can identify that the value of "a" and "b" must be the following:

[tex]a=-2\\\\b=5[/tex]

So the Linear System would be the shown below:

[tex]\left \{ {{y=-2x+5} \atop {y=-2x+5}} \right.[/tex]

Answer:

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Step-by-step explanation:

For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent from other bulbs, 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.

Suppose 30% of the bulbs in the lot are defective.

This means that [tex]p = 0.3[/tex]

A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.

This means that [tex]n = 16[/tex]

What is the probability that exactly 4 bulbs from the sample are defective?

This is P(X = 4).

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

[tex]P(X = 4) = C_{16,4}.(0.3)^{4}.(0.7)^{12} = 0.2040[\tex]

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Answer:

(a) The 5-hour decay factor is 0.5042.

(b) The 1-hour decay factor is 0.8720.

(c) The amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

Step-by-step explanation:

The amount of caffeine in Chase's body decreases exponentially.

The 10-hour decay factor for the number of mg of caffeine is 0.2542.

The 1-hour decay factor is:

[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]

(a)

Compute the 5-hour decay factor as follows:

[tex]5-hour\ decay\ factor=(0.8720)^{5}\\=0.504176\\\approx0.5042[/tex]

Thus, the 5-hour decay factor is 0.5042.

(b)

The 1-hour decay factor is:

[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]

Thus, the 1-hour decay factor is 0.8720.

(c)

The equation to compute the amount of caffeine in Chase's body is:

A = Initial amount × (0.8720)ⁿ

It is provided that initially Chase had 171 mg of caffeine, 1.39 hours after consuming the drink.

Compute the amount of caffeine in Chase's body 2.39 hours after consuming the drink as follows:

[tex]A = Initial\ amount \times (0.8720)^{2.39} \\=[Initial\ amount \times (0.8720)^{1.39}] \times(0.8720)\\=171\times 0.8720\\=149.112[/tex]

Thus, the amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

The 5-hour growth/decay factor is 0.561, the 1-hour growth/decay factor is 0.869, and there are 154 mg of caffeine in Chase's body 2.39 hours after consuming the energy drink.

a. To find the 5-hour growth/decay factor, we raise the 10-hour decay factor to the power of (5/10). So, the 5-hour growth/decay factor is 0.2542^(5/10) = 0.561.

b. To find the 1-hour growth/decay factor, we raise the 10-hour decay factor to the power of (1/10). So, the 1-hour growth/decay factor is 0.2542^(1/10) = 0.869.

c. To find the number of mg of caffeine in Chase's body 2.39 hours after consuming the energy drink, we multiply the initial amount of caffeine by the 2.39-hour decay factor. So, the number of mg is 171 * 0.2542^(2.39/10) = 154.

Answer:

0.0273

Step-by-step explanation:

np  n

10  100

9    100

11   100

7    100

3   100

12  100

8   100

4    100

6    100

11   100

pbar=sumnp/sumn

pbar=10+9+11+7+3+12+8+4+6+11/10+10+10+10+10+10+10+10+10+10

pbar=81/1000

pbar=0.081

nbar=sumn/k=1000/10=100

[tex]Standard deviation for pbar chart=\sqrt{\frac{pbar(1-pbar)}{nbar} }[/tex]

[tex]Standard deviation for pbar chart=\sqrt{\frac{0.081(0.919)}{100} }[/tex]

[tex]Standard deviation for pbar chart=\sqrt{\frac{0.0744}{100} }[/tex]

[tex]Standard deviation for pbar chart=\sqrt{0.0007444 }[/tex]

Standard deviation for p-chart=0.0273

Answer:

Hence the values for MOE and are interval estimate of 19 yr old age drivers in 1983  are 1%(0.01) and 19±0.0196

Step-by-step explanation:

Given: Refer the paper university of Michigan transportation research  Institute website ,April,2012.

It say that 19 year old group age people ,about 87% has driving license .

And of 1200 has a random sample space ,in 1983.

To Find : MOE(margin of error ) and   Interval Estimate  in 1983.

Solution:

Given that ,there is sample space of 1200 19 yr old people  and of which 87% has driving license.

hence , we get that

87% of 1200 = sample  size.

sample size =1044 members has driving license out of 1200.

Consider 95 % of confidence level,

for that Z-score is required.

calculating the Alpha for that=1-confidence level.

=1-0.95=0.05

therefore Z-alpha=Z(0.05)=1.96

1)MOE=margin of error is given by ,

=[tex]Z-alpha*\sqrt{\frac{p(1-p)}{N} }[/tex]

here p=0.87 and N=total size =1200.

MOE=1.96*[tex]\sqrt{\frac{0.87(1-0.87)}{1200} }[/tex]  =1 %.

2) Interval estimate is given by ,

For that we should know mean,standard deviation and sample size,

we are calculating for 19 year old age of entire age of population,

Hence mean will be 19 yr-old age.

Mean=19.

Sample size=1044.

standard deviation given by,

=[tex]p*(\sqrt{(1-p)}[/tex]

=0.87*[tex]\sqrt{0.13}[/tex]=0.3136.

Hence now calculating the interval estimate ,with 95% confidence level,

μ=M±Z(Standard error )

standard error=standard deviation/sqrt(sample size)

=0.3136/32.31=0.01.

μ=19±1.96*0.01

μ=19±0.0196.

The experimental design used in this assessment is a pretest-posttest design where the difference between pre- and posttest scores is calculated.

The experimental design used in this assessment is a pretest-posttest design. A pretest is administered before the students receive the intervention (in this case, the math course), and a posttest is administered after the intervention. The difference between the pre- and posttest scores is then calculated to determine the knowledge-gained score.

This design allows for the comparison of students' scores before and after the math course, giving insight into the effectiveness of the course in improving their knowledge. By comparing the two test scores, the professor can assess how much knowledge students have gained as a result of the course.

Example:

Before the math course, a student's pretest score is 50. After completing the course, the student's posttest score is 75. The knowledge-gained score is calculated as 75 - 50 = 25. This indicates that the student gained 25 points of knowledge between the pre- and posttests.

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The probability that [tex]\( R_2 \)[/tex] exceeds[tex]\( R_1 \)[/tex] by more than 30Ω in series connection is approximately 0.04%, calculated using their normal distributions' means, standard deviations, and the difference's Z-score.

To find the probability that[tex]\( R_2 \)[/tex] exceeds [tex]\( R_1 \)[/tex] by more than 30Ω when they are connected in series, we can use the properties of normal distributions and their differences.

Given:

[tex]\( R_1 \)[/tex] has a mean [tex]\( \mu_1 = 1000 \)[/tex] and standard deviation [tex]\( \sigma_1 = 50 \).[/tex]

[tex]\( R_2 \)[/tex] has a mean[tex]\( \mu_2 = 1202 \)[/tex] and standard deviation [tex]\( \sigma_2 = 10.2 \)[/tex].

The difference [tex]\( R_2 - R_1 \)[/tex] will follow a normal distribution with the mean[tex]\( \mu_2 - \mu_1 = 1202 - 1000 = 202 \)[/tex] and the standard deviation [tex]\( \sqrt{(\sigma_1)^2 + (\sigma_2)^2} = \sqrt{50^2 + 10.2^2} \).[/tex]

Now, we want to find the probability that [tex]\( R_2 - R_1 > 30 \).[/tex]

Using Z-score:

[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

[tex]\[ Z = \frac{30 - 202}{\sqrt{50^2 + 10.2^2}} \][/tex]

[tex]\[ Z = \frac{-172}{\sqrt{2500 + 104.04}} \][/tex]

[tex]\[ Z = \frac{-172}{\sqrt{2604.04}} \][/tex]

Z ≈ -172 / 51.02 ≈ -3.37

Consulting a standard normal distribution table or calculator for the probability associated with[tex]\( Z = -3.37 \)[/tex] we find that the probability is approximately 0.0004 or 0.04%.

Therefore, the probability that \( R_2 \) exceeds \( R_1 \) by more than 30Ω when they are connected in series is approximately 0.04%.

The probability [tex]\( P(R_2 > R_1) \approx 0.9631 \)[/tex], found by standardizing and using the normal distribution table.

To compute [tex]\( P(R_2 > R_1) \)[/tex], we need to find the probability that the resistance [tex]\( R_2 \)[/tex] is greater than [tex]\( R_1 \)[/tex]. Given that [tex]\( R_1 \) and \( R_2 \)[/tex] are normally distributed and independent, we can use properties of the normal distribution.

First, we define the random variables:

- [tex]\( R_1 \sim N(100, 5^2) \)[/tex]

- [tex]\( R_2 \sim N(120, 10^2) \)[/tex]

We are interested in the distribution of [tex]\( R_2 - R_1 \)[/tex].

1. Find the mean and standard deviation of [tex]\( R_2 - R_1 \)[/tex]:

  - The mean of [tex]\( R_2 - R_1 \)[/tex]:

    [tex]\[ \mu_{R_2 - R_1} = \mu_{R_2} - \mu_{R_1} = 120 - 100 = 20 \][/tex]

  - The variance of [tex]\( R_2 - R_1 \)[/tex]:

    [tex]\[ \sigma_{R_2 - R_1}^2 = \sigma_{R_2}^2 + \sigma_{R_1}^2 = 10^2 + 5^2 = 100 + 25 = 125 \][/tex]

  - The standard deviation of [tex]\( R_2 - R_1 \)[/tex]:

    [tex]\[ \sigma_{R_2 - R_1} = \sqrt{125} = 5\sqrt{5} \][/tex]

2. Standardize the variable [tex]\( R_2 - R_1 \)[/tex]:

  We want to find [tex]\( P(R_2 > R_1) \)[/tex], which is equivalent to [tex]\( P(R_2 - R_1 > 0) \)[/tex].

  Standardize [tex]\( R_2 - R_1 \)[/tex] by subtracting the mean and dividing by the standard deviation:

  [tex]\[ Z = \frac{R_2 - R_1 - \mu_{R_2 - R_1}}{\sigma_{R_2 - R_1}} = \frac{R_2 - R_1 - 20}{5\sqrt{5}} \][/tex]

  We want to find:

  [tex]\[ P(R_2 - R_1 > 0) = P\left(\frac{R_2 - R_1 - 20}{5\sqrt{5}} > \frac{0 - 20}{5\sqrt{5}}\right) \][/tex]

  Simplify the right-hand side:

  [tex]\[ P\left(Z > \frac{-20}{5\sqrt{5}}\right) = P\left(Z > \frac{-20}{5 \cdot 2.2361}\right) = P\left(Z > \frac{-20}{11.1803}\right) = P\left(Z > -1.7889\right) \][/tex]

3. Find the probability:

  - Using standard normal distribution tables or a calculator, find the probability [tex]\( P(Z > -1.7889) \)[/tex].

  - The standard normal distribution table provides the cumulative probability for [tex]\( Z \leq z \)[/tex]. For [tex]\( Z > -1.7889 \)[/tex]:

    [tex]\[ P(Z > -1.7889) = 1 - P(Z \leq -1.7889) \][/tex]

  Using standard normal distribution tables or a calculator:

  [tex]\[ P(Z \leq -1.7889) \approx 0.0369 \][/tex]

  Therefore:

  [tex]\[ P(Z > -1.7889) = 1 - 0.0369 = 0.9631 \][/tex]

So, the probability that [tex]\( R_2 \)[/tex] is greater than [tex]\( R_1 \)[/tex] is approximately [tex]\( 0.9631 \)[/tex].

Final answer:

The probability of a student making more than 20 mistakes on the exam is 0.19, making 6 or more mistakes is 0.72, and making at most 20 mistakes is 0.81. Events making at most 20 mistakes and making more than 20 mistakes are complementary.

Explanation:

Given a driving exam consisting of 30 multiple-choice questions, we have the following probabilities:

Now, we can find the probabilities for each scenario using these probabilities:

The two complementary events are:

These events are complementary because their probabilities sum up to 1.

Answer:

The correct option is option C max 10A +15B +7C+8D

Step-by-step explanation:

As the complete question is not given here thus the complete question is found online and is attached herewith.

From the data the profit for 1 unit of A is 10,  B is 15, C is 7 and D is 8, so the profit function is given as 10A+15B+7C+8D. and as profit is to be maximized so the correct option is option C.

Answer:

1) (f + g) (5) = 18, 2) (f + g) (1/2) = - 13/4, 3) (f - g) (-1) = 14, 4) (f * g) (7) = 456, 5) (f/g) (- 2) = - 3/4, 6) (g ○ f)(x) = 6 x - 23, (f ○ g)(x) = 6 x - 6, 7) [tex](g \circ f) (x) = \frac{9}{x} - 24[/tex], [tex](f \circ g) (x) = \frac{1}{x - 2} + 5[/tex], 8) (f ○ g)(5) = 3

Step-by-step explanation:

1) [tex](f + g) (x) = x^{2}-8\cdot x + 3 + 7 \cdot x - 5\\(f + g) (x) = x^{2} - x - 2\\(f + g) (5) = 18[/tex]

2) [tex](f + g) (x) = x^{2} - 7 \cdot x + 4 + 6 \cdot x - 7\\(f + g) (x) = x^{2} - x - 3\\(f + g) (\frac{1}{2} ) = - \frac{13}{4}[/tex]

3) [tex](f - g) (x) = x^{2} - 3 \cdot x + 4 - 4 \cdot x + 2\\(f - g) (x) = x^{2} - 7 \cdot x + 6\\(f - g) (-1) = 14[/tex]

4) [tex](f \cdot g) (x) = (x^{2}-4\cdot x + 3) \cdot (3\cdot x - 2)\\(f \cdot g) (7) = 456[/tex]

5) [tex](f / g) (x) = \frac{x^{2}-3\cdot x + 2}{4 \cdot x - 8} \\(f / g) (-2) = - \frac{3}{4}[/tex]

6) [tex](g \circ f) (x) = 3 \cdot (2 \cdot x - 8) + 1\\(g \circ f) (x) = 6 \cdot x - 23\\(f \circ g) (x) = 2 \cdot (3 \cdot x + 1) - 8\\(f \circ g) (x) = 6 \cdot x - 6[/tex]

7) [tex](g \circ f) (x) = 3 \cdot (\frac{3}{x} - 6) - 6\\(g \circ f) (x) = \frac{9}{x} - 24\\(f \circ g) (x) = \frac{3}{3 \cdot x - 6} + 5\\(f \circ g) (x) = \frac{1}{x - 2} + 5[/tex]

8) [tex](f \circ g) (x) = 2 \cdot (x^{2} - 5 \cdot x) + 3\\(f \circ g) (x) = 2 \cdot x ^{2} - 10 \cdot x + 3\\(f \circ g) (5) = 3[/tex]

1. (f + g)(5) = 18.

2. (f + g)(1/2) = -13/4.

3. (f - g)(-1) = 14.

4. (fg)(7) = 456.

5. (f/g)(-2) = -3/4.

6. (g ○ f)(x) = 6x - 23, (f ○ g)(x) = 6x - 6.

7. (g ○ f)(x) = [tex]\(\frac{9}{x+5} - 6\), (f ○ g)(x) = \(\frac{3}{3x-1}\).[/tex]

8. (f ○ g)(5) = 3.

Let's tackle each problem step by step:

1.Evaluate (f + g)(5):

  Step 1: Find f(5) and g(5).

 [tex]\(f(5) = (5)^2 - 8(5) + 3 = 25 - 40 + 3 = -12\)[/tex]

[tex]\(g(5) = 7(5) - 5 = 35 - 5 = 30\)[/tex]

  Step 2: Add f(5) and g(5).

 [tex]\((f + g)(5) = f(5) + g(5) = -12 + 30 = 18\)[/tex]

2. Evaluate (f + g)(1/2):

  Step 1: Find f(1/2) and g(1/2).

 [tex]\(f(1/2) = (1/2)^2 - 7(1/2) + 4 = 1/4 - 7/2 + 4 = 1/4 - 14/4 + 16/4 = 3/4\)[/tex]

  [tex]\(g(1/2) = 6(1/2) - 7 = 3 - 7 = -4\)[/tex]

  Step 2: Add f(1/2) and g(1/2).

[tex]\((f + g)(1/2) = f(1/2) + g(1/2) = 3/4 - 4 = -13/4\)[/tex]

3.Evaluate (f − g)(−1):

  Step 1: Find f(−1) and g(−1).

 [tex]\(f(-1) = (-1)^2 - 3(-1) + 4 = 1 + 3 + 4 = 8\)[/tex]

  [tex]\(g(-1) = 4(-1) - 2 = -4 - 2 = -6\)[/tex]

  Step 2: Subtract g(-1) from f(-1).

  [tex]\((f - g)(-1) = f(-1) - g(-1) = 8 - (-6) = 8 + 6 = 14\)[/tex]

4.Evaluate (fg)(7):

  Step 1: Find f(7) and g(7).

[tex]\(f(7) = (7)^2 - 4(7) + 3 = 49 - 28 + 3 = 24\)[/tex]

  [tex]\(g(7) = 3(7) - 2 = 21 - 2 = 19\)[/tex]

  Step 2: Multiply f(7) by g(7).

 [tex]\((fg)(7) = f(7) \times g(7) = 24 \times 19 = 456\)[/tex]

5. Evaluate (f/g)(−2):

  Step 1: Find f(-2) and g(-2).

[tex]\(f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12\)[/tex]

  [tex]\(g(-2) = 4(-2) - 8 = -8 - 8 = -16\)[/tex]

  Step 2: Divide f(-2) by g(-2).

  [tex]\((f/g)(-2) = \frac{f(-2)}{g(-2)} = \frac{12}{-16} = -\frac{3}{4}\)[/tex]

6. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 2x − 8, g(x) = 3x + 1:

 [tex]\((g ○ f)(x) = g(f(x)) = g(2x - 8) = 3(2x - 8) + 1 = 6x - 24 + 1 = 6x - 23\)[/tex]

 [tex]\((f ○ g)(x) = f(g(x)) = f(3x + 1) = 2(3x + 1) - 8 = 6x + 2 - 8 = 6x - 6\)[/tex]

7. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 3/x+5, g(x) = 3x − 6:

 [tex]\((g ○ f)(x) = g(f(x)) = g(\frac{3}{x+5}) = 3(\frac{3}{x+5}) - 6 = \frac{9}{x+5} - 6\)[/tex]

 [tex]\((f ○ g)(x) = f(g(x)) = f(3x - 6) = \frac{3}{3x-6+5} = \frac{3}{3x-1}\)[/tex]

8. Evaluate (f ○ g)(5):

  Step 1: Find g(5).

  [tex]\(g(5) = 5^2 - 5 \times 5 = 25 - 25 = 0\)[/tex]

  Step 2: Find f(g(5)).

 [tex]\(f(g(5)) = f(0) = 2 \times 0 + 3 = 3\)[/tex]

Answer:

a) L(x,y,z) = (-4,0,2)+(2,8,-7)*t

b) (2-z)/7= y/8=(x+4)/2 (option B)

Step-by-step explanation:

the parametric equation of the line passing through the point P₀= (-4,0,2) and parallel to the vector v=2i + 8j - 7k is

L(x,y,z)=P₀+v*t

therefore

L(x,y,z) = (-4,0,2)+(2,8,-7)*t

or

x=x₀+vx*t = -4 + 2*t

y=y₀+vy*t = 8*t

z=z₀+vz*t = 2 -7*t

solving for t in the 3 equations we get the symmetric equation of the line:

(2-z)/7= y/8=(x+4)/2

thus the option B is correct

Answer:

a) Parametric equations

x = -4 + 2t

y = 8t

z = 2-7t

b) symmetric equations

[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex]          The answer is the option B

Step-by-step explanation:

For writing the vectorial equation of a line, we need a point in the line and its director vector, thus:

[tex]L: (x_{0},y_{0},z_{0} ) + t(a,b,c)[/tex]

                            Where   [tex](x_{0},y_{0},z_{0})[/tex] is a point in the line

                                           (a,b,c) is the director vector

Then

[tex]L: (-4,0,2) + t(2,8,-7)[/tex]

a) Parametric equations

Since the vectorial equation, we can obtain the parametric equations writing the equation for each component

x = -4 + 2t

y = 8t

z = 2-7t

b)Symmetric equations

Since the parametric equations, we isolate the parameter t

[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex]