If 2 is a factor of n and 3 is a factor of n, then
6 is a factor of n. 2 is not a factor of n or 3 is not a factor of n or 6 is a factor of n.

Answer:

26.76% probability that a randomly chosen golfer's score is above 70 strokes.

Step-by-step explanation:

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.

In this problem, we have that:

[tex]\mu = 68.2, \sigma = 2.91[/tex]

What is the probability that a randomly chosen golfer's score is above 70 strokes?

This is 1 subtracted by the pvalue of Z when X = 70. So

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

[tex]Z = \frac{70 - 68.2}{2.91}[/tex]

[tex]Z = 0.62[/tex]

[tex]Z = 0.62[/tex] has a pvalue of 0.7324.

So there is a 1-0.7324 = 0.2676 = 26.76% probability that a randomly chosen golfer's score is above 70 strokes.

Final answer:

To find the probability that a golfer's score is above 70, calculate the Z-score using the formula Z = (X - μ) / σ, where X is 70, the mean (μ) is 68.2, and the standard deviation (σ) is 2.91. The result, approximately 0.62, corresponds to a cumulative probability that must be subtracted from 1 to find the probability of scoring above 70. This process estimates there's about a 26.8% chance a golfer scores above 70.

Explanation:

The question asks about the probability that a randomly chosen golfer on the PGA tour has a score above 70 strokes, given that the scores are normally distributed with a mean of 68.2 and a standard deviation of 2.91. To find this probability, we use the Z-score formula, which is Z = (X - μ) / σ, where X is the score of interest, μ (mu) is the mean, and σ (sigma) is the standard deviation.

Calculating the Z-score for a score of 70:
Z = (70 - 68.2) / 2.91 ≈ 0.62.

Next, we consult a Z-table or use a calculator to find the probability corresponding to a Z-score of 0.62, which tells us the probability of a score being less than 70. To find the probability of a score being above 70, we subtract this value from 1.

Note: Specific values from the Z-table or calculator are not provided here. Generally, the process would involve looking up the cumulative probability for a Z-score of 0.62, which might be around 0.732. Therefore, the probability of a score above 70 would be 1 - 0.732 = 0.268. This means there's approximately a 26.8% chance that a randomly chosen golfer's score is above 70.

Answer:

19.3795 m

Step-by-step explanation:

If sound travels at 343 m/s and it took Karen 0.113 s to hear the echo from the wall, the distance travelled by the sound is:

[tex]D=343\frac{m}{s} *0.113\ s\\D= 38.759\ m[/tex]

Note that the distance calculated above is the distance travelled from Karen to the wall and then back from the wall to Karen. Therefore, the distance between Karen and the wall is:

[tex]d=\frac{38.759}{2}\\d=19.3795\ m[/tex]

The wall is 19.3795 m away from Karen.

The wall is 19.39 meters away.

To find the distance to the wall based on the echo, we use the speed of sound and the time it takes for the echo to return. The time given includes both the journey to the wall and back, so we need to divide it by 2.

1.   Time for sound to travel to the wall and back: 0.113 s    

2.  Time for sound to travel one way:

0.113 s / 2 = 0.0565 s

3.  Speed of sound: 343 m/s

4.  Distance = Speed x Time

343 m/s * 0.0565 s = 19.39

Therefore, the wall is 19.39 meters away from Karen.

Answer:

Option A) The null hypothesis should not be rejected.

Step-by-step explanation:

We are given the following in the question:

A one sample test for a proportion is being performed.

Critical Value = 2.33

Test statistic = 1.37

Since the critical value is positive and the calculated test statistic is positive, thus it is a right tailed test for a proportion.

The null and alternate hypothesis can be written as:

[tex]H_{0}: p \leq p_0\\H_A: p > p_0[/tex]

Since the calculated test statistic is less than the critical value, we fail to reject the null hypothesis.

We accept the null hypothesis.

Thus, the correct answer is

Option A) The null hypothesis should not be rejected.

The data, the number of days traveled by randomly selected employees, is classified as quantitative, discrete data with a ratio level of measurement.

The data in question, namely, the number of days traveled last month by 100100 randomly selected employees, is considered quantitative data. This is because it deals with numbers that can be quantitatively analyzed. In terms of whether the data is discrete or continuous, it is discrete. The number of days traveled can be counted in whole numbers (you can't travel 2.5 days for example); thus, it is a countable set of data. Lastly, considering the level of measurement, the data falls under the ratio level as it not only makes sense to say that someone traveled more days than someone else (therefore an ordered relationship), but it also makes sense to say someone traveled twice as many days as someone else (giving us a proportion and a well-defined zero point).

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Hi, you haven't provided the concrete and hypothetical populations, I'll be using people with red hair in Texas as our population, but you can use a similar approach for the population you decide to use.

Answer:

Explanation:

As you can see from the questions, the main difference between statics and probability has to do with knowledge, what do we know about our population? For the first question (probability) we talked about a known group of people, and base on our knowledge we ask about the probability of selecting people with red hair. For the second question (statistics) we are asked to  make an inference about all the population based just on a sample (we don't have all the knowledge), also we say that the last question is Inferential because the data come from a large population selected randomly.

Final answer:

An example of a probability question could involve finding the likelihood of a statistics student using a product weekly based on a survey sample. Meanwhile, an inferential statistics question might use a t-test to infer differences in protein value between two lots of hay.

Explanation:

In the context of probability and inferential statistics, an example of a probability question for a concrete population might be: "If a random sample of 25 statistics students was surveyed, and six reported using a certain product within the past week, what is the probability that any given statistics student uses the product weekly?"

An example of an inferential statistics question using hypothetical population data could be: "Using a t-test to compare the mean protein value of two lots of alfalfa hay, can we infer that there is a significant difference between their protein contents?" The t-test determines if the observed difference in means is likely due to chance or if it is statistically significant.

These examples illustrate how probability is used to model uncertainty and random events, while inferential statistics are employed to make educated guesses about a population from sample data, such as calculating a confidence interval or testing a hypothesis.

One angle and its supplement give 180 when summed:

[tex]\alpha+\beta = 180[/tex]

This implies that

[tex]\alpha = 180-\beta[/tex]

we also know that

[tex]\alpha = 2\beta+102[/tex]

So, we wrote [tex]\alpha[/tex] as [tex]180-\beta[/tex], but also as [tex]2\beta+102[/tex]. So, the two expressions must equal each other, because they both equal [tex]\alpha[/tex]:

[tex]180-\beta = 2\beta+102 \iff 78=3\beta \iff \beta = 26[/tex]

This implies that [tex]\alpha[/tex] must complete [tex]\beta[/tex] so that they reach 180 together:

[tex]\alpha = 180-\beta = 180-26 = 154[/tex]

The concept of supplementary angles is used to solve the given problem. By forming a pair of linear equations from the problem and solving them, the measures of the two angles are found to be 154 degrees and 26 degrees respectively.

The subject in question here involves the concept of supplementary angles. Supplementary angles are two angles that add up to 180 degrees. In this case, the problem states that one angle is twice its supplement increased by 102 degrees. This forms a pair of linear equations that we can solve for.

Let's denote the unknown angle as x and its supplement as y. According to the problem, we have the following system of equations:

By substitifying the second equation into the first, we receive: 2y + 102 + y = 180. Solving this equation gives y = 26 degrees. Substituting y back into the first equation gives x = 180 - 26 = 154 degrees.

So the two supplementary angles are 154 degrees and 26 degrees.

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Answer:

Step-by-step explanation:

The length of the half marathon is 13.1 miles. it took Grant 7 minutes and 45 seconds to run from mile marker 1 to mile marker 2. Converting 7 minutes and 45 seconds to minutes, it becomes

7 + 45/60 =7.75 minutes

Speed = distance/time

Therefore, his speed from mile marker 1 to mile marker 2 is

1/7.75 = 0.129 miles per minute

He spent 19 minutes and 15 seconds to run from mile marker 2 to mile marker 4. Converting 19 minutes and 15 seconds to minutes, it becomes

19 + 15/60 =19.25 minutes

Therefore, his speed from mile marker 2 to mile marker 4 is

2/19.255 = 0.104 miles per minute

A) his average speed from miles 1 to 4 would be

(0.129 + 0.104)/2 = 0.1165 miles per minute.

B) after running the 9th mile, distance remaining would be

13.1 - 9 = 4.1 miles

Time left to complete the race would be

110 - 69 = 41 minutes

Average speed needed to complete the race would be

4.1/41 = 0.1 miles per minute.

Final answer:

Grant's average speed from mile 1 to mile 4 was approximately 0.1481 miles per minute. To complete the half marathon in his goal time, he needs to run the last 4.1 miles at an average speed of approximately 0.1 miles per minute.

Explanation:

We're given that Grant took 7 minutes and 45 seconds to run from mile marker 1 to mile marker 2, and 19 minutes and 15 seconds to run from mile marker 2 to mile marker 4. To find Grant's average speed from mile marker 1 to mile marker 4, we first convert the time into minutes. He ran 2 miles in 7.75 minutes and then 2 miles in 19.25 minutes. That's a total of 4 miles in 7.75 + 19.25 = 27 minutes, resulting in an average speed of 4 miles / 27 minutes ≈ 0.1481 miles per minute.

Then, we find out how fast Grant needs to run to complete the half marathon in 110 minutes. Grant is at mile marker 9 after 69 minutes, leaving him with 110 - 69 = 41 minutes to complete the remaining 13.1 - 9 = 4.1 miles. The average speed required for this last stretch is 4.1 miles / 41 minutes ≈ 0.1 miles per minute.

Answer:

The question is incomplete or missing some part, but I have answered it appropriately.

Step-by-step explanation:

The question is incomplete, but I will give a detailed description of where pie chart are appropriate to use. A pie chart as we known is a divided circle containing breakdown d and this are developed into a table from calculations and then a chart is developed. The circle in a pie chart are divided into sectors where the angle of each sector is directly proportional to the frequency of the item or data that is being represented.

The process is done by finding the sum of all the frequencies, the sum is then used to divide 360 degree which is the total degree in a complete circle. thereafter each of the frequencies is used to multiply this result to arrive at the sectoral angle for each item. for example, the breakdown of weekkly expenditures of a house wife. the expensiture are marked with the amount spent on each. this information can be represented or illustrated on pie chart.

It should be noted that a pie chart is only suitable for a single set of data and not more than one date as is the case of histogram where more than one grouped or un-grouped data maybe represented or ilustrated on a histogram. if this is the case, pie chart are best employed when making a description between a single part of a data and not the whole set.

Answer:

Step-by-step explanation:

Answer/Explanation

The complete question is:

Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]

Step-by-step explanation:

If we make the identification For ∅° (x) = 1 and  ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.

Therefore, in the first case, we have

(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx

This will therefore be equal to :

1/n sin nx lim(π) lim(-π) = 1/n  [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)

and in the second case, we have,,

(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx

This will therefore be equal to:

∫ lim(π) lim(-π) cos mx cos nx dx

Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)

1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ]  lim(π) lim(-π) = 0, m ≠ n

Now, to go ahead to find the norms in the given set intervals, we have,

for  ∅°(x) = 1 we have:

//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π

So therefore, //∅°(x)//² = √2π

For ∅°∨n(x)  = cos nx  , n > 0.

It then follows that,

//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π

Thus, for n > 0 , //∅°(x)// = √π

It is therefore ggod to note that,

Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be  normalized—that is, made into an orthonormal set by dividing each function by  its norm. It follows from the above equations that has been set.

Therefore,

{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.

Answer:

[tex] T(t) = M -C_1 e^{-kt}[/tex]

And as we can see that represent the solution for the differential equation on this case.

Step-by-step explanation:

For this case we have the following differential equation:

[tex] \frac{dT}{dt}= k(M-T)[/tex]

We can rewrite this expression like this:

[tex] \frac{dT}{M-T} = k dt[/tex]

We can us the following susbtitution for the left part [tex] u = M-T[/tex] then [tex] du= -dt[/tex] and if we replace this we got:

[tex] \frac{-du}{u} = kdt[/tex]

We can multiply both sides by -1 and we got;

[tex] \frac{du}{u} =-k dt[/tex]

Now we can integrate both sides and we got:

[tex] ln |u| = -kt + C[/tex]

Where C is a contant. Now we can exponetiate both sides and we got:

[tex] u(t) =e^{-kt} *e^C = C_1 e^{-kt}[/tex]

Where [tex] C_1 = e^C[/tex] is a constant. And now we can replace u and we got this:

[tex] M-T = C_1 e^{-kt}[/tex]

And if we solve for T we got:

[tex] T(t) = M -C_1 e^{-kt}[/tex]

And as we can see that represent the solution for the differential equation on this case.

The solution to the differential equation representing Newton's law of cooling is calculated by rearranging the equation and then integrating both sides. The general solution for the temperature of an object over time is given by T = M - Ce^-kt where M is the ambient temperature, T is the temperature of the object, C is a constant, and k is the constant of proportionality.

To solve the given differential equation which represents Newton's law of cooling, we shall proceed with integrating both sides. This is in the form of a separable first order differential equation and can be written in the form as follows:

dT / (M - T) = k dt

By re-arranging and integrating both sides, we can find the solution. When we integrate both sides, we get:

- ln |M - T| = kt + C

Where C is the constant of integration. Simplifying further:

T = M - Ce-kt

This is the general solution to the temperature T of an object over time t according to Newton's law of cooling.

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A single taxi fare is more likely to fall between $21 & $24 or be over $24 compared to the mean of a sample of 10 taxi fares.

The question revolves around understanding the distribution of taxi fares and comparing the probabilities associated with the means of samples versus individual observations from a normally distributed population.

(A) Probability of falling between $21 & $24

A single taxi fare has greater variability and thus a greater probability of falling within the range of $21 & $24 compared to the mean of a random sample of 10 taxi fares. This is due to the Central Limit Theorem, which states that the distribution of the sample means will have a smaller standard deviation than that of individual observations, also known as the standard error. For a sample size of 10, the standard error is the population standard deviation divided by the square root of the sample size, which leads to a narrower distribution for the sample means compared to the distribution of individual fares.

(B) Probability of being over $24

The probability of a single randomly selected taxi fare being over $24 is greater than that of the mean of 10 randomly selected taxi fares. This is because individual observations are more spread out, as indicated by the standard deviation of the population, whereas the distribution of sample means is more concentrated around the mean due to the reduced standard error.

(A) The mean of a random sample of 10 taxi fares should have the greater probability of falling between $21 and $24.

(B) The amount of a single randomly selected taxi fare should have a greater probability of being over $24.

The Central Limit Theorem states that the distribution of the sample means will approach a normal distribution as the sample size increases, with the mean of the sample means being equal to the population mean and the standard deviation of the sample means (also known as the standard error) being equal to the population standard deviation divided by the square root of the sample size.

For a single taxi fare, the probability of falling between $21 and $24 can be calculated using the standard normal distribution. We first find the Z-scores corresponding to $21 and $24:

Z-score for $21: [tex]\( Z = \frac{X - \mu}{\sigma} = \frac{21 - 22.27}{2.20} = -0.58 \)[/tex]

Z-score for $24: [tex]\( Z = \frac{X - \mu}{\sigma} = \frac{24 - 22.27}{2.20} = 0.80 \)[/tex]

Using a standard normal table or calculator, we can find the probabilities corresponding to these Z-scores:

P(-0.58 < Z < 0.80) = P(Z < 0.80) - P(Z < -0.58) ≈ 0.788 - 0.278 ≈ 0.510

For the mean of a random sample of 10 taxi fares, the standard error (SE) is:

[tex]\( SE = \frac{\sigma}{\sqrt{n}} = \frac{2.20}{\sqrt{10}} \approx 0.70 \)[/tex]

Now we calculate the Z-scores for $21 and $24 using the standard error:

[tex]Z-score for $21: \( Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{21 - 22.27}{0.70} \approx -1.75 \)[/tex]

[tex]Z-score for $24: \( Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{24 - 22.27}{0.70} \approx 2.53 \)[/tex]

The probability that the sample mean falls between $21 and $24 is then:

P(-1.75 < Z < 2.53) = P(Z < 2.53) - P(Z < -1.75) ≈ 0.994 - 0.040 ≈ 0.954

Comparing the two probabilities, 0.954 for the sample mean is greater than 0.510 for a single fare.

Explanation for (B):

For a single taxi fare, we already calculated the Z-score for $24, which is 0.80. The probability of a single fare being over $24 is:

P(Z > 0.80) = 1 - P(Z < 0.80) ≈ 1 - 0.788 ≈ 0.212

For the mean of a random sample of 10 taxi fares, we calculated the Z-score for $24 as 2.53. The probability of the sample mean being over $24 is:

P(Z > 2.53) = 1 - P(Z < 2.53) ≈ 1 - 0.994 ≈ 0.006

Comparing the two probabilities, 0.212 for a single fare is greater than 0.006 for the sample mean. Therefore, the amount of a single randomly selected taxi fare has a greater probability of being over $24.

Answer:

The random variables in this case are discrete since they have a Multinomial distribution.

The probability mass function for a discrete random variable X is given by:

[tex]P(X=x_{i} )[/tex]

Where are [tex]x_{i}[/tex] are possible values of X.

The joint probability mass function of two discrete random variables X and Y is defined as

P(x,y) =P(X=x,Y=y).

It follows that, The joint probability mass function of [tex]X_{3} , X_{4}[/tex] is :

[tex]P(X_{3}, X_{4} ) = P( X_{3} = x_{3}, X_{4} = x_{4} ) =\frac{1}{8} +\frac{3}{8} =\frac{1}{2}[/tex]

In a Multinomial Distribution, variables are independent. Hence, the joint mass function of a pair (X3 , X4) is the product of their individual mass functions. Their specific joint mass function equals P(X3=x3)P(X4=x4) = (n choose x3)(1/8)^x3(3/8)^x4 for x3+x4 ≤ n and x3, x4 ≥ 0.

This problem relates to the concept of a Multinomial Distribution in probability theory. The Multinomial Distribution describes the probabilities of potential outcomes from a multinomial experiment.

In this particular case, you are given that (X1, X2, X3, X4) follows a Multinomial Distribution with parameters n (number of trials) and 4 categories, with known probabilities 1/6, 1/3, 1/8, and 3/8 respectively.

You are asked to derive the joint mass function of the pair (X3, X4). This is actually very straightforward. Due to the properties of a multinomial distribution, these two variables are independent and the joint mass function is simply the product of the individual mass functions.

So the joint mass function of X3 and X4 would be P(X3=x3, X4=x4) = P(X3=x3)P(X4=x4) = (n choose x3)(1/8)^x3(3/8)^x4 provided x3+x4 ≤ n and each of x3,x4 are nonnegative.

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Final answer:

The differential equation governing the spread of flu among students on a college campus is modeled by the logistic equation dx/dt = kx(2000 - x), assuming a constant rate of interactions and no outside influences.

Explanation:

The differential equation governing the number of students x(t) who have contracted the flu on an isolated college campus, where the rate of disease spread is proportional to the number of interactions between infected and susceptible students, can be modeled using principles of epidemiology. The total number of students is 2000, and if we use k for the constant of proportionality, we can denote the number of susceptible students as 2000 - x, because x represents the number of infected students. Hence, the rate of change of x, given by dx/dt, is proportional to the product of the number of students who have the flu and the number who do not, which is kx(2000-x). The differential equation is thus:

dx/dt = kx(2000 - x)

This is a standard form of the logistic differential equation, often used in the SIR model in epidemiology to describe how a disease spreads in a population.

Answer:

1.33 x 10⁻⁸ seconds

Step-by-step explanation:

Assuming that the speed of light is 299,792,458 m/s, and that in order to bounce of Bill's forehead, hit the mirror and return back to his eyes, light must travel 4 meters (distance to the mirror and back) the time that it takes for light to travel is:

[tex]t=\frac{4}{299,792,458} \\t=1.33*10^{-8}[/tex]

It takes 1.33 x 10⁻⁸ seconds.

To evaluate the triple integral ∭Tx²dV, we need to integrate over the solid tetrahedron T defined by the vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3).

To evaluate the triple integral ∭Tx²dV, we need to integrate over the solid tetrahedron T defined by the vertices (0,0,0), (3,0,0), (0,3,0), and (0,0,3). Since T is a three-dimensional shape, we need to perform a triple integral.

The limits of integration for each variable are as follows:

x: 0 to 3-y-z

y: 0 to 3-z

z: 0 to 3

Substituting the limits into the integrand Tx², we can then evaluate the triple integral by integrating with respect to x, y, and z in the given limits.

Answer:

The correct option is a) Cluster.

Step-by-step explanation:

Consider the provided information.

Types of sampling:

Here, the population is divided into geographical blocks,

Thus, the type of sampling is cluster.

Therefore the correct option is a) Cluster.

In order to find the percentage of bulbs with a certain lifespan, one must calculate the z-scores for the given values and find the probabilities using the standard normal distribution, converting these into percentages.

The student's question involves using the properties of the normal distribution to determine the probability of light bulb life spans within certain intervals. To solve these problems, the z-score formula is used, which is (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation.

Remember that the answer will be in the form of a percentage representing the likelihood that any given bulb falls within the specified hour range.

Answer:

y(t)= 11/3 e^(-t) - 5/2 e^(-2t) -1/6 e^(-4t)

Step-by-step explanation:

[tex] Y(s)=\frac{s+3}{(s^2+3s+2)(s+4)} + \frac{s+3}{s^2+3s+2} +\frac{1}{s^2+3s+2} [/tex]

We know that [tex] s^2+3s+2=(s+1)(s+2)[/tex], so we have

[tex] Y(s)=\frac{s+3+(s+3)(s+4)+s+4}{(s+1)(s+2)(s+4)}  [/tex]

By using the method of partial fraction we have:

[tex] Y(s)=\frac{11}{3(s+1)} - \frac{5}{2(s+2)} -\frac{1}{6(s+4)} [/tex]

Now we have:

[tex] y(t)=L^{-1}[Y(s)](t) [/tex]

Using linearity of inverse transform we get:

[tex] y(t)=L^{-1}[\frac{11}{3(s+1)}](t) -L^{-1}[\frac{5}{2(s+2)}](t) -L^{-1}[\frac{1}{6(s+4)}](t) [/tex]

Using the inverse transforms

[tex] L^{-1}[c\frac{1}{s-a}]=ce^{at} [/tex]

we have:

[tex] y(t)=11/3 e^{-t} - 5/2 e^{-2t} -1/6 e^(-4t) [/tex]

Answer:

a. 241; b. 206.0; c. 272.0; d. 174 to 232; other answers (see below).

Step-by-step explanation:

First, at all, we need to organize the data from the smallest to the largest value:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

This step is important to find the median, the first quartile, and the third quartile. There are numerous methods to find the median and the other quartiles, but in this case, we are going to use a method described by Tukey, and it does not need calculators or software to estimate it.

In this case, we have 53 values (an odd number of values), the median is the value that has the same number of values below and above it, so what is the value that has 26 values below and above it? Well, in the organized data above this value is the 27th value, because it has 26 values above and below it:

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

The median is 241.

For the first quartile, we need to calculate the median for the lower half of the values from the median previously obtained. Since we have an odd number of values (53), we have to include the median in this calculation. We have 27 values (including the median), so the "median" for these values is the value with 13 values below and above it.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241

Since we are asking to round the answer to one decimal place, the first quartile is 206.0

We use the same procedure used to find the first quartile, but in this case, using the upper half of the values.

241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302

So, the third quartile is 272.0

The second quartile is the median and "50% of the data lies below this point" (Quartile (2020), in Wikipedia). Having this information into account, 50% of the weights lies below the median 241.

Thus, the middle 50% of the weights are from 174 to 232.

174 178 178 184 185 185 185 185 188 190 200 203 205 206 210 212 212 212 212 215 215 220 223 228 230 232 241

If the population were all professional football players, the right option is:

The above data would be a sample of weights because they represent a subset of the population of all football players.

It represents a sample. Supposing Football Team A are all professional, they can be considered a sample from all professional football players.

If the population were Football Team A, the right option is:

The data would be a population of weights because they represent all of the players on Football Team A.

The mean for the population of weights of Football Team A is the sum of all weights (12529 pounds) divided by the number of cases (53).

[tex] \\ \mu = \frac{12529}{53} = 236.39622641509433 [/tex]

[tex] \\ \mu = \frac{12529}{53} = 236.40 [/tex]

The standard deviation for the population is:

[tex] \\ \sigma = \sqrt(\frac{(x_{1}-\mu)^2 + (x_{2}-\mu)^2 + ... + (x_{n}-\mu)^2}{n}) [/tex]

In words, we need to take either value, subtract it from the population mean, square the resulting value, sum all the values for the 53 cases (in this case, the value is 74332.67924), divide the value by 53 (1402.50338) and take the square root of it (37.45001).

Then, the population standard deviation is 37.45.

The weight that is 3 standard deviations below the mean can be obtained using the following formula:

[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]

[tex] \\ \mu= 236.40\;and\;\sigma=37.45[/tex]

Then, in the case of three standard deviations below the mean, z = -3.

[tex] \\ -3 = \frac{x - 236.40}{37.45} [/tex]

[tex] \\ x = -3*37.45 + 236.40 [/tex]

[tex] \\ x = 124.05 [/tex]

For the player that weights 209 pounds:

[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]

[tex] \\ z = \frac{229 - 236.40}{37.45} [/tex]

[tex] \\ z = -0.19 [/tex]

For Team B, we have a mean and a standard deviation of:

[tex] \\ \mu=240.08\;and\;\sigma=44.38[/tex]

And Player B weighed 229 pounds.

[tex] \\ z = \frac{x - \mu}{\sigma} [/tex]

[tex] \\ z = \frac{229 - 240.08}{44.38} [/tex]

[tex] \\ z = -0.70 [/tex]

This value says that Player B is lighter with respect to his team than Player A, because his weight is 0.70 below the mean of Football Team B, whereas Player A has a weight that is closer to the mean of his team. So, the answer is:

Player B, because he is more standard deviations away from his team's mean weight.

Answer:

B

Step-by-step explanation:

617/500 = 1234/1000

Therefore option B is the answer.

Answer: 66 ways

Step-by-step explanation:

Given;

Number of senior math club members = 6

Number of junior math club members = 4

Total number of members of the club = 10

To form a group of 5 members with at least 4 seniors.

N = Na + Nb

Na = number of possible ways of selecting 4 seniors and 1 junior

Nb = number of possible ways of selecting 5 seniors.

Since the selection is does not involve ranks(order is not important)

Na = 6C4 × 4C1 = 6!/4!2! × 4!/3!1! = 15 ×4 = 60

Nb = 6C5 = 6!/5!1! = 6

N = Na + Nb = 60+6

N = 66 ways

Using the combination formula, it is found that there are 66 ways to form the groups.

The order in which the students are selected is not important, hence, the combination formula is used to solve this question.

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

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

In this problem, the possible groups are:

Hence:

[tex]T = C_{4,1}C_{6,4} + C_{6,5} = \frac{4!}{1!3!}\frac{6!}{4!2!} + \frac{6!}{5!1!} = 4(15) + 6 = 60 + 6 = 66[/tex]

There are 66 ways to form the groups.

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Answer:

Sample mean for U=21.5

Sample mean for F=8.4

Step-by-step explanation:

[tex]Sample mean of u=xbar_{u} =\frac{sum(xi)}{n}[/tex]

Where xi are the observations in the urban homes sample and n is the number of observations in the urban homes sample

[tex]sample mean of u=xbar_{u} =\frac{6+5+11+33+4+5+80+18+35+17+23}{11}[/tex]

[tex]sample mean of u=xbar_{u} =\frac{237}{11}=21.545[/tex]

Rounding it to one decimal places

[tex]sample mean of u=xbar_{u}=21.5[/tex]

Now for second sample

[tex]Sample mean of F=xbar_{F} \frac{sumxi}{n}[/tex]

Where xi are the observations in the farm homes sample and n is the number of observations in the farm homes sample

[tex]Sample mean of F=xbar_{F} =\frac{2+15+12+8+8+7+6+19+3+9.8+22+9.6+2+2+0.5)}{15}[/tex]

[tex]Sample mean of F=xbar_{F} =\frac{125.9}{15} =8.393[/tex]

Rounding it to one decimal places

[tex]Sample mean of F=xbar_{F} =8.4[/tex]

The sample mean for urban homes (U) is 21.54 and the sample mean for farm homes (F) is 7.73.

To find the sample mean, you sum up all the data points and then divide by the number of data points. For the Urban homes (U), we first add up all the data points: 6.0 + 5.0 + 11.0 + 33.0 + 4.0 + 5.0 + 80.0 + 18.0 + 35.0 + 17.0 + 23.0 = 237.0. The number of data points is 11, so the sample mean for U is 237.0 / 11 = 21.54 (rounded to two decimal places).

For Farm homes (F), add up all the data points: 2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0 + 3.0 + 9.8 + 22.0 + 9.6 + 2.0 + 2.0 + 0.5 = 115.9. The number of data points is 15, so the sample mean for F is 115.9 / 15 = 7.73 (rounded to two decimal places).

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Answer:

a) [tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+[/tex]

b) [tex]E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2[/tex]

c) [tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]

d) [tex]E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6[/tex]

e) [tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]

f) [tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]

g) [tex]E(u) = E(v) +E(X) = 0+1=1[/tex]

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

[tex] Y = 1+X +u[/tex]

[tex] u = v+X[/tex]

with both Y and u random variables, we also know that:

[tex] [tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2[/tex]

And we want to calculate this:

Part a

[tex] E(u|X=1)= E(v+X|X=1)[/tex]

Using properties for the conditional expected value we have this:

[tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1[/tex]

Because we assume that v and X are independent

Part b

[tex]E(Y| X=1) = E(1+X+u|X=1)[/tex]

If we distribute the expected value we got:

[tex]E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2[/tex]

Part c

[tex] E(u|X=2)= E(v+X|X=2)[/tex]

Using properties for the conditional expected value we have this:

[tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]

Because we assume that v and X are independent

Part d

[tex]E(Y| X=2) = E(1+X+u|X=2)[/tex]

If we distribute the expected value we got:

[tex]E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6[/tex]

Part e

[tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]

Part f

[tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]

Part g

[tex]E(u) = E(v) +E(X) = 0+1=1[/tex]

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

The parametric equations of the position of a particle with constant velocity moving along a straight line path from points P(7,2) to Q(2,7) are x(t) = 7 - 5t and y(t) = 2 + 5t.

In this context, the movement of a particle can be represented on a plane using a usual 2D Cartesian coordinate system. The constant velocity of the particle dictates it will always move along a straight line. The straight line path can be found by determining the slope between points P(7,2) and Q(2,7).

The slope m of the line is given by:

m = (y2 - y1) / (x2 - x1)

Where P = (x1, y1) and Q= (x2, y2). Applying these coordinates gives us:

m = (7 - 2) / (2 - 7) = -1

So, the line equation we have is something like y - y1 = m(x - x1), and substituting in all the values gives:

y - 2 = -1 * (x - 7) which simplifies to y = -x + 9

To get the parametric equations, we can consider the particle moving along the straight line path from P to Q in time t = 1 second. The parametric equations of the position of the particle is therefore:

x(t) = 7 - 5t and y(t) = 2 + 5t

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Answer:

a) [tex]2.05[/tex]

b) [tex]z = 2.01[/tex]

c) No, we cannot conclude that a larger proportion of graduates have jobs than reported in the article.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 200

p = 0.333

Alpha, α = 0.02

Number of graduates had jobs , x = 80

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.333\\H_A: p > 0.333[/tex]

This is a one-tailed(right) test.  

b) Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{80}{200} = 0.4[/tex]

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

Putting the values, we get,

a) [tex]z = \displaystyle\frac{0.4-0.333}{\sqrt{\frac{0.333(1-0.333)}{200}}} = 2.01[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]

c) Since, the calculated z statistics less than the critical value, we fail to reject the null hypothesis and accept it.

Thus, same proportion of graduates have jobs as compared to previously reported.

Thus, we conclude that there is not enough evidence to support the claim that a larger proportion of graduates have jobs than previously reported.              

In the absence of all forces except gravity, the ball is approximately 0.28 feet below the ground when it crosses the home plate.

(a) In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of 130i +0j-3k ft/s (roughly 90mi/hr). How far above the ground is the ball when it cross the home plate.

To determine the vertical distance the ball is above the ground when it crosses home plate, we need to find the time it takes to reach the home plate using the vertical component of the initial velocity. The initial vertical velocity is -3 ft/s, and the vertical acceleration due to gravity is -32 ft/s2. Using the equation y = y0 + v0yt + 0.5at2, where y is the vertical distance above the ground, y0 is the initial vertical position, v0y is the initial vertical velocity, t is the time, and a is the acceleration due to gravity, we can solve for t. Plugging in the given values, we get:

y = 0 + (-3)t + 0.5(-32)t2

Simplifying the equation, we get 32t2 - 3t = 0. Factoring out t, we get t(32t - 3) = 0. So, either t = 0 (which is the initial time when the ball is released) or 32t - 3 = 0. Solving for t, we find that t = 3/32 seconds. Since the ball is in the air for this amount of time, we can substitute this value back into the equation to find the vertical distance above the ground when the ball crosses the home plate:

y = 0 + (-3)(3/32) + 0.5(-32)(3/32)2

Simplifying, we get y = -9/32 feet, which is approximately -0.28 feet above the ground. Therefore, the ball is 0.28 feet below the ground when it crosses the home plate.

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Answer:

See explanation below.

Step-by-step explanation:

Definition

The cross product is a binary operation between two vectors defined as following:

Let two vectors [tex] a = (a_1 ,a_2,a_3) , b=(b_1, b_2, b_3)[/tex]

The cross product is defined as:

[tex] a x b = (a_2 b_3 -a_3 b_2, a_3 b_1 -a_1 b_3 ,a_1 b_2 -a_2 b_1)[/tex]

The last one is the math definition but we have a geometric interpretation as well.

We define the angle between two vectors a and b [tex]\theta[/tex] and we assume that [tex] 0\leq \theta \leq \pi[/tex] and we have the following equation:

[tex] |axb| = |a| |b| sin(\theta)[/tex]

And then we conclude that the cross product is orthogonal to both of the original vectors.

Some properties

Let a and b vectors

If two vectors a and b are parallel that implies [tex] |axb| =0[/tex]

If [tex] axb \neq 0[/tex] then [tex]axb[/tex] is orthogonal to both a and b.

Let u,v,w vectors and c a scalar we have:

[tex] uxv =-v xu[/tex]

[tex] ux (v+w) = uxv + uxw[/tex] (Distributive property)

[tex] (cu)xv = ux(cv) =c (uxv)[/tex]

[tex] u. (vxw) = (uxv).w[/tex]

Other application of the cross product are related to find the area of a parallelogram for two dimensions where:

[tex] A = |axb|[/tex]

And when we want to find the volume of a parallelepiped in 3 dimensions:

[tex] V= |a. (bxc)|[/tex]

Answer:

There is enough evidence to support the claim that  the true proportion is in fact different from 0.40  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 120

p = 0.4

Alpha, α = 0.05

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.4\\H_A: p \neq 0.4[/tex]

This is a two-tailed test.  

Formula:

[tex]\hat{p} = 0.3[/tex]

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

Putting the values, we get,

[tex]z = \displaystyle\frac{0.3-0.4}{\sqrt{\frac{0.4(1-0.4)}{120}}} = -2.236[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

Since,

The calculated z-statistic does not lies in the acceptance region, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there is enough evidence to support the claim that  the true proportion is in fact different from 0.40

Answer:

See proof below.

Step-by-step explanation:

If we assume the following linear model:

[tex] y = \beta_o + \beta_1 X +\epsilon[/tex]

And if we have n sets of paired observations [tex] (x_i, y_i) , i =1,2,...,n[/tex] the model can be written like this:

[tex] y_i = \beta_o +\beta_1 x_i + \epsilon_i , i =1,2,...,n[/tex]

And using the least squares procedure gives to us the following least squares estimates [tex] b_o [/tex] for [tex]\beta_o[/tex] and [tex] b_1[/tex] for [tex]\beta_1[/tex]  :

[tex] b_o = \bar y - b_1 \bar x[/tex]

[tex] b_1 = \frac{s_{xy}}{s_xx}[/tex]

Where:

[tex] s_{xy} =\sum_{i=1}^n (x_i -\bar x) (y-\bar y)[/tex]

[tex] s_{xx} =\sum_{i=1}^n (x_i -\bar x)^2[/tex]

Then [tex] \beta_1[/tex] is a random variable and the estimated value is [tex]b_1[/tex]. We can express this estimator like this:

[tex] b_1 = \sum_{i=1}^n a_i y_i [/tex]

Where [tex] a_i =\frac{(x_i -\bar x)}{s_{xx}}[/tex] and if we see careful we notice that [tex] \sum_{i=1}^n a_i =0[/tex] and [tex]\sum_{i=1}^n a_i x_i =1[/tex]

So then when we find the expected value we got:

[tex] E(b_1) = \sum_{i=1}^n a_i E(y_i)[/tex]

[tex] E(b_1) = \sum_{i=1}^n a_i (\beta_o +\beta_1 x_i)[/tex]

[tex] E(b_1) = \sum_{i=1}^n a_i \beta_o + \beta_1 a_i x_i[/tex]

[tex] E(b_1) = \beta_1 \sum_{i=1}^n a_i x_i = \beta_1[/tex]

And as we can see [tex]b_1[/tex] is an unbiased estimator for [tex]\beta_1[/tex]

In order to find the variance for the estimator [tex]b_1[/tex] we have this:

[tex] Var(b_1) = \sum_{i=1}^n a_i^2 Var(y_i) +\sum_i \sum_{j \neq i} a_i a_j Cov (y_i, y_j) [/tex]

And we can assume that [tex] Cov(y_i,y_j) =0[/tex] since the observations are assumed independent, then we have this:

[tex] Var (b_1) =\sigma^2 \frac{\sum_{i=1}^n (x_i -\bar x)^2}{s^2_{xx}}[/tex]

And if we simplify we got:

[tex] Var(b_1) = \frac{\sigma^2 s_{xx}}{s^2_{xx}} = \frac{\sigma^2}{s_{xx}}[/tex]

And with this we complete the proof required.

Final answer:

β1^ in simple linear regression is normal because it is a ratio of linear combinations of normal variables Yi. It is unbiased as its expected value equals the true parameter β1. The variance of β1^ is σ2/sxx, derived from properties of variance and the independent nature of errors ϵi.

Explanation:

The student's question pertains to the properties of the estimated coefficient β1^ in a simple linear regression model. To show that β1^ is a normal random variable, we consider the linear combination of the normal random variables Yi, because a linear combination of normal random variables is also normally distributed. Since each Yi is normal and given by Yi=β0+β1xi+ϵi, and ϵi is N(0,σ2), the estimator β1^=sxy/sxx becomes a ratio of linear combinations of these normal variables and hence, normal.

Next, to prove that β1^ is an unbiased estimator, we take the expectation of β1^ and show that E[β1^]=β1. It's implied by calculating the expected value of the numerator sxy and denominator sxx separately and showing the ratio equals β1.

The variance of β1^, Var(β1^), can be shown to be σ2/sxx by leveraging properties of variance of the linear combinations of Yi and noting that ϵi's are independent random variables with variance σ2. The calculations involve squaring the deviations and utilizing expectations.