In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 775 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses.
1. Welcome a white person ( , )
2. Welcome a Latino ( , )
3. Welcome a Black person ( , )

Answer:

a) Probability of no broken bats in a game P(X=0) = 0.3678

b) Probability of at-least broken bats in a game P(X≥2) = 0.2644

Step-by-step explanation:

we will use Poisson distribution

P(X=x) = e^(-λ) λ^ r/r!

a) Probability of no broken bats in a game P(X=0) = e^-1

                                                                                   = 0.3678

b) Probability of at-least two broken bats in a game

                                                                     P(X≥2) = 1-([p(x=0)+P(x=1)]

                                                                               = 1- (e^(-1)+e^(-1))

                                                                                  = 1-(0.3678+0.3678)

                                                                   P(X≥2)= 1- 0.7356

                                                                  P(X≥2)= 0.2644

Answer: option 3 is the correct answer.

Step-by-step explanation:

The formula for determining the area of a sector is expressed as

Area = θ/360 × πr²

Where

θ represents the central angle.

r represents the radius of the circle.

π is a constant whose value is 3.14

From the information given,

r = 4 miles

θ = 120°

Area of sector = 120/360 × 3.14 × 4²

= 16.75 square miles

To determine the area of ∆ABC, we would apply the formula,

Area of triangle = 1/2abSinC

From the information given,

a = 4

b = 4

C = 120°

Therefore,

Area = 1/2 × 4 × 4 × Sin120

Area = 8 × Sin120 = 6.92 square miles. Therefore area of the segment is

16.75 - 6.92 = 9.83 square miles

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.

Final answer:

Explanation of expected total times for Carol and the last customer, along with the probability of being last.

Explanation:

(a) Expected total time for Carol to complete her businesses:

Carol will go to the first available teller, who has a service time of 3 minutes (or service rate of 20 per hour).

Hence, the expected total time for Carol to complete her businesses is 3 minutes.

(b) Expected total time until the last customer leaves:

Calculating the expected total time for each customer to complete their businesses gives Anne: 3 minutes, Betty: 6 minutes.

Therefore, the expected total time until the last customer leaves is 6 minutes.

(c) Probability of being the last to leave:

The last to leave will be the customer with the highest expected service time, which is Betty (6 minutes).

The probability for Anne, Betty, and Carol to be the last one to leave is: Anne 0, Betty 1, Carol 0.

(a) Expected total amount of time for Carol to complete her business: the expected total amount of time for Carol to complete her business is 2 minutes.

(b) Expected total time until the last of the three customers leaves:the expected total time until the last of the three customers leaves is 8 minutes.

c) Probability for Anne, Betty, and Carol to be the last one to leave:the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Let's solve each part of the problem:

(a) Expected total amount of time for Carol to complete her business:

Since Carol waits for the first available teller, we need to consider the minimum of the service times of the two tellers. Let's denote:

- ( X ) as the service time for the first teller (with mean three minutes).

- ( Y ) as the service time for the second teller (with mean six minutes).

The minimum of two exponentially distributed random variables with means [tex]\( \mu_1 \) and \( \mu_2 \)[/tex] is also exponentially distributed with mean \( \frac{1}[tex]\( \frac{1}[/tex][tex]{\lambda_1 + \lambda_2} \), where \( \lambda_1 \) and \( \lambda_2 \)[/tex] are the rates.

Given:

- The rate for the first teller is [tex]\( \lambda_1 = \frac{1}{3} \)[/tex] per minute.

- The rate for the second teller is [tex]\( \lambda_2 = \frac{1}{6} \)[/tex] per minute.

So, the rate for the minimum service time for Carol is [tex]\( \lambda = \lambda_1 + \lambda_2 = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \)[/tex] per minute.

The expected total amount of time for Carol to complete her business is the reciprocal of the rate, which is [tex]\( \frac{1}{\lambda} = \frac{1}{\frac{1}{2}} = 2 \) minutes.[/tex]

Therefore, the expected total amount of time for Carol to complete her business is 2 minutes.

(b) Expected total time until the last of the three customers leaves:

The total time until the last customer leaves is the maximum of the service times of all three customers. Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time.

Let's denote:

- ( Z ) as the service time for Carol (which we found to be 2 minutes).

- ( W ) as the maximum of the service times for Anne and Betty.

Since ( W ) is the maximum of two exponentially distributed random variables with the same rate, it follows that ( W ) is also exponentially distributed with the same rate.

So, the expected total time until the last of the three customers leaves is the sum of the expected service time for Carol and the expected maximum service time for Anne and Betty, which is [tex]\( 2 + 6 = 8 \)[/tex]minutes.

Therefore, the expected total time until the last of the three customers leaves is 8 minutes.

(c) Probability for Anne, Betty, and Carol to be the last one to leave:

Since Anne and Betty go directly into service, their service times are independent of each other and of Carol's service time. Therefore, each of them has an equal chance of being the last one to leave.

So, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Therefore, the probability for Anne, Betty, and Carol to be the last one to leave is [tex]\( \frac{1}{3} \)[/tex] for each of them.

Answer:

6435 different communities.

Step-by-step explanation:

This is a combination problem,

We need to know how many ways 8 species out of 15 can be selected for the new small island

15C8 = 15!/(15-8)!(8!) = 15!/(8!)(7!) = 6435 different communities.

Answer:

A) The brand of cars driven by Harvard instructors

C) The types of plants growing in yellowstone  

Step-by-step explanation:

Categorical and numerical variables:

A) The brand of cars driven by Harvard instructors

Since the brands cannot be expressed in numerical, it is a categorical variable.

B) The age of randomly selected biology students.

The age of student are expressed in whole numbers or decimals. Thus, it is a numerical data.

C) The types of plants growing in yellowstone

Again, the type of plants cannot be expressed with the help of numerical. Thus, it is a categorical variable.

Variables A ('The brand of cars driven by Harvard instructors') and C ('The types of plants growing in Yellowstone') are categorical. Variable B ('The age of randomly selected biology students') is numerical.

In statistical terms, variables are characteristics or properties that can take different values or categories. They can be either categorical (qualitative) or numerical (quantitative).

For the variables provided:

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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:

Pat's score therfore was the 20th percentiles

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

In this problem, we have that:

34 scores lower than Pat's.

[tex]\frac{34}{170} = 0.20[/tex]

So Patrick's score was the 20th percentile.

Answer:

1. 90% 2. 10% 3. 50%

Step-by-step explanation:

Standard Deviation (σ) = 50 days

Average/Mean (μ) = 300days

Probability that it would last more than 300 days = P(Bulb>300 days)

We will assume there are 365 days in a year.

P(Bulb>300 days)  implies that the bulb would

Using the normal equation;

z = standard/normal score = (x-μ)/σ where x is the value to be standardized

P(Bulb>300 days) implies x = 365 days

Therefore z = (365-300)/50 = 1.3

Using the normal graph for z=1.3, probability = 90%

2. P(Bulb<300 days) = 1 - P(Bulb>300 days)\

P(Bulb<300 days) = 1 - 0.9

P(Bulb<300 days)  = 10%

3. P(Bulb=300 days) implies z=0 since x=300

Using the normal graph for z=0, probability =50%

1. The probability that an Acme light bulb will last more than 300 days is 50%. 2. The probability that an Acme light bulb will last less than 300 days is 50%. 3. The probability that an Acme light bulb will last exactly 300 days is zero.

1. To find the probability that an Acme light bulb will last more than 300 days, we need to determine the area under the normal distribution curve to the right of 300 days. We use the z-score formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Substituting the values, we get: z = (300 - 300) / 50 = 0. The area to the right of 300 days is equal to the area to the left of 0. Using a standard normal distribution table, we find that the area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last more than 300 days is 0.5 or 50%.

2. To find the probability that an Acme light bulb will last less than 300 days, we need to determine the area under the normal distribution curve to the left of 300 days. Using the same z-score formula, we get: z = (300 - 300) / 50 = 0. The area to the left of 0 is 0.5. Therefore, the probability that an Acme light bulb will last less than 300 days is 0.5 or 50%.

3. To find the probability that an Acme light bulb will last exactly 300 days, we need to determine the area under the normal distribution curve at 300 days. Since the normal distribution is continuous, the probability of any single value is zero. Therefore, the probability that an Acme light bulb will last exactly 300 days is zero.

Answer: 804.25cm

Step-by-step explanation:

V=πr2h/3

pi= 3.14

r=8cm

h= 12cm

Slot the values

3.14× (8x8) × 12/3

3.14 × 64 x 4

V= 804.25cm

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: 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.

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].

Answer:

The null hypothesis was rejected.

Conclusion: The bartender uses incorrect proportions in more than 50% of margaritas.

Step-by-step explanation:

The hypothesis for this test can be defined as:

H₀: The bartender uses incorrect proportions in less than 50% of margaritas, i.e. p < 0.50.

Hₐ: The bartender uses incorrect proportions in more than 50% of margaritas, i.e. p > 0.50.

Given:

[tex]\hat p=\frac{10}{30} =0.33\\n=30[/tex]

The test statistic is:

[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}} }=\frac{0.33-0.50}{\sqrt{\frac{0.50(1-0.50)}{30}}} =-1.862[/tex]

Decision Rule:

If the p-value of the test statistic is less than the significance level, α = 0.05, then the null hypothesis is rejected.

The p-value of the test is:

[tex]p-value=P(Z<-1.86) = 0.0314[/tex]

*Use the z-table.

The p-value = 0.0314 < α = 0.05.

The null hypothesis will be rejected.

Conclusion:

As the null hypothesis was rejected it can be concluded that the bartender uses incorrect proportions in more than 50% of margaritas.

Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.

At the null hypothesis, we test if the manager's suspicion is incorrect, that is, the proportion is correct in at least 50% of the margaritas.

[tex]H_0: p \geq 0.5[/tex]

At the alternative hypothesis, we test if the suspicion is correct, that is, the proportion is less than 50%.

[tex]H_1: p < 0.5[/tex]

The test statistic is given by:

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

In which:

For this problem, the parameters are: [tex]p = 0.5, n = 30, \overline{p} = \frac{10}{30} = 0.3333[/tex]

Then, the value of the test statistic is:

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

[tex]z = \frac{0.3333 - 0.5}{\sqrt{\frac{0.5(0.5)}{30}}}[/tex]

[tex]z = -1.83[/tex]

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05, is [tex]z^{\ast} = -1.645[/tex].

Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the manager's suspicion is correct.

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Answer:

(a) The total number of combinations that can be applied for making potato chips is 36.

(b) The number of combinations that will be used for each type of oil is 12.

(c) Permutations are not an issue because order does not matter.

Step-by-step explanation:

The effects of cooking temperature, cooking time and cooking oil is studied for making potato chips.

The number of different temperatures applied is, n (T) = 3.

The number of different times taken is, n (t) = 4.

The number of different oils used is, n (O) = 3.

If an assignment can be done in n₁ ways and if for this assignment another assignment can be done in n₂ ways then these two assignments can be performed in (n₁ × n₂) ways.

(a)

Compute the total number of combinations that can be applied for making potato chips as follows:

Total number of combinations for making chips = n (T) × n (t) × n (O)

                                                                               [tex]=3\times4\times 3\\=36[/tex]

Thus, the total number of combinations that can be applied for making potato chips is 36.

(b)

To make potato chips 4 different temperatures are used and 3 different oils are used.

Each of the oil type is cooked in 4 different temperatures.

So the number of ways to select each oil type is,

n (T) × n (O) = [tex]4\times3=12[/tex]

Thus, the number of combinations that will be used for each type of oil is 12.

(c)

Permutation is the arrangement of objects in a specified order.

Since in this case ordering of the the three effects, i.e. temperature. time and oil type is not important, permutations are not an issue.

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:

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:

Since it's a removable discontinuity, we will remove the discontinuity by creating a new function defined by x=3;

So we have;

F(x) = {[x² - 2x - 3]/(x-3); x ≠ 3

         {4, x = 3

Step-by-step explanation:

I have attached my explanation as the system here is not allowing me to save my answer.

To remove the discontinuity at x=3 in the function f(x) = (x^2 - 2x - 3) / (x - 3), simplify the function to f(x) = x + 1, then calculate f(3) = 3 + 1 = 4. Defining f(3) = 4 makes the function continuous at x=3.

To remove the discontinuity of a function such as f(x) = (x^2 - 2x - 3) / (x - 3), you need to find a value for f(3) that would make the function continuous at x = 3. First, let's manipulate the function:

The function f(x) = (x^2 - 2x - 3) / (x - 3) simplifies to f(x) = (x - 3)(x + 1) / (x - 3).

As you can see, the (x - 3) terms cancel out, leaving f(x) = x + 1.

However, this is only valid for x ≠ 3. To find f(3), you can now simply substitute 3 into the simplified function:

f(3) = 3 + 1 = 4.

So, if we define f(3) = 4, then the function becomes continuous at x = 3.

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Answer:

343 pieces of wood.

Step-by-step explanation:

In order to determine the starting number of pieces of wood that yield 330 railings, the easiest course of action is to work your way back through step 3, step 2, and step 1. The number of pieces that must arrive at each step are given by:

[tex]n_f=330\\n_3 = \frac{330}{1-0.016}\\n_2 = \frac{n_3}{1-0.011}\\n_1 = \frac{n_2}{1-0.011} \\\\n_3 = 335.3658\\n_2=339.09591\\n_1=342.86745=343[/tex]

Rounding up to the next whole piece, 343 pieces are needed.

Therefore, you must start with 343 pieces of wood to have 330 railings that can be sold.

Part(a):

Then the outcomes can be,

[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]

Part(b):

The correct option is (a).

Part(c):

The outcomes are,

[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]

Experimental probability, also known as Empirical probability, is based on actual experiments and adequate recordings of the happening of events.

Let the ordered pair (a,b,c) denote the outcome with a,b,c taking either 1 if the position is offered

Or 0  if the position is not offered.

Then the outcomes can be,

[tex]\{(1,1,1)(1,1,0)(1,0,1)(0,1,1)(1,0,0)(0,1,0)(0,0,1)(0,0,0) \}[/tex]

Let, [tex]x[/tex] is number probability function offers made.

The variable is discrete taking 0 or 1 or 2 or 3 as values.

So, the correct option is (a)

The outcomes are,

[tex](1,1,1):x=3\\(1,1,0):x=2\\(1,0,1):x=2\\(0,1,1):x=2\\(1,0,0):x=1\\(0,1,0):x=1[/tex]

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There are 8 experimental outcomes for the interviews. The variable representing the number of offers is discrete, taking values between 0 and 3. For the given subsets of experimental outcomes, the value of this variable is the number of 'Y' present.

A) The total number of experimental outcomes is calculated by the rule of product. There are two possible results (Yes or No) for each of the three interviews. So the number of experimental outcomes is 2*2*2, which is 8.

B) The variable x, which is the number of students receiving an offer, is discrete. A variable is discrete if it can only take on a finite or countable number of values. In this case, x can take on only four possible values (0, 1, 2, or 3), depending on the number of students receiving an offer.

C) The value of the random variable X for the subset of experimental outcomes is as follows:
(Y,Y,Y) - X = 3
(Y,N,Y) - X = 2
(N,Y,Y) - X = 2
(N,N,Y) - X = 1
(N,N,N) - X = 0

The above experimental outcomes show the different possible results of the three interviews, with N signifying no job offer and Y signifying a job offer after the interview.

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Answer:

5.96%

7.6%

38 days

Step-by-step explanation:

15 months = 1.25 years

5138.5 = 4780 × R^1.25

R^1.25 = 43/40

1.25 lgR = lg(43/40)

lg R = 0.0251267714

R = 1.059562969

R - 1 = 0.059562969

Interest: 5.96%

(150/365) × (r/100) × 12000 = 375

r/100 = 0.0760416667

r = 7.6%

40 = (n/365) × (6/100) × 6400

(n/365) = 5/48

n = 38.02083333 = 38

Answer:

α= 133.6 degrees

(a)Sin(α/2)=0.9191

(b)cos(α/2)=0.3939

(c)Tan(α/2)=2.3332

Step-by-step explanation:

If Tan α= [tex]-\frac{21}{20}[/tex]

90<α<180

We determine first the value of α in the first quadrant

α=[tex]Tan^{-1}\frac{21}{20}[/tex]

=46.4

Since 90<α<180

α=180-46.4=133.6 degrees

(a)Sin(α/2)=Sin(133.6/2)=Sin 66.8 =0.9191

(b)cos(α/2)=cos(133.6/2)=cos 66.8 =0.3939

(c)Tan(α/2)=Tan(133.6/2)=Tan 66.8 =2.3332

Answer:

6/8

Step-by-step explanation:

The Fraction of mowers that fail the functional performance test is the ratio of the number of mowers that fail to the total number of mowers. Hence the fraction of mowers that fail [tex] \frac{9}{500}[/tex]

The probability of failure can be defined thus :

Therefore, the probability of having x failures in the next 100 mowers is (0.018x)

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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:

a) Total 16 possibilities

MMMM

FFFF

MMMF

MMFM

MFMM

FMMM

FFFM

FFMF

FMFF

MFFF

MMFF

MFMF

MFFM

FFMM

FMMF

FMFM

b) P(MMMM) = 1/16

Final answer:

The equally likely gender combinations for 4 children are listed by considering all possibilities, including MMMM, MMMF, and so on. The probability that all 4 children are male is 0.0625, or 1/16 as a fraction. The probability of having at least one female child is the complement, 0.9375 or 15/16 as a fraction.

Explanation:

Listing the Equally Likely Gender Combinations

To list the equally likely events for the gender of the 4 children in a family where 'M' represents a male child and 'F' represents a female child, consider all possible combinations. These combinations are: MMMM, MMMF, MMFM, MMFF, MFMM, MFMF, MFFM, MFFF, FMMM, FMMF, FMFM, FMFF, FFMM, FFMF, FFMM, FFFF.

Probability of All Male Children

To calculate the probability that all 4 children are male, note that the events are independent, and the probability of each child being male is 0.5. Since the events are independent, multiply the probabilities for each child: 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 or 1/16 as a fraction.

Alternatively, the complement of having all male children is having at least one female child. The probability of at least one female child can be found by subtracting the probability of all male children from 1: 1 - 0.0625 = 0.9375 or 15/16 as a fraction.

Answer: the due date would be 92 weeks

Step-by-step explanation:

Since the time required to complete a project is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = number of weeks.

µ = mean

σ = standard deviation

From the information given,

µ = 80 weeks

σ = 10 weeks

If a construction company bidding on this contract wishes to be 90 percent sure of finishing by the due date, the z score corresponding to 90%(90/100 = 0.9) is 1.29

Therefore,

1.29 = (x - 80)/10

x - 80 = 1.2 × 10

x - 80 = 12

x = 80 + 12 = 92

To determine the due date the construction company should negotiate for a 90 percent confidence level of finishing on time, we use the z-score formula with the mean and standard deviation.

The negotiated due date is approximately 93 project week .

To determine the due date the construction company should negotiate, we need to find the project week number that corresponds to being 90 percent sure of finishing by that time. To do this, we use the z-score formula to find the z-score associated with a 90 percent confidence level. We then use this z-score to find the corresponding project week number using the mean and standard deviation of the project time.

The z-score formula is: z = (X - μ) / σ, where X is the project week number, μ is the mean (80 weeks), and σ is the standard deviation (10 weeks).

By substituting the values into the formula, we get: z = (X - 80) / 10.

Next, using a z-table or a calculator, we find the z-score associated with a 90 percent confidence level, which is approximately 1.2816.

Substituting this value for z in the formula, we get: 1.2816 = (X - 80) / 10.

Now, we can solve for X by multiplying both sides of the equation by 10 and then adding 80 to both sides. This gives us: 12.816 = X - 80.

Finally, adding 80 to both sides of the equation, we find that the negotiated due date (project week #) should be approximately 92.816. However, since project week numbers are typically whole numbers, we can round the negotiated due date up to the nearest whole number, which is 93.

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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:

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:

The First-In, First-Out (FIFO) inventory costing method assumes that the inventory items ordered first are the first ones sold.

Step-by-step explanation:

The First-In, First-Out inventory costing method assumes that the inventory items ordered first are the first sold. This is ideal for goods that are highly perishable, for example fresh milk. Since no figures or dates are given, we will assume that the month is March 2019 and use any figures to make the example.

Date Item      Quantity of stock Cost Price

01  Opening stock bought on Feb 28  10   100

05  Sale of 5 goods (cost is $10 each)  (5)   50

15 Purchase of stock (20 goods at $20 each) 20   400

25 Sale of 15 goods                     (15)   250  

(5 at $10 each & 10 at $20 each)

31 Closing Stock               10   200

       (20 goods bought on 15th - 10 goods sold on 25th)

The quantity on hand at the end of the month is 10 units.  

Total cost of goods on hand at end of the month = 10 units * $20 = 200.

Total cost of goods purchased during the month = $20 * 20 units = $400

Total cost of goods sold during the month = [($10 *5) + ($10 * 5)+ ($20 * 10)] = $200