Given the system of equations
(D+5)x+D2y=cost
D2x+(D+9)y=t2+3t
(a) Write the system as a differentialequation for x , in operator form, by filling thepolynomial operator in D (on the left-hand-sideof the equation) and the corresponding right-hand-side of theequation:
x=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)

(b) Note that we could equally well solve fory. Do this here, obtaining an equivalent differentialequation for y.
y=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)

Answer:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

The estimated proportion is [tex] \hat p =0.5[/tex]. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

Answer:

Avis Company is a car rental company that is located three miles from the Los Angeles airport (LAX). Avis is dispatching a bus from its offices to the airport every 2 minutes. The average traveling time (round-trip) is 20 minutes.

a. How many Avis buses are traveling to and from the airport?

b. The branch manager wants to improve the service and suggests dispatching buses every 0.5 minutes. She argues that this will reduce the average traveling time from the airport to Avis offices to 2.5 minutes. Is she correct? If your answer is negative, then what will the average traveling time be?

The answers to the question are

a. 10 buses

b. Yes, increasing the fleet by a factor of 4 will reduce the apparent average travel time from the airport to Avid offices from 10 minutes to 2.5 minutes

Step-by-step explanation:

Number of buses dispatched every 20 minutes= 20÷2=10

Therefore 10 Avid buses are travelling to and from the Los Angeles airport

b. By dispatching every 0.5 minutes we have 20÷0.5 = 40 buses.

Since with 10 buses it takes 10 minutes to make the journey from the airport to Avid offices, then it would take 40 buses 10÷4 = 2.5 minutes time to deliver what it took 10 buses, 10 minutes to deliver.

The Avis Company has 10 buses traveling to and from LAX airport with an average round trip time of 20 minutes and a dispatch frequency of 2 minutes. Increasing the dispatch frequency to every 0.5 minutes does not reduce the average travel time; it remains at 20 minutes as it is dependent on the route's distance and traffic conditions, not the dispatch frequency.

Part A: Number of Avis Buses

To calculate the number of buses traveling to and from the airport, we use the average traveling time for a round trip and the dispatch frequency. With a round trip time of 20 minutes and buses leaving every 2 minutes, we divide the round trip time by the dispatch interval:

Total Buses = Round Trip Time / Dispatch Frequency

Total Buses = 20 minutes / 2 minutes

Total Buses = 10

Therefore, Avis has 10 buses traveling to and from the airport.

Part B: Average Traveling Time with Increased Dispatch Frequency

Increasing the dispatch frequency to every 0.5 minutes does not affect the average traveling time for a round trip. The travel time is determined by the distance between the Avis office and the airport and the traffic conditions, not merely the frequency of bus dispatches. If Avis dispatches buses every 0.5 minutes, there will be more buses in operation, but the average round trip time remains the same at 20 minutes, assuming no changes in traffic conditions or route speeds.

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

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

In this problem, we have that:

[tex]\mu = 98.6, \sigma = 0.6, n = 36, s = \frac{0.6}{\sqrt{36}} = 0.1[/tex]

If 36 adults are randomly​ selected, find the probability that their mean body temperature is greater than 98.4degrees° F.

This is 1 subtracted by the pvalue of Z when X = 98.4. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{98.4 - 98.6}{0.1}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Answer:

We conclude that 4095 different teams can be selected.

Step-by-step explanation:

We know that a company has 24 employees: 10 women and 14 men.

A team of 4 employees must be selected. The team must have two women and two men.

So from 10 women we choose 2 women:

[tex]C_2^{10}=\frac{10!}{2!(10-2)!}=45\\[/tex]

So from 14 men we choose 2 men:

[tex]C_2^{14}=\frac{14!}{2!(14-2)!}=91\\[/tex]

We get: 45 · 91 = 4095.

We conclude that 4095 different teams can be selected.

Final answer:

The total number of different teams that can be selected from the department to represent at the companywide meeting is 4095.

Explanation:

To select a team of 4 employees with 2 women and 2 men from a department with 10 women and 14 men, we can use the concept of combinations. We need to choose 2 women out of 10 and 2 men out of 14.

The number of ways to choose 2 women from 10 is given by 10C2 = 10! / (2! * (10-2)!) = 45.

The number of ways to choose 2 men from 14 is given by 14C2 = 14! / (2! * (14-2)!) = 91.

Therefore, the total number of different teams that can be selected is 45 * 91 = 4095.

Answer:

Step-by-step explanation:

The first two answers could not possibly be correct.

"Skewed to the right" implies that the mean is larger than the median.  This is the correct answer.

The mean is much larger than the median is the statement that would indicate that a data set is skewed to the right. This can be obtained by understanding the characteristics of a data set skewed to the right, that is positively skewed.

⇒Characteristics of a positively skewness

Hence the mean is much larger than the median is the statement that would indicate that a data set is skewed to the right. Therefore option 3 is correct.

Learn more about skewness here:

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Answer:

95.4% of family vehicles is between 1 and 3 years old.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2

Standard Deviation, σ = 6 months = 0.5 year

We are given that the distribution of age of cars is a bell shaped distribution that is a normal distribution.

Formula:

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

P(family vehicles is between 1 and 3 years old)

[tex]P(1 \leq x \leq 3)\\\\ = P(\displaystyle\frac{1 - 2}{0.5} \leq z \leq \displaystyle\frac{3-2}{0.5}) = P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.977 -0.023 = 0.954= 95.4\%[/tex]

[tex]P(1 \leq x \leq 3) = 95.4%[/tex]

95.4% of family vehicles is between 1 and 3 years old.

Final answer:

Approximately 95.4% of family vehicles are between 1 and 3 years old based on the normal distribution with a mean of 2 years and a standard deviation of 6 months.

Explanation:

To determine the percentage of family vehicles that are between 1 and 3 years old when the age is normally distributed with a mean of 2 years and a standard deviation of 6 months, we can use the properties of the normal distribution.

First, convert the years to a z-score, which is the number of standard deviations away from the mean a data point is. The formula to calculate a z-score is: z = (X - μ) / σ

Where X is the value, μ is the mean, and σ is the standard deviation.

For 1 year: z = (1 - 2) / 0.5 = -2

For 3 years: z = (3 - 2) / 0.5 = 2

By looking up these z-scores on a z-table or using a calculator with a normal distribution function, we find that roughly 95.4% of the data falls between z-scores of -2 and 2. Therefore, approximately 95.4% of family vehicles are between 1 and 3 years old.

Answer:

[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]  

[tex]p_v =P(z>1.01)=0.156[/tex]  

So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

Step-by-step explanation:

Data given and notation

n=195 represent the random sample taken

X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy

[tex]\hat p=\frac{65}{195}=0.333[/tex] estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

[tex]p_o=0.3[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:  

Null hypothesis:[tex]p\leq 0.3[/tex]  

Alternative hypothesis:[tex]p > 0.3[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>1.01)=0.156[/tex]  

So the p value obtained was a very low value and using the significance level asumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.

To find the p-value for testing the hypothesis that the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman, we can use the binomial probability formula.

The formula is:

P(X ≥ k) = 1 - P(X < k)

where X is the number of women in the erythromycin group who complain of nausea, and k is the number of complaints of nausea in the control group. In this case, X = 65 and k = 0, because the question specifies that there are no complaints of nausea in a typical pregnant woman. By plugging these values into the formula, we can calculate the p-value.

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The initial reactant concentration is 0.150 M. Using the first order kinetics equation, we converted the time to seconds and plugged in the given values for initial concentration, time and rate constant. The final concentration of the reactant after 19.0 minutes is 4.92 x 10^-5 M.

The initial concentration of the reactant is 0.150 M and the rate constant, k = 8.70×10−3 s−1. We use the formula for first order kinetics, [A] = [A]0 * e^-kt, where [A] is the final concentration, [A]0 is the initial concentration, k is the rate constant and t is the time.

First, convert the time from minutes to seconds (since the rate constant is in s^-1). Therefore, t = 19 min * 60 s/min = 1140 s.

Then, substitute the given values into the equation: [A] = 0.150 M * e^-((8.70 x 10^-3 s^-1) * (1140 s)) = 0.150 M * e^-9.918 = 4.92 x 10^-5 M.

So, the concentration of the reactant after 19.0 minutes will be 4.92 x 10^-5 M.

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Part A: Concentration ≈ 0.186 MM after 8.00 minutes.

Part B: Initial concentration ≈ 5.90 MM for zero-order reaction.

Part A:

Using the first-order integrated rate law equation:

[tex]\[ [A] = [A]_0 \times e^{-kt} \][/tex]

[tex]\[ [A] = 0.650 \times e^{-(3.70 \times 10^{-3} \times 8 \times 60)} \][/tex]

[tex]\[ [A] \approx 0.186 \, MM \][/tex]

Part B:

Using the zero-order integrated rate law equation:

[tex]\[ [A] = -kt + [A]_0 \][/tex]

[tex]\[ [A]_0 = -kt + [A] \][/tex]

[tex]\[ [A]_0 = -(3.00 \times 10^{-4} \times 80) + 3.50 \times 10^{-2} \][/tex]

[tex]\[ [A]_0 = 3.50 \times 10^{-2} + 2.40 \times 10^{-2} \][/tex]

[tex]\[ [A]_0 \approx 5.90 \, MM \][/tex]

The correct question is:

Part A

The rate constant for a certain reaction is [tex]$k k k=3.70 \times 10^{-3} \mathrm{~s}-1 \mathrm{~s}-1$[/tex]. If the initial reactant concentration was [tex]$0.650 \mathrm{MM}$[/tex], what will the concentration be after 8.00 minutes?

Part B

A zero-order reaction has a constant rate of [tex]$3.00 \times 10^{-4} \mathrm{M} / \mathrm{sM} / \mathrm{s}$[/tex]. If after 80.0 seconds the concentration has dropped to [tex]$3.50 \times 10^{-2} \mathrm{MM}$[/tex], what was the initial concentration?

Answer:

Step-by-step explanation:

You can find your answer in attached documents.

The student's question involves calculating compensating variation, equivalent variation, and change in consumer surplus due to a price increase in one of two goods, given a quasilinear utility function. These calculations require the application of microeconomic principles related to utility functions, indifference curves, and consumer choices, but cannot be completed without specific mathematical expressions and functions provided in the coursework.

The question relates to a consumer's preference relation and utility maximization in the context of an increase in the price of one of the goods. The compensating variation, equivalent variation, and change in consumer surplus are economic measures used to quantify the impacts of price changes on consumer welfare. These concepts involve understanding indifference curves, marginal rates of substitution, and consumer budget constraints.

Computing the compensating and equivalent variations requires setting up and solving expenditure function problems before and after the price change, whereas the change in consumer surplus can be represented graphically and calculated as the area under the demand curve and above the price level. Unfortunately, without specific details or mathematical expressions for utility maximization and budget constraints, which are typically provided in academic coursework, it is not possible to provide explicit calculations for these measures.

Understanding the utility function's form and how it translates into demand can help derive these economic measures. For instance, the quasilinear utility function is notable for its constant elasticity of demand and specific responses to income and price changes which can be used to predict consumer behavior and welfare changes.

Answer:

DE = 3 units

Step-by-step explanation:

The image is attached.

There are 2 secant lines in the circle. We can use secant theorem to solve this easily.

It states that "if 2 secants are drawn to a circle from an outside point, then product of 1 secant and its "outside" part is equal to product of other secant and its "outside" part.

From the figure, we can say:

AX * BX = EX * DX

We let the length to find , DE, be "x".

Thus, we can write:

[tex]AX * BX = EX * DX\\(7+2)(2)=(x+3)(3)[/tex]

Now, we solve this for x:

[tex](7+2)(2)=(x+3)(3)\\(9)(2)=(3)(x+3)\\18=3x+9\\3x=18-9\\3x=9\\x=3[/tex]

Thus,

DE = 3 units

Answer:

the answer is b 3 units

Step-by-step explanation:

The statement 'All insects have six legs' is converted to 'If x is an insect, then x has six legs' using the principle of Universal Generalization.

The process used here is known as Universal Generalization. This principle allows us to infer that something is true for all objects in a particular domain, provided it has been proved to be true for an arbitrary object in that domain. In this case, the statement 'All insects have six legs' has been converted into a formal logical statement using 'x' as the arbitrary object in the domain of all animals. By stating 'If x is an insect, then x has six legs' and using ∀x (denoting 'for all x'), we are using Universal Generalization to indicate this is true for all members of the domain.

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Answer:

The value of the test statistic is 1.667

Step-by-step explanation:

We are given that the sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. For this a random sample of 64 days of sales was selected. It was found that the average was $8,250 per day. From past information, it is known that the standard deviation of the population is $1,200.

We have to determine whether or not the advertising campaigns have been effective in increasing sales.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = $8,000 {means that the advertising campaigns have not been effective in increasing sales}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > $8,000 {means that the advertising campaigns have been effective in increasing sales}

The test statistics that will be used here is One sample z-test statistics;

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

where, Xbar = sample mean = $8,250

                [tex]\sigma[/tex]  = population standard deviation = $1,200

                 n = sample size = 64

So, test statistics = [tex]\frac{8,250-8,000}{\frac{1,200}{\sqrt{64} } }[/tex]

                            = 1.667

Therefore, the value of test statistics is 1.667 .

Final answer:

The sample space S for a. is a set of all possible denominations of cards; the sample space S for b. is a set of all possible combinations of two cards with color recorded; event H is defined as the first card being hearts in c.; event DC is defined as the first card being diamonds and second card being red in d.; events C and B are disjoint in e.

Explanation:

a. One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?

The sample space S for this scenario is the set of all possible denominations of cards in a standard deck, which includes the numbers 1 through 10, as well as the face cards (J, Q, K) and the Ace (A).

b. Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?

The sample space S for this scenario is the set of all possible combinations of two cards from a standard deck, with the colors (red or black) of each card being recorded. Since there are 26 red cards and 26 black cards, the number of possible outcomes is 26 * 26 = 676.

c. Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?

Event H is defined as the event where the first card selected is a heart. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a heart.

d. Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?

Event DC is defined as the event where the first card selected is a diamond and the second card selected is red. In terms of the sample space, it can be defined as the subset of sample space S that includes all outcomes where the first card is a diamond and the second card is red.

e. Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?

Events C and B are disjoint because the first card cannot be both a club and black at the same time. However, events R and B are not disjoint because there are red clubs in the deck, which would satisfy both events.

The sample spaces for the following are as follows:  a) Hearts, Diamonds Clubs, Spades. b) Red Red,  Red Black, Black Red, Black Black.

c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.

d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.

e) The events C and R are disjoint.

A standard deck of cards has 52 cards, with 26 red and 26 black cards across four suits: hearts, diamonds, clubs, and spades, each containing 13 cards.

a. One card is selected at random and the denomination is recorded.

The sample space for the denomination of a single card is:

Hearts

Diamonds

Clubs

Spades

b. Two cards are selected at random and the color is recorded.

The sample space for the color of two cards selected can be:

Red, Red

Red, Black

Black, Red

Black, Black

c. Two cards are selected at random and the denomination is recorded. T

Event H is defined as drawing a heart for the first card. The possible outcomes for the second card can be any of the four denominations: hearts, diamonds, clubs, or spades. So, the event H sample space is:

Hearts - Hearts

Hearts - Diamonds

Hearts - Clubs

Hearts - Spades

d. Two cards are selected at random and the denomination is recorded.

Event D is defined as drawing a diamond for the first card and a red card for the second card. Since a red card can only be a diamond or heart, event DC (complement of event D) includes any draw combinations not fitting this pattern:

Diamond - Black

Black - Black

Black - Red

Hearts - Any

e. Two cards are selected at random.

Events C (first card is clubs) and R (first card is red) are disjoint since a card cannot simultaneously be clubs (black) and red. However, either of these events can be paired with event B (second card is black). Therefore, events C and R are disjoint.

The sample spaces for the following are as follows:  a) Hearts, Diamonds Clubs, Spades. b) Red Red,  Red Black, Black Red, Black Black.

c) Hearts - Hearts, Hearts - Diamonds, Hearts - Clubs, Hearts - Spades.

d) Diamond - Black, Black - Black, Black - Red, Hearts - Any.

e) The events C and R are disjoint.

Answer:

73.30% probability that the sample mean score will be within 10 of the population mean

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

Central limit theorem:

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

In this problem, we have that:

[tex]\mu = 510, \sigma = 90, n = 100, s = \frac{90}{\sqrt{100}} = 9[/tex]

What is the probability that the sample mean score will be (a) within 10 of the population mean

This is the pvalue of Z when X = 510 + 10 = 520 subtracted by the pvalue of Z when X = 510 - 10 = 500.

X = 520

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

By the Central Limit Theorem

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

[tex]Z = \frac{520 - 510}{9}[/tex]

[tex]Z = 1.11[/tex]

[tex]Z = 1.11[/tex] has a pvalue of 0.8665

X = 500

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

[tex]Z = \frac{500 - 510}{9}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sample mean score will be within 10 of the population mean

Answer:

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

Based on the defnitions above we can conclude that the best answer for this case is:

B. rejecting the null hypothesis when it is true.

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

Based on the defnitions above we can conclude that the best answer for this case is:

B. rejecting the null hypothesis when it is true.

Answer:

The probability that the result is positive is P=0.04475=4.475%.

Step-by-step explanation:

We have the events:

D: disease present

ND: disease not present

P: test positive

F: test false

Then, the information we have is:

P(D)=0.005

P(P | D)=0.99

P(P | ND)=0.04

The total amount of positive test are the sum of the positive when the disease is present and the false positives (positive tests when the disease is not present).

[tex]P(P)=P(P | D)*P(D)+P(P | ND)*(1-P(D))\\\\P(P)=0.99*0.005+0.04*0.995\\\\P(P)=0.00495+0.0398=0.04475[/tex]

The probability that the result is positive is P=0.04475.

Final answer:

The overall probability of a diagnostic test delivering a positive result, given its sensitivity and false positive rate, alongside the prevalence of the disease in the population, is calculated to be 4.475%.

Explanation:

The question revolves around calculating the probability that a diagnostic test for a given disease is positive. Given that the disease is present in 0.5% of the population, the test has a 99% sensitivity (true positive rate) and a 4% false positive rate (when the disease is not present, the test incorrectly indicates disease 4% of the time).

Steps to Calculate the Probability of a Positive Test Result

This calculation shows that the overall probability of getting a positive test result, regardless of actually having the disease, is 4.475%.

the length of the diagonal of the blue tablecloth is five times the length of the diagonal of the red tablecloth. So , Aaron is not correct .

Step-by-step explanation:

Here we have , The red tablecloth has a diagonal of V 10 feet. The blue tablecloth has a diagonal  V 50 feet .  Aaron says that the length of the diagonal of the blue tablecloth is three  times the length of the diagonal of the red tablecloth. Let's find out what Aaron says is correct or not :

Length of red tablecloth = 10 feet = p feet

Length of blue tablecloth = 50 feet

⇒ Length of blue tablecloth = 50 feet

⇒ Length of blue tablecloth = 5(10) feet

⇒ Length of blue tablecloth = 5p feet

That means the length of the diagonal of the blue tablecloth is five times the length of the diagonal of the red tablecloth. So , Aaron is not correct .

Aaron's claim that the length of the diagonal of the blue tablecloth is three times that of the red one is incorrect because the square root of 50 is not equal to three times the square root of 10.

No, Aaron is not correct. The length of the diagonal of the blue tablecloth is not three times the length of the diagonal of the red tablecloth. This is because the square root of 50 (√50) is not equal to three times the square root of 10 (3 * √10). In reality, the square root of 50 is approximately 7.07 and three times the square root of 10 is approximately 9.49.

Indeed:√50 ≈ 7.07 and 3 * √10 ≈ 9.49. These are not the same, hence Aaron's claim is incorrect.

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Answer:

31

Step-by-step explanation:

116 - 27 -27 = 62 / 2 = 31

Answer:

The length is 31 in

Step-by-step explanation:

Perimeter = 2(Length + width)

say p = perimeter, l = length and w = width

perimeter = 166, w = 27

166 = 2(l  + 27)

116 = 2l + 2×27

2l + 54 = 116

2l = 116 - 54

2l = 62

l = 62/2

l = 31 in

say

Answer:

x = 1   ,or   x = -7

Obviously, the true answer is not in the options given.

Step-by-step explanation:

x² + 6x - 6 = 10

The above equation can not be factorized, hence the use of Almighty Formula.

x = [-b +- √b² - 4ac] / 2a

Where a = 1, b = 6, c = -6

x = [-6 +- √6² - (4*1*-6)] / 2*1

x = [-6 +- √36 - (-24)] / 2

x = [-6 +- √36 + 24] / 2

x = [-6 +- √60] / 2

x = [-6 +- 7.75] / 2

x = [-6 + 7.75] / 2   ,or   x = [-6 - 7.75] / 2

x = 1.75/2   ,or   x = -13.75/2

x = 0.875   ,or   x = -6.875

Approximately

x = 1   ,or   x = -7

Answer:D

Step-by-step explanation:

Answer:C

Step-by-step explanation:

Argument against the person, abusive as the US government should completely eradicate dictatorship and also curb the incident of abusive things from the dictatorship.

Answer:

This is No Fallacy ( A )

Step-by-step explanation:

This is a no Fallacy statement because it embodies the truth in its argument against dictatorship and its evil deeds/effects on the society that is been ruled by a dictator.

while a fallacy is an argument that is based on invalid reasoning or unjust reasoning while creating an argument in other to win a judgement.in most cases Fallacious argument deceive the audience into believing that the argument is true and passing blind judgement on the subject matter , but with the argument found in the question every part of the argument is a valid reason for the U.S to eradicate dictatorship.

Answer:

Moving average MSE =1.235

Exponential smoothing MSE =3.555

forecast for next month =(83+84+83)/3=83.33

Step-by-step explanation:

a)

Select your answer - Plot (iii)

What type of pattern exists in the data? -Horizontal Pattern

b)

for Moving Average MSE =1.235

for Exponential smoothing MSE =3.555

Moving Average is better

c)

forecast for next month =(83+84+83)/3=83.33

It can be deduced from the graph that the type of pattern that exists in the data is a horizontal pattern.

From the complete question, when comparing the three-month moving average forecast with an exponential smoothing forecast, the forecast for moving average is 1.235 while that of exponential smoothing is 3.555. Therefore, moving average is better.

In conclusion, the forecast for next month will be:

= (83 + 84 + 83)/3

= 83.33

Learn more about graphs on:

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Answer:

The probability that exactly 2 of the next four customers order turkey entries is 6 * 0.0393359 = 0.2360

Step-by-step explanation:

There are a total of 8+12+10 = 30 entries, 8 of them being turkey. Lets compute the probability that the first 2 customers order turkey entries and the remaining 2 do not.

For the first customer, he has 8 turkey entries out of 30, so the probability for him to pick a turkey entry is 8/30. For the second one there will be 29 entries left, 7 of them being turkey. So the probability that he picks a turkey entry is 7/29. The third one has 28 entries left, 6 of them which are turkey and 22 that are not. The probability that he picks a non turkey entry is 22/28. And for the last one, the probability that he picks a non turkey entry is 21/27. So the probability of this specific event is

8/30*7/29*22/28*21/27 = 0.0393359

Any other event based on a permutation of this event will have equal probability (we are just swapping the order of the numerators when we permute). The total number of permutations is 6, and since each permutation has equal probability, then the probability that exactly 2 of the next four customers order turkey entries is 6 * 0.0393359 = 0.2360

Answer:

Correct option: As the sample size decreases, the margin of error increases.

Step-by-step explanation:

The (1 - α) % confidence interval for population proportion is:

[tex]CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]

The margin of error in this confidence interval is:

[tex]\\ MOE=z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]

The sample size n is inversely related to the margin of error.

An inverse relationship implies that when one increases the other decreases and vice versa.

In case of MOE also, when n is increased the MOE decreases and when n is decreased the MOE increases.

Compute the new margin of error for n = 125 as follows:

[tex]\\ MOE=z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }=1.96\times \sqrt{\frac{0.40(1-0.40)}{125} }=0.086[/tex]

*Use z-table for the critical value.

For n = 125 the MOE is 0.086.

And for n = 275 the MOE was 0.058.

Thus, as the sample size decreases, the margin of error increases.

To approximate the probability, we can use the normal approximation to the binomial distribution. We need to calculate the mean and standard deviation, and then use the normal distribution to find the cumulative probability for each number of errors. Finally, we add up the probabilities to get the probability of at least 2 errors.

To approximate the probability, we can use the normal approximation to the binomial distribution. We know that there are 50 pages with errors out of 1000, so the probability of success is p = 50/1000 = 0.05. The probability of failure is q = 1 - p = 1 - 0.05 = 0.95.

The sample size is 100 pages. To find the probability that at least 2 of the pages in error are in the sample, we need to find the probability of getting 2 errors, 3 errors, 4 errors, and so on, up to 100 errors.

We can use the normal approximation to the binomial distribution by calculating the mean and standard deviation. The mean is given by np = 100 * 0.05 = 5, and the standard deviation is given by sqrt(npq) = sqrt(100 * 0.05 * 0.95) ≈ 2.179.

Next, we can use the normal distribution to find the probability. We need to calculate the z-scores for each number of errors, and then find the cumulative probability from the z-score table or using a calculator. Finally, we add up the probabilities for each number of errors to get the probability of at least 2 errors.

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Answer:

D.The events are disjoint. They cannot occur at the same time.

Step-by-step explanation:

Event A = randomly selecting a french horn from the instrument assembly line and getting one that is free of defects

Event B = randomly selecting a french horn from the instrument assembly line and getting one with a dented bell

The two Events are disjoint because they cannot happens at the same time. It is either a french horn is free of defects or it has defects.

Disjoint event is when two event cannot occur at the same time.

Answer:

$1 USD = $1.20 Canadian dollars

Step-by-step explanation:

Use proportions of Canadian dollars over US dollars so you get

300 CAD/250 USD = x CAD / 1 USD

solve for x to get 1.2

So $1 USD is equal to $1.20 Canadian dollars.

Answer:

$1.20

Step-by-step explanation:check work by any of these ways.

300/1.20=250,

300/250=1.20,

Or 1.20*250=300

In this question, we cannot provide most of the answer, as it requires the use of Powerpoint and the personal experiences of each participant. However, we are able to talk about some of the personal conflicts and work conflicts that students might have faced in their lives. Some examples that you could use in your text are:

Answer:

The number of complete award announcements possible are 19,554,575,040.

Step-by-step explanation:

Combination is the number of ways to select k items from n distinct items when the order of selection does not matters.

Whereas permutation is the number of ways to select k item from n items when order of selection matters.

The number of people entering this year is 22.

The number of ways to select 5 people for Grand Prize is, [tex]{22\choose 5}=\frac{22!}{5!(22-5)!} =26334[/tex].

The remaining number of people is, 22 - 5 = 17.

It is provided that the other 5 are selected according to an order.

The number of ways to select other 5 winners is,

[tex]^{17}P_{5}=\frac{17!}{(17-5)!} =742560[/tex]

The total number of ways to select 10 winners of 22 is:

Total number of ways = 26334 × 742560 = 19,554,575,040.

Answer:

The PMF of B is given by

P(B=0) = 0.7

P(B=1) = 0.16

P(B=2) = 0.08

P(B=3) = 0.04

P(B=4) = 0.02

Step-by-step explanation:

Let x denote P(B=1), we know that

P(B=0) = 1-0.3 = 0.7

P(B=1) = x

P(B=2) = x/2

P(B=3) = x/4

P(B=4) = x/8

Also, the probabilities should sum 1, thus

0.7+x+x/2+x/4+x/8 = 1

15x/8 = 0.3

x = 0.16

As a result, the PMF of B is given by

P(B=0) = 0.7

P(B=1) = 0.16

P(B=2) = 0.08

P(B=3) = 0.04

P(B=4) = 0.02

Based on given conditions, the probability of a single (B=1) is 0.160, a double (B=2) is 0.080, a triple (B=3) is 0.040, and a home run (B=4) is 0.020. The probability of player making an out (B=0) is 0.700.

Let's denote the probability of a home run as p. Then, the probability of a triple would be 2p, the double would be 4p, and the single would be 8p, all because of the twice-as-likely condition. As these are all the situations in which the player can get a hit, their sum should equal the player's batting average, i.e., 0.300.

So, 8p + 4p + 2p + p = 0.300. Solving this equation, we get that p = 0.020.

Therefore, using the same notations, the PMF of B (probability mass function) would be as follows: P(B=0) = 0.700, P(B=1) = 0.160, P(B=2) = 0.080, P(B=3) = 0.040, and P(B=4) = 0.020.

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Answer:

The type of sampling is stratified.

Step-by-step explanation:

Samples types are classified as:

Random: Basically, put all the options into a hat and drawn some of them.

Systematic: Every kth element is taken. For example, you want to survey something on the street, you interview every 5th person, for example.

Cluster: Divides population into groups, called clusters, and each element in the cluster is surveyed.

Stratified: Also divides the population into groups. However, then only some elements of the group are surveyed.

In this problem, we have that:

The drivers are divided into two groups according to their ages.

There are thousands and thousand of drivers both under and over 30 years of age, and 500 for each group(some elements of the group) are selected.

So the type of sampling is stratified.

The type of sampling used in the example is stratified sampling. The researcher divided the total population of drivers into two groups (under 30 and over 30), and selected 500 drivers from each group.

In this example, the type of sampling used by the market researcher is called stratified sampling. In stratified sampling, the population is divided into different subgroups, or 'strata', and a sample is taken from each stratum. In this case, the researcher has divided the population of drivers into two strata based on their age: drivers under 30 and drivers over 30. 500 drivers are then selected from each of these strata.

Stratified sampling is a common method used in market research because it allows for a more accurate representation of the population, as each subgroup is adequately represented in the sample.

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