Cheating: For a statistics project a community college student at Diablo Valley College (DVC) decides to investigate cheating in two popular majors at DVC: business and nursing. She selects a random sample of nursing and business courses and convinces the professors to distribute a short anonymous survey in their classes. The question about cheating is one of many other questions about college life. When the student summarizes the data, she finds that 42 of the 50 business students and 38 of the 70 nursing students admitted to cheating in their courses. True or false? The counts suggest that the normal model is a good fit for the sampling distribution of sample differences. (a) a·False o b.True

Answer: value of the speed, llv(t)ll = 4.0

Step-by-step explanation:

Given;

time interval t1 =5, t2= 7

Length of path ( distance) L = 8

Speed = distance travelled/ time taken

Speed = dL/dt

Speed = 8/(t2-t1) = 8/(7-5)

Speed = 8/2

Speed = 4.0

Since it's moving at constant speed the speed = 4.0

The null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

Here, it is specified that the scientist wants to study if students political beliefs change as they go through college. He wants to test if there are changes in the proportions of people who are liberal( or conservative).

Given that:

100 students before and after were asked their policial belief, as shown in table:

                               Liberal                Conservative

After college              80                       20

Before college           85                       15

Proportion of liberal = 1 - proportion of conservatives

So we will symbolize the hypotheses in terms of one of them, let it be proportions of liberals.

Sample size = 100

Favorable cases = Number of people from sample who consider themselves liberal. = X (say)

Sample proportion is: [tex]\hat{p}_1 = \dfrac{X}{N} = \dfrac{80}{100} = 0.8[/tex]

Sample proportion is: [tex]\hat{p}_2 = \dfrac{X}{N} = \dfrac{85}{100} = 0.85[/tex]

Let p1 and p2 be the population proportion of people believing themself liberal after and before college respectively.

Then, the null hypothesis will assume that the claim of difference in belief the scientist wants to test is false, and therfore,

[tex]H_0: p_1 = p_2[/tex] or [tex]H_0: p_1 - p_2 = 0[/tex] (no difference, proportions are same, indicating that belief of students doesn't change much, as d)

Thus, the null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.

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Answer: B. 264

Step-by-step explanation:

Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:

[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex]

, where z = Critical z-value corresponds to the given confidence interval

E=  margin of error

Let p be the population proportion of clear days.

As per given , we have

Prior sample size : n= 150

Number of clear days in that sample = 117

Prior estimate of the population proportion of clear days = [tex]p=\dfrac{117}{150}[/tex]

E= 0.05

The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)

Then, the required sample size will be :

[tex]n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2[/tex]

Simplify ,

[tex]n= (0.1716)(39.2)^2[/tex]

[tex]n= 263.687424\approx264[/tex]

Hence, the sample size necessary to construct this interval =264

Thus the correct option is B. 264

Answer:

Descriptive

Step-by-step explanation:

Statistics can be broadly classified into two main branches

i) Descriptive and ii) inferential

Descriptive statistics deal with values such as mean, standard deviation from the data.

Inferential statistics is used to predict unknown values from the observed values.

Applied statistics mainly analyses the data according to the needs of business or industries.

Hence the average expenditure on different items such as food, clothes, fuel comes under

Descriptive Statistics

Final answer:

Average expenditure on different items like food, clothes, and fuel falls under descriptive statistics, essential for summarizing trends in expenditure data. The correct option is a.

Explanation:

Descriptive statistics involve summarizing trends in data, such as calculating averages and standard deviations, making them suitable for analyzing average expenditures on items like food and clothes.

Inferential statistics are used to make inferences about populations based on sample data, aiming to find cause and effect relationships or correlations, which is essential when studying average expenditures across different categories.

Therefore, analyzing average expenditures on various items falls under the domain of descriptive statistics as it involves summarizing and interpreting trends in expenditure data.

Answer: 1.7 x 10^6

Step-by-step explanation:

6^8 = 1,679,616

1,679,616 = 1.7 x 10^6

Answer:

619.13 feet

Step-by-step explanation:

Please find the attachment.

Let x represent the distance between doghouse to the foot of the tower.

We have been given that from the top of a vertical tower, 331 feet above the surface of the earth, the angle of depression to a doghouse is 28 degrees 8'. We are asked to find the distance between doghouse to the foot of the tower.

First of all, we will convert our given angle into degrees as it is given in degrees and minutes.

We will divide 8 by 60 to convert 8 minutes into degrees as:

[tex]\frac{8}{60}=0.13333[/tex]

The doghouse, tower and angle of depression forms a right triangle with respect to ground, where, 331 feet is opposite side and x is adjacent side to angle 28.13 degrees.

[tex]\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}[/tex]

[tex]\text{tan}(28.13^{\circ})=\frac{331}{x}[/tex]

[tex]x=\frac{331}{\text{tan}(28.13^{\circ})}[/tex]

[tex]x=\frac{331}{0.534623339864}[/tex]

[tex]x=619.12747783[/tex]

Upon rounding to nearest hundredth, we will get:

[tex]x\approx 619.13[/tex]

Therefore, the doghouse is 619.13 feet far from the foot of the tower.

Answer:

a) [tex]k = 0.000124[/tex]

b) According to these data, the Shroud of Turin has around 760 years.

Step-by-step explanation:

The amount of carbon-14 is modeled by the following equation:

[tex]C(t) = C_{0}e^{-kt}[/tex]

In which [tex]C_{0}[/tex] is the initial amount and k is the rate of decrease.

(a) Find the value of the constant k in the differential equation.

Half-life of 5595 years.

So [tex]C(5595) = 0.5C_{0}[/tex]

[tex]C(t) = C_{0}e^{-kt}[/tex]

[tex]0.5C_{0} = C_{0}e^{-5595k}[/tex]

[tex]e^{-5595k} = 0.5[/tex]

Applying ln to both sides

[tex]-5595k = -0.69[/tex]

[tex]k = 0.000124[/tex]

b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?

This is t when [tex]C(t) = 0.91C_{0}[/tex]

[tex]C(t) = C_{0}e^{-kt}[/tex]

[tex]0.91C_{0} = C_{0}e^{-0.000124t}[/tex]

[tex]e^{-0.000124t} = 0.91[/tex]

Applying ln to both sides

[tex]-0.000124t = -0.094[/tex]

[tex]t = 760.57[/tex]

According to these data, the Shroud of Turin has around 760 years.

Answer:

d. It can be biased by one or two extremely small or large values

Step-by-step explanation:

Specially in a small sample, a single measure can cause a large difference.

For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.

So the correct answer is:

d. It can be biased by one or two extremely small or large values

The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.

The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.

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Answer:

B.75%

Step-by-step explanation:

Answer:

A. 50%

Step-by-step explanation:

about 50% b/c over half the people who play sports all get injuries

Answer: a) 84 and b ) 2084

Step-by-step explanation:

Given : Sample standard deviation : s= $22.82

(Population standard deviation is unknown ) , so we use t-test.

Critical value or the​ 95% confidence intervals :[tex]t_{n-1}*=2.0[/tex]

Formula to find the sample size :

[tex]n=(\dfrac{t^*\cdot s}{E})^2[/tex]

a) E = 5

[tex]n=(\dfrac{(2)\cdot 22.82}{5})^2[/tex]

[tex]n=(9.128)^2=83.320384\approx84[/tex]

i.e. Required sample size : n= 84

b)  E = 1

[tex]n=(\dfrac{(2)\cdot 22.82}{1})^2[/tex]

[tex]n=(45.64)^2=2083.0096\approx2084[/tex]

i.e. Required sample size : n= 2084

Mutliply his hourly rate by 1.5 to find his night pay:

30.40 x 1.5 = $45.60 per hour at night.

Multiply his rate by number of hours:

45.60 x 15 = $684

Answer:

The 90% confidence interval would be given by (57.006;62.994)  

Step-by-step explanation:

Previous concepts

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".  

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".

[tex]\bar X=60[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=10[/tex] represent the population standard deviation  

n=30 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=10)[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]  

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]  

Using the normal standard table, excel or a calculator we see that:  

[tex]z_{\alpha/2}=1.64[/tex]

Since we have all the values we can replace:

[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]  

[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]  

So on this case the 90% confidence interval would be given by (57.006;62.994)  

Final answer:

To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.

Explanation:

To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:

CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)

Where:

\(\bar{x}\) is the sample mean (60 days)

\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)

\(\sigma\) is the population standard deviation (10 days)

\(n\) is the sample size (30 customers)

Plugging the values into the formula gives:

CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})

Calculating the margin of error:

ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00

Now compute the confidence interval:

CI = [60 - 3.00, 60 + 3.00]

CI = [57.00, 63.00]

So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.

Answer:

a) i) This is a right-tailed test.

b)

[tex]\text{Test statistic} = 1.08[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 23 pounds

Sample mean, [tex]\bar{x}[/tex] = 23.6 pounds

Sample size, n = 40

Alpha, α = 0.10

Sample standard deviation, s = 3.5 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 23\text{ pounds}\\H_A: \mu > 23\text{ pounds}[/tex]

This is a one tailed(right).

Formula:

[tex]\text{Test statistic} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]\text{Test statistic} = \displaystyle\frac{23.6 - 23}{\frac{3.5}{\sqrt{40}} } = 1.08[/tex]

Answer:

The equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

Step-by-step explanation:

The polynomial has roots at places where f(x) cuts the x axis.

The function cuts the x-axis at 3 points: 4, -4, and 0.

It cuts the axis at the point, x=0; x=4; x=-4

Therefore, the points equation has to be of the form, k*x*(x-4)*(x+4)*p(x)

where k is any arbitrary constant and p(x) is a polynomial of any degree depending on what the equation does in the region and (5,∞) and (-∞,-5).

Therefore, the equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

Answer:

This information indicates that the mean level of arsenic in this well is less than 8 ppbat  0.01 level of signifcance..

Step-by-step explanation:

Given that a well in Texas is used to water crops.

This well is tested on a regular basis for arsenic.

A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb

H0: Sample mean = 8

Ha: Sample mean <8

(left tailed test at 1% level)

Mean difference =-0.70

Std error of mean = [tex]\frac{1.9}{\sqrt{36} } \\=0.3167[/tex]

Test statistic t = -2.204

df = 35

p value = 0.0171

Since p >0.01 we accept H0

This information indicates that the mean level of arsenic in this well is less than 8 ppbat  0.01 level of signifcance..

Answer:

The girl will have $335,544.32

Step-by-step explanation:

2^25 = 33,554,432

Divide by 100 to turn the amount of pennies into dollars:

33,554,432/100

$335,544.32

Answer:

Option A) any numerical value in an interval or collection of intervals

Step-by-step explanation:

Continuous Random Variable:

Option A) any numerical value in an interval or collection of intervals

A continuous random variable can take on any numerical value within a given range or collection of ranges, and it's a characteristic feature of it to take an infinite number of values in any interval. Some examples of this can be a person's height, time, temperature, and weight in physics.

A continuous random variable is a type of random variable that can assume any numerical value in a given interval or collection of intervals, making option a correct. This is in contrast to a discrete random variable, which can only take on a finite number of values. A key characteristic of continuous random variables is that they can take on an infinite number of values in any interval.

For example, the height of a person can be treated as a continuous variable, since it can theoretically take any value within a certain range, not just whole number values. The same applies to variables such as time, temperature, and weight in physics.

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If a generic interval [tex][a,b][/tex] is partitioned into [tex]n[/tex] subintervals, each one has a length:

[tex]\Delta x = \dfrac{b-a}{n}.[/tex]

In this case, [tex]a = 4[/tex], [tex]b=6[/tex] and [tex]n=4[/tex], so:

[tex]\Delta x = \dfrac{6-4}{4} = \dfrac{2}{4} = \dfrac{1}{2} = 0.5.[/tex]

The grid points are given by:

[tex]x_k = a + k\Delta x, \quad\textrm{with } k \in \{0,1,2,3,4\}.[/tex]

Since [tex]a = 4[/tex] and [tex]\Delta x = 0.5[/tex], we have:

[tex]x_0 = 4 + 0 \times 0.5 = 4 \\x_1 = 4 + 1 \times 0.5 = 4 + 0.5 = 4.5\\x_2 = 4 + 2 \times 0.5 = 4 + 1 = 5\\x_3 = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5\\x_4 = 4 + 4 \times 0.5 = 4 + 2 = 6[/tex]

The [tex]4[/tex] subintervals are of the form [tex]I_k = [x_{k-1}, x_k], \quad\textrm{with } k \in \{1, 2, 3,4\}[/tex]:

[tex]I_1 = [x_0, x_1] = [4,4.5]\\I_2 = [x_1, x_2] = [4.5,5]\\I_3 = [x_2, x_3] = [5,5.5]\\I_4 = [x_3, x_4] = [5.5, 6][/tex]

For the left Riemann sums we will use the left-handed points, namely:

[tex]\{x_0, x_1, x_2, x_3\} = \{4,4.5,5,5.5\}.[/tex]

For the right Riemann sums we will use the right-handed points, namely:

[tex]\{x_1, x_2, x_3, x_4\} = \{4.5,5,5.5,6\}.[/tex]

For the midpoint Riemann sums we will use the average between the two extrema of each subinterval, given by

[tex]I_k \to \tilde{x}_k = \dfrac{x_{k-1}-x_k}{2}, \quad \textrm{with } k \in\{1,2,3,4\}.[/tex]

This gives the midpoints:

[tex]I_1 = [x_0, x_1] = [4,4.5] \to \tilde{x}_1 = \dfrac{x_0 + x_1}{2} = \dfrac{4+4.5}{2} = \dfrac{8.5}{2} = 4.25\\\\I_2 = [x_1, x_2] = [4.5,5] \to \tilde{x}_2 = \dfrac{x_1 + x_2}{2} = \dfrac{4.5+5}{2} = \dfrac{9.5}{2} = 4.75\\\\I_3 = [x_2, x_3] = [5,5.5] \to \tilde{x}_3 = \dfrac{x_2 + x_3}{2} = \dfrac{5+5.5}{2} = \dfrac{10.5}{2} = 5.25\\\\I_4 = [x_3, x_4] = [5.5,6] \to \tilde{x}_4 = \dfrac{x_3 + x_4}{2} = \dfrac{5.5+6}{2} = \dfrac{11.5}{2} = 5.75[/tex]

The points used for the midpoint Riemann sums are therefore:

[tex]\{\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\tilde{x}_4\} =\{4.25,4.75,5.25,5.75\}.[/tex]

The interval and sub intervals are illustrations of Riemann sums

The given parameters are:

[tex]\mathbf{Interval = [4,6]}[/tex]

[tex]\mathbf{n =4}[/tex] --- sub intervals

(a) The sub interval length

This is calculated as:

[tex]\mathbf{\Delta x = \frac{b - a}{n}}[/tex]

Where:

[tex]\mathbf{[a,b] =[4,6]}[/tex]

So, we have:

[tex]\mathbf{\Delta x = \frac{6 - 4}{4}}[/tex]

[tex]\mathbf{\Delta x = \frac{2}{4}}[/tex]

[tex]\mathbf{\Delta x = 0.5}[/tex]

Hence, the length of the sub-intervals is 0.5

(b) The grid points

This is calculated as:

[tex]\mathbf{x_k = a + k\Delta x}[/tex]

So, we have:

[tex]\mathbf{x_0 = 4 + 0 \times 0.5 = 4}[/tex]

[tex]\mathbf{x_1 = 4 + 1 \times 0.5 = 4.5}[/tex]

[tex]\mathbf{x_2 = 4 + 2 \times 0.5 = 5}[/tex]

[tex]\mathbf{x_3 = 4 + 3 \times 0.5 = 5.5}[/tex]

[tex]\mathbf{x_4 = 4 + 4 \times 0.5 = 6}[/tex]

So, the grid points are 4, 4.5, 5, 5.5 and 6

(c) The left, right and midpoint Riemann sums

The midpoint is the average of the left and right points.

So, we have:

[tex]\mathbf{x_0 = 0.5 \times (4 + 4.5) = 4.25}[/tex]

[tex]\mathbf{x_1 = 0.5 \times (4.5 + 5) = 4.75}[/tex]

[tex]\mathbf{x_2 = 0.5 \times (5 + 5.5) = 5.25}[/tex]

[tex]\mathbf{x_3 = 0.5 \times (5.5 + 6) = 5.75}[/tex]

So, the midpoints are 4.25, 4.75, 5.25 and 5.75

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Answer:

P-value = 0.032794

Step-by-step explanation:

We are given the following information in the question:

Population mean, μ = 98.6 degrees

Sample size, n = 9

Alpha, α = 0.05

Test t-statistic = 2.132

The null and the alternate hypothesis :

[tex]H_{0}: \mu = 98.6\text{ degrees}\\H_A: \mu > 98.6\text{ degrees}[/tex]

We have to find the p-value for degree of freedom 8 and significance level 0.05

The calculated p-value is 0.032794

Answer:

The probability that there are 8 occurrences in ten minutes is

option B. 0 .0771

Step-by-step explanation:

Given:

Random Variable = x

Mean number of occurrences in ten minutes is 5.3.

The probability of an occurrence is the same in any two time periods of an equal length

To Find:

The probability that there are 8 occurrences in ten minutes  = ?

Solution:

Let X be the number of  occurrences of the event X

[tex]X \sim {Pois} (\lambda)[/tex]

[tex]\lambda = E(X) = 5.3[/tex]

Possion of distribution is given by ,

[tex]P(X=x) = \frac{e^{- \lambda} \lambda^{x}}{x!}[/tex]

Substituting the values,

[tex]P(X=8) = \frac{e^{- 5.3} 5.3^{8}}{8!}[/tex]

[tex]P(X=8) = \frac{(0.004994) ( 622596.904)}{40320}[/tex]

[tex]P(X=8) = \frac{(3109.24894)}{40320}[/tex]

P(X=8) = 0.0771

To find the probability of 8 occurrences in ten minutes with a mean of 5.3, use the Poisson distribution formula to calculate the probability, resulting in approximately 0.0771.

Given: Mean number of occurrences in ten minutes = 5.3

Formula: Probability mass function of a Poisson distribution is given by: P(x) = (e^-λ * λ^x) / x!

Calculations: Plugging in the values, P(X=8) = (e^(-5.3) * 5.3^8) / 8! ≈ 0.0771

Step-by-step explanation:

I took 3 indicated as 5 and its adjacent angle to be 3

<2 = <4

As <2 =<3( corresponding angle)

And <3 = <4 ( Vertically opp.angle)

hence <2 = <4

<2 =>110

so , <4 = 110

So, <4 =>110

As <4 and <5 form linear pair

So <4 + <5 =>180

<5 = 180 -110 =>70

As i took <5 as replacing angle to <3

So According to Question fig

<3 =>70

Hence proved

Answer:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Step-by-step explanation:

1) 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]

2) 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 0.025[/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)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

[tex]1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}[/tex]

Now, differentiate to obtain

[tex]\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}[/tex]

[tex]=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}[/tex]

To obtain the required form, extract a factor in both the numerator and denominator:

[tex]\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)[/tex]

Answer:

Step-by-step explanation:

The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]

a)

The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month

[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}

(24)\\\\=8[/tex]

(Here. t = 24)

For Poisson process mean and variance are same,

[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828

[/tex]

b)

For the Poisson process the intervals between events(finding a new planet) have  independent  exponential  distribution with parameter [tex]\lambda[/tex]. The  sum  of [tex]K[/tex] of these  independent exponential has distribution Gamma [tex](K, \lambda)[/tex].

From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}

[/tex]

Calculate the expected value.

[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}

\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]

(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])                                                                      

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г  (k)λ^k)(x)^(k-1).(e)^(-x/λ)

[tex]=\frac{1}{Г  (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]

Hence, the required probability is 0.2149 or 21.49%

We can calculate the expected value and standard deviation of the number of planets Audrey will find in the next two years using probability and statistics. Using geometric distribution, we can also calculate the expected value and standard deviation of the amount of time until Audrey is eligible for a prize after finding her sixth new planet. Finally, we can calculate the probability of Audrey becoming eligible for the prize within one year using the binomial distribution formula.

To find the expected value and standard deviation of the number of planets Audrey will find in the next two years, we can use the concepts of probability and statistics. Since she finds one planet every three months on average, the expected value can be calculated by multiplying the average number of planets found per month (1/3) by the number of months in two years (24). The standard deviation can be calculated using the formula sqrt(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.

To find the expected value and standard deviation of the amount of time until Audrey is eligible for the prize after finding her sixth new planet, we can use the concept of geometric distribution. The expected value can be calculated by taking the reciprocal of the probability of success (1/6) and the standard deviation can be calculated using the formula sqrt((1-p) / (p^2)).

To find the probability that Audrey will become eligible for the prize within one year, we can calculate the cumulative probability of finding six or more new planets in one year using the binomial distribution formula.

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Washington High School won the meet.

Johnson High School came in second with 159 points.

Difference between first and second was 3 points.

Step-by-step explanation:

At a particular swim meet, the details of the points which were awarded to the first three places finish are given.

To find the total number of points scored by each school is equal to sum of multiplying the respective points which were awarded to number of respective places finish.  

i.e. Washington High School had 10 first places finish, so total points for first place finish [tex]= 10 \times10=100[/tex] points.

Similarly, for second places finish = [tex]6\times 8=48[/tex]

and for third places finish = [tex]=2\times 7=14[/tex]

Therefore, Washington High School had total points = 100 + 48 + 14 = 162 points

By using this method, we can make the matrices (Please refer the below attachment).

From the matrices,

Total number of points:

Among these scores of four schools, 162 is the highest score. So, Washington High School won the meet.

Johnson High School came in second with 159 points.

Difference between first and second = 162 - 159 = 3 points.

Answer:

[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]  

[tex]p_v =2*P(Z<-1.364)=0.173[/tex]  

The p value obtained was a very high value and using the significance level assumed [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 graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.  

Step-by-step explanation:

1) Data given and notation

n=116 represent the random sample taken

X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace

[tex]\hat p=0.28[/tex] estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace

[tex]p_o=0.34[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is 0.34 or no.:  

Null hypothesis:[tex]p=0.34[/tex]  

Alternative hypothesis:[tex]p \neq 0.34[/tex]  

When we conduct a proportion test we need to use the z statistic, 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].

3) Calculate the statistic  

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

[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]  

4) 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 next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z<-1.364)=0.173[/tex]  

The p value obtained was a very high value and using the significance level assumed [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 graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.  

The probability that at least one route to work will still be open is [tex]\( \frac{209}{216} \)[/tex] when leaving.

To find the probability that at least one route to work will still be open when you leave for work, we need to find the probability that none of the streets will be closed within the next hour.

Let [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] represent the events that Alabaster, Brillantine, and Clancy Streets are closed due to accidents within the next hour, respectively.

The probability of each street being closed within the next hour is:

[tex]\[ P(A) = \frac{25}{60} = \frac{5}{12} \][/tex]

[tex]\[ P(B) = \frac{50}{60} = \frac{5}{6} \][/tex]

[tex]\[ P(C) = \frac{40}{60} = \frac{2}{3} \][/tex]

Since the accidents on each street are independent events, the probability that all three streets remain open is the product of the probabilities that each street remains open:

[tex]\[ P(\text{})[/tex] = [tex]P(\neg A \cap \neg B \cap \neg C) = P(\neg A) \times P(\neg B) \times P(\neg C) ][/tex]

where [tex]\( \neg A \)[/tex], [tex]\( \neg B \)[/tex], and[tex]\( \neg C \)[/tex] represent the complementary events that the streets are not closed.

So, we have:

[tex]\[ P(\text{})[/tex] = [tex](1 - P(A)) \times (1 - P(B)) \times (1 - P(C)) ][/tex]

[tex]\[ = \left(1 - \frac{5}{12}\right) \times \left(1 - \frac{5}{6}\right) \times \left(1 - \frac{2}{3}\right) \][/tex]

[tex]\[ = \left(\frac{7}{12}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{3}\right) \][/tex]

[tex]\[ = \frac{7}{12} \times \frac{1}{6} \times \frac{1}{3} \]\[ = \frac{7}{216} \][/tex]

Therefore, the probability that at least one route to work will still be open when you leave for work is [tex]\( 1 - \frac{7}{216} = \frac{209}{216} \)[/tex].

Answer:

a) P(24) = 1025.

b) P(24) = 191.

Step-by-step explanation:

This population can be modeled by the following exponential model.

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.

The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].

Population after 24 hours.

(a) No antibiotic is present, so the relative growth rate is 15%.

So r = 0.15.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]

(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.

So r = 0.08.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]

Answer:

D. 348

Step-by-step explanation:

The volume of the square prisma is given by the following formula:

[tex]V = s^{2}h[/tex]

In which h is the height, and s is the side of the base.

Let's use implicit derivatives to solve this problem:

[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]

In this problem, we have that:

[tex]\frac{ds}{dt} = 5, \frac{dh}{dt} = -2, h = 7, s = 6[/tex]

So

[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]

[tex]\frac{dV}{dt} = 2*6*7*5 + (6)^{2}*(-2) = 348[/tex]

So the correct answer is:

D. 348