In the beginning of the study, a randomly selected group of 121 students scored an average of 252 words per minute on the reading speed test. Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M

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) 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:

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:

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:

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.

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:

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: 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: 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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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:

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.

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:

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:

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: 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) 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:

  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:

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: 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: 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

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

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].

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.

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:

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

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 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.

A similar problem is given at https://brainly.com/question/23265899

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]

Learn more about the topic experimental outcomes:

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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:

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