Rob owns 1/8 of the stock in a local bread company. His sister Talia owns half as much stock as Rob. What part of the stock is owned by neither Rob nor​ Talia?


Please show work so I can use it as a visual for my next question on homework please!

Answer:

Step-by-step explanation:

Figuring it out rn

Scott needs to run 5/8 miles more.

Step-by-step explanation:

Distance covered = 11/8 miles

Total distance = 11/8 miles

Distance left to cover = Total distance - Distance covered

[tex]Distance\ left\ to\ cover= \frac{16}{8}-\frac{11}{8}\\Distance\ left\ to\ cover=\frac{16-11}{8}\\Distance\ left\ to\ cover=\frac{5}{8}[/tex]

Scott needs to run 5/8 miles more.

Keywords: distance, subtraction

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Answer:

a) 2π

b) 0

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

[tex]\large \int_{C}[P(x,y)dx+Q(x,y)dy]=\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt[/tex]

Where P, Q are scalar functions

a)

C(t) = (x(t), y(t)) = (cos(t), sin(t)), 0 ≤ t ≤ 2π

P(x,y) = -y ==> P(x(t),y(t)) = -y(t) = -sin(t)

Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = cos(t)

x'(t) = -sin(t)

y'(t) = cos(t)

[tex]\large \int_{C}[-ydx+xdy]=\int_{0}^{2\pi}[(-sin(t))(-sin(t))+cos(t)cos(t)]dt=\\\\\int_{0}^{2\pi}[sin^2(t)+cos^2(t)]dt=\int_{0}^{2\pi}dt=2\pi[/tex]

b)

C(t) = (x(t), y(t)) = (2cos(πt), 2sin(πt)), 2 ≤ t ≤ 4

P(x,y) = y ==> P(x(t),y(t)) = y(t) = 2sin(πt)

Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = 2cos(πt)

x'(t) = -2πsin(πt)

y'(t) = 2πcos(πt)

[tex]\large \int_{C}[ydx+xdy]=\int_{2}^{4}[(2sin(\pi t))(-2\pi sin(\pi t))+2cos(\pi t)2\pi cos(\pi t)]dt=\\\\4\int_{2}^{4}[cos^2(\pi t)-sin^2(\pi t)]dt=4\int_{2}^{4}cos(2\pi t)dt=\\\\4\left[\frac{sin(2\pi t)}{2\pi}\right]_2^4=\frac{2}{\pi}(sin(8 \pi)-sin(4\pi))=0[/tex]

[tex]y=2x^2\implies\mathrm dy=4x\,\mathrm dx[/tex]

Then in the integral we have

[tex]\displaystyle\int_Cxy\,\mathrm dx+(x+y)\,\mathrm dy=\int_1^2(2x^3+4x(x+2x^2))\,\mathrm dx=\int_1^210x^3+4x^2\,\mathrm dx=\boxed{\frac{281}6}[/tex]

Answer: No, Y1 and Y2 are not independent

Step-by-step explanation:

Because they don't satisfy this condition:

FsubscriptY1Y2(Y1,Y2) = FsubscriptY1(y1) × FsubscriptY2(y2)

... for all given values of Y1 and Y2

This is the condition for independence.

How do we know that Y1 and Y2 don't satisfy this condition?

We use the information in the Joint Probability Distribution Table.

Let's see if the condition stands when Y1 is zero and Y2 is zero

FsubscriptY1Y2(0,0) = 0.38

FsubscriptY1(0) × FsubscriptY2(0) = 0.76×0.55 = 0.418

We can see that 0.38 is not equal to 0.418

Doing the test for any other combination of Y1 and Y2 values will give unequal figures as well.

a.

[tex]f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}[/tex]

is a proper joint density function if, over its support, [tex]f[/tex] is non-negative and the integral of [tex]f[/tex] is 1. The first condition is easily met as long as [tex]C\ge0[/tex]. To meet the second condition, we require

[tex]\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}[/tex]

b. Find the marginal joint density of [tex]X[/tex] and [tex]Y[/tex] by integrating the joint density with respect to [tex]z[/tex]:

[tex]f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz[/tex]

[tex]\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}[/tex]

Then

[tex]\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy[/tex]

[tex]\approx\boxed{0.12886}[/tex]

c. This probability can be found by simply integrating the joint density:

[tex]\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]

[tex]\approx\boxed{0.012262}[/tex]

(a) We determined the constant C by integrating the joint density function over the entire space, finding = 1/10.

​(b) We calculated P(X≤1.375,Y≤1.5) by integrating the joint density function over the specified region, resulting in approximately 0.1286.

(c) For P(X≤1.375,Y≤1.5,Z≤1), we utilized the results from part (b) and integrated over, yielding approximately 0.1163.

The problem involves determining the constant C and calculating certain probabilities for the given joint density function f(x, y, z).

Below is the step-by-step solution:

Part (a): Finding the Constant C

First, we need to find the value of C such that the total probability is 1.

This means we need to evaluate the integral of the joint density function over the entire space.

[tex]\int \int \int C e^{-(0.5x + 0.2y + 0.1z)} \, dx \, dy \, dz = 1[/tex], where x, y, z ≥ 0

[tex]\int_{0}^{\infty} Ce^{-0.5x} \, dx = \frac{C}{0.5}[/tex]

[tex]\int_{0}^{\infty} e^{-0.5x} \, dx = \frac{1}{0.5}[/tex] from 0 to ∞ = 2C

[tex]\int_{0}^{\infty} e^{-0.2y} \, dy = \frac{1}{0.2} = 5[/tex]

[tex]\int_{0}^{\infty} e^{-0.1z} \, dz = \frac{1}{0.1} = 10[/tex]

2C * 5C * 10C = 1

100C³ = 1

C = 1 / 10

Part (b): Finding P(X ≤ 1.375 , Y ≤ 1.5)

P(X ≤ 1.375 , Y ≤ 1.5) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} 0.1 e^{-0.5x} e^{-0.2y} \, dy \, dx[/tex]

[tex]\int_{0}^{1.5} e^{-0.2y} \, dy = -\frac{1}{0.2} (e^{-0.3} - 1)[/tex]

= [tex]\frac{1}{0.2} (1 - e^{-0.3}) = \frac{5}{3} (1 - e^{-0.3})[/tex]

= 5 * 0.2592

≈ 1.296

[tex]\int_{0}^{1.375} 0.1 \cdot 1.296 \cdot e^{-0.5x} \, dx = 0.1 \cdot 1.296 \cdot \left( 2 - 2 e^{-0.6875} \right)[/tex]

= [tex]0.1296 \cdot \left[ -2 e^{-0.5x} \right]_{0}^{1.375}[/tex]

= [tex]0.1296 \cdot 2 \cdot \left( 1 - e^{-0.6875} \right)[/tex]

= 0.1296 * 2 * (1 - 0.5038)

≈ 0.1296 * 2 * 0.4962

≈ 0.1286

Part (c): Finding P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1)

P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = [tex]\int_{0}^{1.375} \int_{0}^{1.5} \int_{0}^{1} 0.1 e^{-0.5x} e^{-0.2y} e^{-0.1z} \, dz \, dy \, dx[/tex]

[tex]\int_{0}^{1} e^{-0.1z} dz = \left[ -\frac{1}{0.1} e^{-0.1z} \right]_{0}^{1}[/tex]

= [tex](1/0.1)(1 - e^{-0.1})[/tex]

= [tex]10 (1 - e^{-0.1})[/tex]

≈ 0.905

Combining all parts:

P(X ≤ 1.375 , Y ≤ 1.5 , Z ≤ 1) = 0.1286 * 0.905

≈ 0.1163

Answer:

The 98% confidence interval would be given (0.616;0.686).  

We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).

Step-by-step explanation:

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".  

Description in words of the parameter p

[tex]p[/tex] represent the real population proportion of people that they were planning to pursue a graduate degree

[tex]\hat p[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree

n=1010 is the sample size required  

[tex]z_{\alpha/2}[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

[tex]\hat p =\frac{X}{n}[/tex] where X represent the number of people with a characteristic and n the total sample size selected.

[tex]\hat p=\frac{658}{1010}=0.651[/tex] represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

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

For the 98% confidence interval the value of [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2=0.01[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=2.33[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.651 - 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.616[/tex]  

[tex]0.651 + 2.33 \sqrt{\frac{0.651(1-0.651)}{1010}}=0.686[/tex]  

And the 98% confidence interval would be given (0.616;0.686).  

We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).  

Answer:

(-3.0486, -0.2848)

Step-by-step explanation:

Let the number of hours per day spent using electronic media from 1999 be the first population and the number of hours per day spent using electronic media from 1999 the second population.  

We have small sample sizes [tex]n_{1} = 15[/tex] and

[tex]n_{2} = 15[/tex].

[tex]\bar{x}_{1} = 5.9333[/tex] and [tex]\bar{x}_{2} = 7.6[/tex]; [tex]s_{1} = 1.0998[/tex] and [tex]s_{2} = 1.5946[/tex].  

The pooled estimate is given by  

[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(1.0998)^{2}+(15-1)(1.5946)^{2}}{15+15-2} = 1.8762[/tex]

The 99% confidence interval for the true mean difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is given by  

[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex], i.e.,

[tex](5.9333-7.6)\pm t_{0.005}1.3697\sqrt{\frac{1}{15}+\frac{1}{15}}[/tex]

where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (15+15-2) = 28 degrees of freedom. So

[tex]-1.6667\pm(-2.7633)(1.3697)(0.3651)[/tex], i.e.,

(-3.0486, -0.2848)

######

Using R

time1999 <- c(4, 5, 7, 7, 5, 7, 5, 6, 5, 6, 7, 8, 5, 6, 6)

time2009 <- c(5, 9, 5, 8, 7, 6, 7, 9, 7, 9, 6, 9, 10, 9, 8)

n1 <- length(time1999)

n2 <- length(time2009)

(mean(time1999)-mean(time2009))+qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)

(mean(time1999)-mean(time2009))-qt(0.005, df = 28)*sqrt(((n1-1)*var(time1999)+(n2-1)*var(time2009))/(n1+n2-2))*sqrt(1/n1+1/n2)

To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula: CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2)). Using R, the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours.

To find the 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999, we can use the formula:

CI = (x1 - x2) ± Z * sqrt((s1^2/n1) + (s2^2/n2))

where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Z is the critical value for a 99% confidence level.

Using R, we can calculate the confidence interval as follows:

The 99% confidence interval for the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is (0.4733, 2.1933) hours. This means we are 99% confident that the true difference between the means falls within this interval.

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Answer:

We conclude that the sonnets were written by by a certain Elizabethan poet.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8.9

Sample mean, [tex]\bar{x}[/tex] =10.2

Sample size, n = 6

Alpha, α = 0.05

Population standard deviation, σ = 2.5

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 8.88\\H_A: \mu > 8.88[/tex]

We use One-tailed z test to perform this hypothesis.

a) Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{10.2 - 8.9}{\frac{2.5}{\sqrt{6}} } = 1.28[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

b) We calculate the p value with the help of z-table.

P-value = 0.1003

The p-value is greater than the significance level which is 0.05

c) Since the p-value is greater than the significance level, there is not enough evidence to reject the null hypothesis and accept the null hypothesis.

Thus, we conclude that the sonnets were written by by a certain Elizabethan poet.

The z-score is 1.86 and the p-value is 0.0314. As the p-value is less than the level of significance α (0.05), we reject the null hypothesis and conclude that the new sonnets were likely written by another author.

In this statistical testing scenario for authorship of literary works, we need to find out the z-score or z test statistic and then determine the p-value to check if the new sonnets could be the works of the known Elizabethan poet or not.

For calculating the z score, you use the formula z = (x~ - μ) / (σ / √n) = (10.2 - 8.9) / (2.5/ √6) = 1.86 to two decimal places. The p-value is determined from the standard normal distribution table which for a z-score of 1.86 is 0.0314.

Given that α = 0.05, since the p-value is less than α, we reject the null hypothesis H0 (that the works were by the Elizabethan poet). Therefore, we accept the alternative hypothesis Ha (the sonnets were written by another author).

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Answer:Test 1 < Test 4

Step-by-step explanation: Test 1 was easier so it took less time to complete the questions. As the tests continue they get harder meaning they will take more time to answer.

Answer:

the answer is a

Step-by-step explanation:

i just took the test

Answer: There is sufficient evidence to support the claim that the bags are under-filled.

Step-by-step explanation:

Since we have given that

[tex]H_0:\mu=418\\\\H_a:\mu<418[/tex]

Sample mean = 411

Standard deviation = 20

n = 9

So, the test statistic value is given by

[tex]z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\z=\dfrac{411-418}{\dfrac{20}{\sqrt{9}}}\\\\\\z=\dfrac{-7}{\dfrac{20}{3}}\\\\\\z=-1.05[/tex]

At 0.025 level of significance,

critical value z = -2.306

since -2.306<-1.05

so, we will reject the null hypothesis.

Yes, there is sufficient evidence to support the claim that the bags are underfilled.

Answer:

Sample size should be atleast 55320

Step-by-step explanation:

Given that a  group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week.

population standard deviation [tex]\sigma = 6 hrs[/tex]

Std error = [tex]\frac{6}{\sqrt{n} }[/tex]

Margin of error for 95%

= [tex]1.96*\frac{6}{\sqrt{n} } <0.5\\\sqrt{n} >235.2\\n \geq 55319.04[/tex]

Since sample size is number of items it cannot be indecimal.

So we can round off to the next high integer on a safer side.

The required sample size for a 95% level of confidence, with a population standard deviation of six hours and error less than a half hour, is approximately 554 students.

The question is about finding the required sample size for a statistics project. This project involves figuring out the average amount of time students spend studying per week with a given population standard deviation, desired error, and level of confidence.

To solve this, we use the formula for sample size: n = [(Z_α/2*σ)/E]^2. Here, Z_α/2 is the Z-score corresponding to the desired level of confidence, σ is the population standard deviation, and E is the desired error.

For a 95% level of confidence, the Z-score is about 1.96 (you can find this value in Z-tables). Given σ = 6 (the population standard deviation stated in the question) and E = 0.5 (the desired error less than a half hour), we substitute these into the formula to get:

n = [(1.96*6)/0.5]^2

So, n ≈ 553.29. Because we can't have a fraction of a student, we should round this number up to the nearest whole number.

Therefore, "the required sample size is 554 students".

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Answer:

21, 144, 12

Step-by-step explanation:

Given that a sample of 30 distance scores measured in yards has a mean of 7, a variance of 16, and a standard deviation of 4.

Let X be the distance in yard.

i.e. each entry of x is multiplied by 3.

New mean variance std devition would be

E(3x) = [tex]3E(x) = 21[/tex]

Var (3x) = [tex]3^2 Var(x) = 9(16) =144[/tex]

Std dev (3x) = [tex]\sqrt{144 } =12[/tex]

Thus we find mean and std devition get multiplied by 3, variance is multiplied by 9

The new mean after converting the distances from yards to feet is 21, the median remains unchanged at 7, the new variance is 144, and the new standard deviation is 12.

To find the new mean, we multiply the original mean by the conversion factor. Since there are 3 feet in a yard, we have:

New Mean = Original Mean × Conversion Factor

 = 7 yards × 3 feet/yard

 = 21 feet

The median is a measure of central tendency that represents the middle value of a data set. When each data point is multiplied by a constant factor, the median is also multiplied by that factor. However, since the median is the middle value of an ordered list of data, and we are not changing the order or adding or removing any data points, the median in terms of yards and feet is the same numerical value, even though the units are different:

 New Median = [tex]Original Median Ã[/tex]— Conversion Factor

 = 7 yards (since the median is the same numerical value)

For the variance, when each data point is multiplied by a constant, the variance is multiplied by the square of that constant:

New Variance = [tex]Original Variance × (Conversion Factor)^2[/tex]

 = [tex]16 yards^2 × (3 feet/yard)^2[/tex]

 = [tex]16 yards^2 × 9 (feet^2/yard^2)[/tex]

 = [tex]144 feet^2[/tex]

Finally, the standard deviation is the square root of the variance, so to find the new standard deviation, we take the square root of the new variance:

New Standard Deviation = [tex]√New Variance[/tex]

 = [tex]√144 feet^2[/tex]

 = 12 feet

There is no significant difference in mean lead concentration for the medicines.

Hypothesis means an assumption or a claim which is needed to be tested.

To test the following hypothesis:

[tex]Null\ hypothesis,H_0: \mu=17\\Alternative\ Hypothesis,\ H_1: \mu < 17[/tex]

For the given data to test the hypothesis, t statistics can be used as follows:

[tex]t=\dfrac{\bar{X}-\mu}{\sigma}[/tex]

Given, Mean, μ=17

We are required to sample mean and  σ in the excel table shown

below:

t=(17-13.05)/6.23

t=0.634

Using excel function to calculate p value of the t statistics as follows:

=T.DIST(0.634,9,1)

p=0.729

Since, p value is greater than level of significance 0.01. We fail to reject the null hypothesis and conclude that there is no significant difference in mean.

Learn more about Hypothesis testing, here:

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To test the claim, calculate the mean and standard deviation of your sample data, set up your null and alternative hypotheses, calculate your test statistic t, reference a t-distribution table and compare your calculated t with the critical t-value.

To test the claim that the mean lead concentration for all such medicines is less than 17 mu​g/g, we'll use a one-sample t-test because we have one sample present and we're comparing it to a known mean. Here are the steps:

Remember, rejecting the null hypothesis supports our claim that the mean concentration is less than 17 mu​g/g. Not rejecting the null hypothesis means that our data does not support the claim that the mean concentration is less than 17 mu​g/g, but it isn't proof that the concentration exactly equals to or exceeds 17 mu​g/g.

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Answer: (16x - 2)inches

Step-by-step explanation:

The shape of Jake's window is a rectangle. It means that the two opposite sides are equal. The perimeter of the rectangular window is the distance round it.

The perimeter of a rectangle is expressed as 2(length + width)

From the dimensions given

Length = height of the window

= 5x - 3 inches

Width = 3x + 2 inches

Perimeter of the window =

2(5x - 3 + 3x + 2) = 10x - 6 + 6x + 4

= 10x + 6x + 4 - 6

Perimeter of the window =

(16x - 2)inches

Answer:The perimeter of window can be calculated by the following formula;

Step-by-step explanation:

Answer:

(b) 4,445

Step-by-step explanation:

If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?

Given a=0.05, |Z(0.025)|=1.96 (check standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.96/0.015)^2*0.5*0.5

=4268.444

Take n=4269

Answer:(b) 4,445

Answer:

  300 g

Step-by-step explanation:

There are 1000 grams in one kilogram. Dividing 7200 grams into 24 equal parts makes each part ...

  (7200 g)/(24 persons) = 300 g/person

Each person get 300 grams.

Answer:

The 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex]  

Step-by-step explanation:

1) Data given and notation  

n=400 represent the random sample taken    

X=320 represent the people drivers claimed they always buckle up

[tex]\hat p=\frac{320}{400}=0.8[/tex] estimated proportion of people drivers claimed they always buckle up

[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)    

Confidence =95% or 0.95

p= population proportion of people drivers claimed they always buckle up

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".  

2) Calculating the interval for the proportion

The confidence interval would be given by this formula  

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

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex]0.8 - 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.761[/tex]  

[tex]0.8 + 1.96 \sqrt{\frac{0.8(1-0.8)}{400}}=0.839[/tex]  

And the 95% confidence interval would be given (0.761;0.839). The error bound is [tex]Me=\pm 0.0392[/tex]  

We are confident that about 76.1% to 83.9% of people drivers that they always buckle up  at 95% of confidence

Answer:

Since, in the population function,

[tex]P = ab^t[/tex]

a = initial population,

b = population change factor,

If 0 < b < 1, then population will shrink,

While, if b > 1, then the population will grow,

(a) Since, 1.18, 1.08 and 1.191 is greater than 1,

Thus, town 4), 5) and 6) are growing.

(b) Since, 0.8, 0.77 and 0.98 are less than 1,

Thus, town 1), 2) and 3) are shrinking.

(c) An exponential growth function with highest change factor grows fastest.

∵ 1.191 > 1.18 > 1.08

⇒ town 6) is growing fastest.

(d) An exponential decay function with lowest change factor shrinks fastest,

∵ 0.77 < 0.8 < 0.98 < 1.08 < 1.18 < 1.191,

⇒ Town 2) shrinks fastest.

(e) Since,

2400 > 2100 > 1700 > 1100 > 900 > 600

⇒ town 1) has the largest initial population.

(f) Similarly,

Town 4) has the smallest initial population.

Answer:

a.

i, ii, and iv

Step-by-step explanation:

Answer:

0.6

Step-by-step explanation:

Given:

P( Person chosen at random has a B.S. degree), P(C) = 0.5

P( Person chosen at random does not have a B.S. degree), P(C') = 1 - 0.5 = 0.5

P(Student earns more than $100,000) = P(E)

P(Student earns more than $100,000, without going college) = P(E | C') = 0.4

P(Student earns more than $100,000, with college degree) = P(E | C) = 0.6

Now,

P(at least a B.S. degree | earns more than $100,000), P(C | E)

using Baye's theorem

we have

P(C | E) = [tex]\frac{P(C)\timesP(E | C)}{P(C)\timesP(E | C)+P(C')\timesP(E | C')}[/tex]

or

P(C | E) = [tex]\frac{0.5\times0.6}{0.5\times0.6+0.5\times0.4}[/tex]

or

P(C | E) = [tex]\frac{0.3}{0.5}[/tex]

or

P(C | E) = 0.6

To find the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000, we can use Bayes' Theorem and the given probabilities.

To find the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000, we can use Bayes' Theorem. Let A be the event that the person has at least a B.S. degree, and let B be the event that the person earns more than $100,000. The probability of A given B can be calculated as:

P(A|B) = (P(B|A) * P(A)) / P(B)

Given that P(B|A) = 0.6, P(A) = 0.5, and P(B) = 0.4, we can substitute the values into the formula:

P(A|B) = (0.6 * 0.5) / 0.4 = 0.75

Therefore, the probability that a person has at least a B.S. degree if it is known that he or she earns more than $100,000 is 0.75.

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Answer:

The cylinder is given straight away by x^2+y^2=r^2=16\implies r=4. To get the cylinder, we complete one revolution, so that 0\le\theta\le2\pi. The upper limit in z is a spherical cap determined by

x^2+y^2+z^2=144\iff z^2=144-r^2\implies z=\sqrt{144-r^2}

So the volume is given by

\displaystyle\iiint_{\mathcal D}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=0}^{z=\sqrt{144-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta

and has a value of \dfrac{128(27-16\sqrt2)\pi}3 (not that we care)

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Step-by-step explanation:

Answer:

The answer is: y = 2/3x - 3

Step-by-step explanation:

Given point: (3, -1)

Given equation: y = 2/3x - 5, which is in the form y = mx + b where m is the slope and b is the y intercept.

Parallel lines have the same slope. Use the point slope form of the equation with the point (3, -1) and substitute:

y - y1 = m(x - x1)

y - (-1) = 2/3(x - 3)

y + 1 = 2/3x - 6/3

y + 1 = 2/3x - 2

y = 2/3x - 3

Proof:

f(3) = 2/3(3) - 3

= 6/3 - 3

= 2 - 3

= -1, giving the point (3, -1)

Hope this helps! Have an Awesome Day!!  :-)

Answer: The critical value is: insufficient.

Step-by-step explanation: Refer to Exhibit 9-2. The p-value is:

a. .0500  

b. .0228  

c. .0456  

d. .0250

In order to test the hypotheses H0: μ ≤ 100 and Ha: μ > 100 at an α level of significance, the null hypothesis will be rejected if the test statistic z is:

The practice of concluding "do not reject H0" is preferred over "accept H0" when we:

a. have an insufficient sample size.  

b. have not controlled for the Type II error.  

c. are testing the validity of a claim.  

d. are conducting a one-tailed test.

In hypothesis testing, the critical value is: insufficient

Answer:

 h /r  =  2.55

Step-by-step explanation:

Area of a can:

 Total area of the can = area of (top + bottom)  + lateral area

lateral area  2πrh    without waste

area of base (considering that you use 2r square)  is 4r²

area of bottom      ( for same reason )                           4r²

Then Total area  =  8r² + 2πrh

Now can volume is   1000 = πr²h        h = 1000/πr²

And  A(r)  =   8r² + 2πr(1000)/πr²

A(r)  =   8r² + 2000/r

Taking derivatives both sides

A´(r)   =  16 r    - 2000/r²

If A´(r)  =  0              16 r    - 2000/r²   = 0

(16r³  - 2000)/ r²  =  0        16r³  - 2000  =  0

r³  =  125      

r = 5 cm      and      h  = ( 1000)/ πr²      h  = 1000/ 3.14* 25

h = 12,74  cm

ratio   h /r  =  12.74/5         h /r  =  2.55

To find the ratio of h to r for the most economical can, we need to minimize the amount of aluminum used. The ratio is 0.

To find the ratio of h to r for the most economical can, we need to minimize the amount of aluminum used. The total amount of aluminum used is given by the equation A = 8r^2 + 2πrh. To minimize A, we can take the derivative of A with respect to h and set it equal to 0. Solving this equation will give us the value of h in terms of r, and we can then find the ratio of h to r.

Taking the derivative of A with respect to h, we get dA/dh = 2πr. Setting this equal to 0 and solving for h, we find that h = 0. To find the ratio of h to r, we divide both sides of the equation by r, giving us h/r = 0.

Therefore, the ratio of h to r for the most economical can is 0.

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Answer:

You are expected to lose $0.05 (or win -$0.05)

Step-by-step explanation:

Since the roulette wheel has the numbers 1 through 36, 0, and 00, there are 38 possible outcomes.

In this bet, you are allowed to pick 3 out of the 38 numbers. Thus, your chances of winning (P(W)) and losing (P(L)) are:

[tex]P(W)=\frac{3}{38}\\P(L) = 1 - P(W)\\P(L) = \frac{35}{38}\\[/tex]

The expected value of the bet is given by the sum of the product of each outcome pay by its probability. Winning the bet means winning $11 while losing the bet means losing $1. The expected value is:

[tex]EV = (11*\frac{3}{38}) -(1*\frac{35}{38})\\EV = -\$0.0525[/tex]

Therefore, with a $1 bet, you are expected to lose roughly $0.05

Answer:

  x = 55°

Step-by-step explanation:

The sum of the angles in a triangle is 180°, so you have ...

  x + 50° +75° = 180°

  x = 55° . . . . . . . . . . . . subtract 125°

Answer:

p value = 0.3122

Step-by-step explanation:

Given that Pedro thinks that he has a special relationship with the number 3. In particular,

Normally for a die to show 3, probability p = [tex]\frac{1}{6} =0.1667[/tex]

Or proportion p = 0.1667

Pedro claims that this probability is more than 0.1667

[tex]H_0 = P =0.1667______H_a = P>0.1667_____[/tex]

where P is the sample proportion.

b) n=30 and [tex]P = 6/30 =0.20[/tex]

Mean difference = [tex]0.2-0.1667=0.0333[/tex]

Std error for proportion = [tex]\sqrt{\frac{p(1-p)}{n} } \\=0.0681[/tex]

Test statistic Z = p difference/std error =  0.4894

p value = 0.3122

p >0.05

So not significant difference between the two proportions.

The null hypothesis is that Pedro's claim is not true, while the alternative hypothesis is that Pedro's claim is true. The P-value is approximately 0.082.

(a) The null hypothesis states that Pedro's claim is not true, so the null hypothesis is H0: p = 1/6. The alternative hypothesis states that Pedro's claim is true and he rolls a 3 more often than expected, so the alternative hypothesis is Ha: p > 1/6.

(b) To determine the P-value, we can use the binomial distribution. The probability of rolling a 3 with a fair 6-sided die is 1/6. Using the binomial distribution, we can calculate the probability of rolling a 3 six times out of 30 rolls. This gives a P-value of approximately 0.082, which is the probability of observing a result as extreme as the one obtained or more extreme, assuming the null hypothesis is true.

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Answer:

92

Step-by-step explanation:

She is taking four tests. She wants to average a 95 on those tests. Therefore, she must score 95*4=380 total. She scored 97 and 91, or 188 total. Therefore, she must score 380-188 = 192 total on her next two tests. Her maximum score possible on the fourth test is 100. Therefore, the lowest score possible for the third test would be 192-100=92

The lowest possible score Isabella could have made on the third test

is 92 marks.

A numerical expression is a mathematical statement written in the form of numbers and unknown variables. We can form numerical expressions from statements.

Given, Isabella must take four 100-point tests in her math class and her goal is to achieve an average grade of 95 on the tests.

So, The total marks she needs to score in four games is (4×95).

= 380.

Now, She scores 97 and 91 in her first and second tests.

So, She needs to score

380 - (97 + 91)

= 380 - 188.

= 192 in her two tests.

Now, The highest marks she can score in the fourth test is 100.

Therefore, The lowest possible marks she can score is,

= 192 - 100.

= 92.

learn more about numerical expressions here :

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Answer: 0.547

Step-by-step explanation:

As per given , we have

Population mean = [tex]\mu=20[/tex]

Population standard deviation= [tex]\sigma=4[/tex]

Sample size : n= 36

We assume that number of apps used per month by smartphone owners is normally distributed.

Let [tex]\overline{x}[/tex] be the sample mean.

Formula : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

The probability that the sample mean is between 19.5 and 20.5 :-

[tex]P(19.5<\overline{x}<20.5)\\\\=P(\dfrac{19.5-20}{\dfrac{4}{\sqrt{36}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{20.5-20}{\dfrac{4}{\sqrt{36}}})\\\\=P(-0.75<z<0.75)\\\\=P(z<0.75)-P(z<-0.75)\ \ [\because P(z_1<z<z_2)=P(z<z_2)-P(z<z_1)]\\\\=P(z<0.75)-(1-P(z<0.75))\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=2P(z<0.75)-1=2(0.7734)-1=0.5468\approx0.547[/tex]

[using standard normal distribution table for z]

Hence, the required probability = 0.547

To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.

To find the probability that the sample mean is between 19.5 and 20.5 for a random sample of 36 smartphone owners, we need to calculate the z-scores for both values and find the area under the standard normal curve between those z-scores.

The formula for calculating the z-score is: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

Using the given information, the z-scores for 19.5 and 20.5 are: z1 = (19.5 - 20) / (4 / √36) = -0.75 and z2 = (20.5 - 20) / (4 / √36) = 0.75.

Now, we can use a standard normal table or a calculator to find the area under the curve between -0.75 and 0.75. The probability is approximately 0.467.

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Answer: It takes 15.8 seconds to slide down a 2000 ft slope.

Step-by-step explanation:

Since we have given that

[tex]t=\sqrt{\dfrac{d}{16\sin\theta}}[/tex]

where, t is the time taken,

d is the length to slide down a slope

sin θ is the angle at which it inclined.

So, we have d = 2000 ft

θ = 30°

So, the time taken is given by

[tex]t=\sqrt{\dfrac{2000}{16\sin 30^\circ}}\\\\t=\sqrt{\dfrac{2000}{8}}\\\\t=\sqrt{250}\\\\t=15.81\ seconds[/tex]

Hence, it takes 15.8 seconds to slide down a 2000 ft slope.

The time it takes to slide down a 2000 ft slope inclined at 30° is 250 seconds, calculated using the equation t = d / (16 sin θ).

We're given the equation for calculating time of a sled sliding down a slope as t = d / (16 sin θ). Here, d is the length of the slope, and θ is the incline of the hill slope.

To calculate the time it takes to slide down a 2000 ft slope inclined at 30°, we substitute the respective values into the formula: t = 2000 / (16 sin 30). The sin value for 30° is 0.5, so the formula simplifies to t = 2000 / (16 * 0.5), which yields t = 2000 / 8. By performing that division, we see that t = 250 seconds. Therefore, it takes 250 seconds to slide down a 2000 ft slope inclined at 30°.

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