Ten experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:
34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.
1. What is the mean deviation of the ratings?
Select one:
a. 8.00
b. 4.12
c. 12.67
d. 0.75

The tangent to [tex]r(\theta)=e^\theta[/tex] has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], where

[tex]\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}[/tex]

By the chain rule, we have

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}[/tex]

and by the product rule,

[tex]\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta[/tex]

so that with [tex]\frac{\mathrm dr}{\mathrm d\theta}=e^\theta[/tex], we get

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}[/tex]

The tangent line is horizontal when the slope is 0; this happens for

[tex]-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi[/tex]

where [tex]n[/tex] is any integer. In the interval [tex]0\le\theta\le2\pi[/tex], this happens for [tex]n=1,2[/tex], or

[tex]\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4[/tex]

i.e at the points

[tex](r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)[/tex]

and

[tex](r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)[/tex]

Answer:

The obtained value of the appropriate statistic is ____.

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]  

d. 3.06

[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]  

d. reject H0; the class appeared to improve study habits

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.  

Let put some notation  

x=before method , y = after method  

x: 10 12 14 16 12

y: 15 14 17 17 20

The system of hypothesis for this case are:  

Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]  

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]  

The first step is define the difference [tex]d_i=y_i-x_i[/tex], that is given so we have:

d: 5,2,3,1,8

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=3.8[/tex]  

The third step would be calculate the standard deviation for the differences, and we got:  

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =2.775[/tex]  

The fourth step is calculate the statistic given by :  

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]  

The next step is calculate the degrees of freedom given by:  

[tex]df=n-1=5-1=4[/tex]  

Now we can calculate the p value, since we have a two tailed test the p value is given by:  

[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]  

The p value is less than the significance level given [tex]\alpha=0.05[/tex], so then we can conclude that we reject the null hypothesis.

d. reject H0; the class appeared to improve study habits

Answer:

[tex]p_v =2*P(Z>2.4)=0.016[/tex]  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"

Step-by-step explanation:

1) Data given and notation

n represent the random sample taken

X represent the business students who have personal computers (PC's) at home

[tex]\hat p[/tex] estimated proportion of business students who have personal computers (PC's) at home

[tex]p_o=0.25[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.25:  

Null hypothesis:[tex]p=0.25[/tex]  

Alternative hypothesis:[tex]p \neq 0.25[/tex]  

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

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

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

3) Calculate the statistic  

For this case the value of the statistic is given by z=2.4 and that's given.

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>2.4)=0.016[/tex]  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"

Answer:

There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes (P-value=0.009).

Step-by-step explanation:

We have to perform an hypothesis test on the mean.

The null and alternative hypothesis are:

[tex]H_0: \mu=10\\\\H_1: \mu \neq 10[/tex]

The significance level is [tex]\alpha=0.05[/tex].

The test statistic t can be calculated as:

[tex]t=\frac{M-\mu}{s/\sqrt{N} } =\frac{11.2-10}{3.1/\sqrt{50} }=2.737[/tex]

The degrees of freedom are:

[tex]df=N-1=50-1=49[/tex]

The P-value (two-tailed test) for t=2.737 and df=49 is P=0.00862.

This P-value (0.009) is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes.

Final answer:

A hypothesis test can be conducted to test the claim made in the commercial for the new Meat Man Barbecue. The test results indicate that there is evidence to support that the population mean assembly time is not equal to 10 minutes.

Explanation:

To test the claim made in the commercial for the new Meat Man Barbecue, a hypothesis test can be conducted. The null hypothesis, H0, states that the population mean assembly time is 10 minutes, while the alternative hypothesis, Ha, states that it is not equal to 10 minutes. The test statistic to use in this case is the t-test, as we are comparing the sample mean to a known value. The p-value for this test is 0.009, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that there is evidence to support that the population mean assembly time is not equal to 10 minutes.

Answer:

x=5/7

Step-by-step explanation:

3(x-2)+7×=1/2×(6×-2)

3x-6+7×=1/2×2(3×-1)

3×-6+7×=3×-1

-6+7×=-1

7×=-1+6

7×=5

Answer: [tex]\dfrac{7}{8}[/tex]

Step-by-step explanation:

Given : When a coin is tossed twice , the total outcomes will become

HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT

Total outcomes = 8

Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT

Number of favorable outcomes = 7

Probability that the number of tails obtained will be between 1 and 3 inclusive = [tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]

[tex]=\dfrac{7}{8}[/tex]

Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive [tex]=\dfrac{7}{8}[/tex]

Answer:

B. Goodness-of-Fit Test

Step-by-step explanation:

Given that a professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10.

She records the actual distribution of grades and wants to compare it to the normal distribution

For this expected values are to be got assuming normal and the observed and expected are used to calculate chi square.

Depending on the chi square statistic conclusion is made

Here the test to be done is

B) Goodness of fit test.

A is wrong because test of independence is done when there are more than one categorical variable in the rows.

C is wrong because here homogeneity is not tested

Only B is right.

Answer:

5.63

Step-by-step explanation:

The expected value of the given probability distribution is calculated by multiplying each value by its probability and summing the results, yielding an expected value of 3.25.

To find the expected value of the given probability distribution, you multiply each value of X by its corresponding probability P(X) and then sum those products. Mathematically, the expected value E(X) is calculated as:

The expected value is 3.25, rounded to two decimal places as instructed.

Answer:

Step-by-step explanation:

At the significance level given, the conclusion is B. Reject the null hypothesis.

The first thing to do is to calculate the test statistic. This will be 1.98 from the z table.

In this case, the rejection region will be when the number is more than 1.28. Here, 1.98 is more than 1.28.

Therefore, the best thing to do is to simply reject the null hypothesis.

Learn more about hypothesis on:

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Answer:

A sample size of at least 328 students is required.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the width M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

[tex]M = 10, \sigma = 110[/tex]

So:

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]10 = 1.645*\frac{110}{\sqrt{n}}[/tex]

[tex]10\sqrt{n} = 180.95[/tex]

[tex]\sqrt{n} = 18.095[/tex]

[tex]n = 327.43[/tex]

A sample size of at least 328 students is required.

Answer:

A sample size of at least 328 students is required.

Step-by-step explanation:

We have that to find our  level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the width M as such

In which  is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

So:

A sample size of at least 328 students is required.

Answer:

[tex]P(\bar X <m)=0.844[/tex]

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mass of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

And the z score is defined as:

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

We know that for some amount  m we have this:

[tex]P(X>m)=0.4[/tex]   (a)

[tex]P(X<m)=0.6[/tex]   (b)

We can use the z table or excel in order to find a quantile that satisfy the two conditions. The excel code would be:

"=NORM.INV(0.6,0,1)" and using condition (b) we have that Z=0.253

So we have this:

[tex]0.253=\frac{m-\mu}{\sigma}[/tex]

If we solve for m from the last equation we got:

[tex]m=0.253\sigma +\mu [/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

And we want this probability:

[tex]P(\bar X<m)[/tex]

We can apply the z score formula for this case given by:

[tex]z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{m-\mu}{\frac{\sigma}{\sqrt{16}}})[/tex]

[tex]P(Z<\frac{4(m-\mu)}{\sigma}[/tex]

If we use the expression obtained for m we got:

[tex]P(Z<\frac{4(0.253\sigma +\mu-\mu)}{\sigma}[/tex]

[tex]P(Z<\frac{1.012 \sigma}{\sigma})=P(Z<1.012)=0.844[/tex]

Answer: x = 4 , y = -3

Step-by-step explanation:

Going by the Cramer's rule , we first determine the determinant by dealing with only  the coefficients of x and y in the 2 x 2 matrix.

2x  -  3y = 17

5x  + 4y = 8

2        -3

5         4,   going by the rule now, we now have

(2 x 4) - (5 x -3)

8 + 15

= 23.

Now to find the value of x , replace the constants with the coefficient of x and divide by he determinants.

 17        -3

 8          4

---------------  

2           -3

5            4

( 17 x 4 ) - ( 8 x -3 )

---------------------------

         23

     =  68 + 24

         ------------

              23

    =       92/23

    =           4.

 x =   4

Now  to find y, just repeat the process by replacing the coefficient of y with the constants.

2        17

5        8

-----------

2        -3

5         4

( 2 x 8 ) - ( 5 x 17 )

-----------------------

            23

       16 - 85

       ---------

          23

  =  -69/23

  =   -3

y =  -3.

check

substitute for the values in any of the equations above.

2(4)  - 3(-3)

8  + 9

= 17

Answer:

(B) 0.2843

Step-by-step explanation:

Let d be the difference in proportions from Country X and Country Y who were first married between the ages of 18 and 32.

Then hypotheses are

[tex]H_{0}[/tex]: d=0.15

[tex]H_{a}[/tex]: d<0.15

Test statistic can be found using the equation

[tex]z=\frac{p1-p2-0.15}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

Then [tex]z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}[/tex] ≈ 0.5646

p-value of test statistic is ≈ 0.2843

p-value states the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is at least 0.15

To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions. The answer is (E) 0.7157.

To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions.

The standard error of the difference in proportions can be calculated as:

SE = sqrt((p1 * (1 - p1)) / n1 + (p2 * (1 - p2)) / n2)

where p1 and p2 are the proportions of first married adults from Country X and Country Y respectively, and n1 and n2 are the sample sizes for Country X and Country Y.

Let's plug in the values:

SE = sqrt((0.36 * (1 - 0.36)) / 60 + (0.26 * (1 - 0.26)) / 50)

SE = sqrt(0.0024 / 60 + 0.0016 / 50)

SE = sqrt(0.00004 + 0.00003)

SE = sqrt(0.00007)

SE ≈ 0.00836660027

Now, we can use the standard normal distribution to find the probability that the difference in proportions is greater than 0.15 by calculating the z-score and looking it up in the z-table or using a calculator.

Z = (0.15 - 0) / 0.00836660027

Z ≈ 17.914253

The probability that the difference in proportions is greater than 0.15 is very close to 1, which means it is highly likely that the difference in proportions will be greater than 0.15 in a random sample. Therefore, the answer is (E) 0.7157.

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Answer:

99% CI: [45.60; 58.00]min

Step-by-step explanation:

Hello!

Your study variable is:

X: Time a customer stays in a certain restaurant. (min)

X~N(μ; σ²)

The population standard distribution is σ= 17 min

Sample n= 50

Sample mean X[bar]= 51.8 min

Sample standard deviation S= 27.68

You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:

X[bar] ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)

[tex]Z_{1-\alpha /2} = Z_{0.995}= 2.58[/tex]

Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min

Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min

With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.

I hope you have a SUPER day!

PS: Missing Data in the attached files.

Answer:

44.89 - 57.27

Answer:

See proof below.

Step-by-step explanation:

Remember that a collection of subsets B of a set X is a base for a topology of X if the following conditions hold

In this case, X=R and B={[a,b]: a is rational and b is irrational}. Let's prove the statements above

Final answer:

To calculate the probability that Kim wins the match, we sum the probabilities of winning in 2 sets, 3 sets, and 4 sets. The probability that Kim wins the match is 0.6553. The probability that Kim wins the match in exactly 2 sets is 0.2339. The probability that 3 sets are played is 0.1728.

Explanation:

To calculate the probability that Kim wins the match, we need to consider the different possible outcomes. Kim can win the match in two sets, which happens with a probability of [tex](0.64)^2 = 0.4096[/tex] . Kim can also win the match in three sets, which happens with a probability of [tex](0.64)^2 * (1-0.64) * (1-0.64) = 0.1728[/tex]. And finally, Kim can win the match in four sets, which happens with a probability of [tex](0.64)^2 * (1-0.64)^2 = 0.0729.[/tex] Therefore, the probability that Kim wins the match is the sum of these probabilities: 0.4096 + 0.1728 + 0.0729 = 0.6553.

To calculate the probability that Kim wins the match in exactly 2 sets, we use the probability of winning a set twice and the probability of losing a set once:[tex](0.64)^2 * (1-0.64) = 0.2339.[/tex]

To calculate the probability that 3 sets are played, we need to consider the different possible outcomes. Kim can win the match in three sets, which happens with a probability of [tex](0.64)^2[/tex]* (1-0.64) * (1-0.64) = 0.1728. Therefore, the probability that 3 sets are played is 0.1728.

The first three nonzero nodal points are at[tex]\( \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2} \),[/tex] where the factors multiplying [tex]\( \lambda \)[/tex] are [tex]\( \frac{1}{2}, 1, \frac{3}{2} \).[/tex]

To find the first three nodal points of the displacement of the string [tex]\( y(x, t) \)[/tex], where x > 0, we need to consider the standing wave pattern formed on the string. In a standing wave, the displacement is zero at certain points along the string for all times.

For a standing wave on a string fixed at both ends, the displacement y(x, t) can be expressed as:

[tex]\[ y(x, t) = A \sin(kx) \cos(\omega t) \][/tex]

Where:

- A is the amplitude of the wave.

- k is the wave number.

- [tex]\( \omega \)[/tex] is the angular frequency.

The nodal points occur where y(x, t) = 0. Since [tex]\( \cos(\omega t) \)[/tex] oscillates between -1 and 1, we need [tex]\( \sin(kx) = 0 \)[/tex] for the entire expression to be zero.

For [tex]\( \sin(kx) = 0 \)[/tex], kx must be a multiple of [tex]\( \pi \)[/tex] (since[tex]\( \sin(\pi n) = 0 \)[/tex] for integer n ). So, we have:

[tex]\[ kx = n \pi \][/tex]

Solving for x, we get:

[tex]\[ x = \frac{n \pi}{k} \][/tex]

Since [tex]\( k = \frac{2\pi}{\lambda} \)[/tex], where [tex]\( \lambda \)[/tex] is the wavelength, we can rewrite x as:

[tex]\[ x = \frac{n \lambda}{2} \][/tex]

Where n is an integer.

The first three nonzero nodal points occur when n = 1, 2, and 3. Substituting these values into the equation for x, we get:

1. [tex]\( x_1 = \frac{\lambda}{2} \)[/tex]

2. [tex]\( x_2 = \lambda \)[/tex]

3. [tex]\( x_3 = \frac{3\lambda}{2} \)[/tex]

So, the factors that multiply [tex]\( \lambda \)[/tex] in increasing order are [tex]\( \frac{1}{2}, 1, \) and \( \frac{3}{2} \)[/tex].

Let's summarize the steps:

1. Express displacement: Write down the displacement expression for a standing wave on a string.

2. Identify nodal points: Find where the displacement is zero by setting [tex]\( \sin(kx) = 0 \)[/tex].

3. Express nodal points: Use [tex]\( k = \frac{2\pi}{\lambda} \)[/tex] to express the nodal points in terms of the wavelength [tex]\( \lambda \)[/tex].

4. Determine factors of [tex]\( \lambda \)[/tex]: Identify the factors that multiply [tex]\( \lambda \)[/tex] in the nodal points.

5. List the factors: Arrange the factors in increasing order and list them.

The first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are

[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]

[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]

[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]

To find the first three nodal points closest to [tex]\(x = 0\)[/tex] with [tex]\(x > 0\)[/tex], where the displacement of the string [tex]\(y(x, t)\)[/tex] is zero for all times, we can use the formula for the nth nodal point of a standing wave on a string fixed at both ends:

[tex]\[ x_n = \frac{n}{2} \lambda \][/tex]

where \( n \) is the nodal number ([tex]1, 2, 3[/tex], ...), [tex]\( \lambda \)[/tex] is the wavelength of the wave, and [tex]\( x_n \)[/tex] is the nth nodal point.

Since we are looking for the first three nonzero nodal points, we will calculate [tex]\( x_1 \), \( x_2 \)[/tex], and [tex]\( x_3 \)[/tex] using [tex]\( n = 1 \), \( n = 2 \),[/tex] and [tex]\( n = 3 \),[/tex] respectively.

The factors that multiply [tex]\( \lambda \)[/tex] in increasing order for the first three nonzero nodal points are [tex]1[/tex], [tex]2[/tex], and [tex]3[/tex].

Therefore, the first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are:

[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]

[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]

[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]

These nodal points are located at [tex]\( \frac{1}{2} \)[/tex], [tex]1[/tex], and [tex]\( \frac{3}{2} \)[/tex] times the wavelength away from the origin, respectively.

The probability that the number in a sample of 130 who do so differs from the expected value is 0.6198.

From the information, the critical value is given as 84. Therefore, the mean will be:

= np = 130 × 2/3 = 86.67

The standard deviation is also 5.3748. The corresponding z score will be:

= (84 - 86.67)/5.3748

= -0.50.

Therefore, the left tailed area will be:

P(z < -0.50) = 0.3099

Since, it's two tailed, we'll multiply by 2. This will be:

= 2 × 0.3099

= 0.6198

In conclusion, the probability is 0.6198.

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The probability that the number differs from the expected value by at least as much is approximately 0.620 (rounded to three decimal places).

To determine the probability that the number in a sample of 130 kissing couples who lean to the right differs from the expected value by at least as much as observed, we will conduct a hypothesis test using the binomial distribution. We can then approximate this with a normal distribution since the sample size is large.

Given:

- Sample size [tex](\( n \)) = 130[/tex]

- Observed number of right-leaning couples [tex](\( X \)) = 84[/tex]

- Expected proportion of right-leaning couples [tex](\( p \)) = \(\frac{2}{3}\)[/tex]

- Significance level [tex](\( \alpha \))[/tex] = 0.05

Step 1: Calculate the expected number and standard deviation

The expected number of right-leaning couples is:

[tex]\[ \mu = n \cdot p = 130 \cdot \frac{2}{3} \approx 86.67 \][/tex]

The standard deviation is:

[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{130 \cdot \frac{2}{3} \cdot \frac{1}{3}} \approx \sqrt{130 \cdot \frac{2}{9}} \approx \sqrt{28.89} \approx 5.375 \][/tex]

Step 2: Calculate the z-score for the observed value

The z-score measures how many standard deviations the observed value is away from the expected value:

[tex]\[ z = \frac{X - \mu}{\sigma} = \frac{84 - 86.67}{5.375} \approx \frac{-2.67}{5.375} \approx -0.497 \][/tex]

Since we are interested in the probability that the number differs from the expected value by at least as much as observed, we need to consider both tails of the distribution.

Step 3: Calculate the probability using the standard normal distribution

The probability for a z-score of [tex]\( \pm 0.497 \)[/tex] can be found using the standard normal distribution table or a calculator. For z = 0.497 :

[tex]\[ P(Z \leq 0.497) \approx 0.690 \][/tex]

Since we are considering both tails:

[tex]\[ P(|Z| \geq 0.497) = 2 \cdot (1 - 0.690) = 2 \cdot 0.310 = 0.620 \][/tex]

Step 4: Conclusion

The probability that the number of right-leaning couples in a sample of 130 differs from the expected value by at least as much as observed is approximately 0.620.

So, the rounded answer to three decimal places is:

[tex]\[ \boxed{0.620} \][/tex]

Answer:

The best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864

Step-by-step explanation:

Consider the provided information.

he linear correlation coefficient is r= 0.765 and the regression equation is [tex]\hat y= 3.46 + 1.01\times x[/tex]

The mean of the 16 amounts of sugar is 0.295 grams and the mean of the 16 calorie counts is 3.76.We need to find the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g

Substitute x = 0.40 in the given equation.

[tex]\hat y= 3.46 + 1.01\times 0.40[/tex]

[tex]\hat y= 3.46 + 0.404[/tex]

[tex]\hat y= 3.864[/tex]

Hence, the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864

The solution is price of 1 slice of pizza  is $ 1.75 and price of 1 bottle of water is $ 1.25

Solution:

Let "p" be the price of 1 slice of pizza

Let "b" be the price of 1 bottle of water

Given that Abby bought two slices of pizza and  three bottles of water for $7.25

So we can frame a equation as:

two slices of pizza x price of 1 slice of pizza  + three bottles of water x price of 1 bottle of water = $ 7.25

[tex]2 \times p + 3 \times b = 7.25[/tex]

2p + 3b = 7.25 ------- eqn 1

Cameron bought four slices of pizza and  one bottle of water for $8.25

So we can frame a equation as:

four slices of pizza x price of 1 slice of pizza  + one bottles of water x price of 1 bottle of water = $ 8.25

[tex]4 \times p + 1 \times b = 8.25[/tex]

4p + 1b = 8.25 ----- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "p" and "b"

Multiply eqn 1 by 2

4p + 6b = 14.5 ----- eqn 3

Subtract eqn 2 from eqn 3

4p + 6b = 14.5

4p + 1b = 8.25

( - ) ------------------

5b = 6.25

Substitute b = 1.25 in eqn 1

2p + 3b = 7.25

2p + 3(1.25) = 7.25

2p + 3.75 = 7.25

2p = 3.5

Thus the solution is:

price of 1 slice of pizza  = $ 1.75

price of 1 bottle of water = $ 1.25

The cost of one slice of pizza  is $1.75. The cost of one bottle of water  $1.25.

To find the cost of one slice of pizza and one bottle of water, we can set up a system of linear equations based on the information given:

Let p be the cost of one slice of pizza and w be the cost of one bottle of water.

1. Abby's purchase: 2p + 3w = 7.25

2. Cameron's purchase:  4p + 1w = 8.25

We can solve this system using either substitution or elimination. Let's use elimination.

Step 1: Eliminate \w

First, let's multiply the second equation by 3 to align the coefficients of w

3(4p + 1w) = 3(8.25)

12p + 3w = 24.75

Now our system of equations looks like this:

1. 2p + 3w = 7.25

2. 12p + 3w = 24.75

Step 2: Subtract the first equation from the second

12p + 3w) - (2p + 3w) = 24.75 - 7.25

10p = 17.50

Step 3: Solve for P

[tex]\[ p = \frac{17.50}{10} \][/tex]

p = 1.75

So, one slice of pizza costs $1.75

Step 4: Solve for w

Now substitute p = 1.75 back into the first equation:

2(1.75) + 3w = 7.25

w = 1.25

So, one bottle of water costs $1.25

Answer:

The average cost function is [tex]\bar C(x)=-\frac{1200}{x}+34[/tex].

The cost function is [tex]C(x)=-1200+34x[/tex].

The fixed costs are -$1200.

Step-by-step explanation:

If [tex]y = C(x)[/tex] is the total cost of producing x items, then the average cost, or cost per unit, is [tex]\bar C(x)=\frac{C(x)}{x}[/tex] and the marginal average cost is [tex]\bar C'(x)=\frac{d}{dx}\bar C(x)[/tex].

We know that the marginal average cost is given by

[tex]\bar C'(x)=\frac{1200}{x^2}[/tex]

[tex]\bar C'(x)=\frac{1200}{x^2}\\\\\bar C(x)=\int {\frac{1200}{x^2}} \, dx[/tex]

[tex]\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\1200\cdot \int \frac{1}{x^2}dx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\1200\cdot \frac{x^{-2+1}}{-2+1}\\\\-\frac{1200}{x}\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\-\frac{1200}{x}+C[/tex]

Because [tex]\bar C(100)=22[/tex] C is

[tex]22=-\frac{1200}{100}+C\\C=34[/tex]

And the average cost function is

[tex]\bar C(x)=-\frac{1200}{x}+34[/tex]

[tex]x\cdot \bar C(x)=C(x)\\\\C(x)=(-\frac{1200}{x}+34)\cdot x\\\\C(x)=-1200+34x[/tex]

Cost functions are express as

[tex]C=(fixed \:costs)+(variable \:costs)\\C=a+bx[/tex]

According with this definition and the cost function that we have the fixed costs are -$1200.

To find the average cost function, use the given formula and substitute the value of x. To find the cost function, integrate the average cost function. The fixed costs are $22.

To find the average cost function, we will substitute the value of C^overbar(x) into the formula. The average cost function is given by C^overbar(x) = 1,200/x^2.

To find the cost function, we will integrate the average cost function. The cost function m C(x) is the integral of C^overbar(x) with respect to x. F

inally, to find the fixed costs, we will substitute the value of C^overbar(100) into the average cost function.

The average cost function is C^overbar(x) = 1,200/x^2.

The cost function m C(x) = 1,200/x.

The fixed costs are $22.

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Answer: (0.881, 0.919)

Step-by-step explanation:

Let p be the population proportion of respondents who say the Internet has been a good thing for them personally.

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where z* = Critical value ,

[tex]\hat{p}[/tex] = Sample proportion

n = sample size.

As per given , we have

n= 957

[tex]\hat{p}=0.90[/tex]

By z-table , Critical value for 95% confidence interval : z*= 1.96

Then, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. will be :

[tex]0.90\pm 1.96\sqrt{\dfrac{0.90(1-0.90)}{957}}[/tex]

[tex]0.90\pm 1.96\sqrt{0.0000940438871473}[/tex]

[tex]0.90\pm 1.96(0.00969762275753)[/tex]

[tex]\approx0.90\pm 0.019\\\\=(0.90-0.019,\ 0.90+0.019)[/tex]

[tex](0.881,\ 0.919)[/tex]

Hence, the required 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally = (0.881, 0.919)

To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula: CI = p ± Z * sqrt((p * (1 - p)) / n). Plugging in the values, we can calculate a 95% confidence interval of [0.8808, 0.9192].

To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula:

CI = p ± Z * sqrt((p * (1 - p)) / n)

Where:

Plugging in the values, we can calculate:

CI = 0.90 ± 1.96 * sqrt((0.90 * (1 - 0.90)) / 957)

This gives us a 95% confidence interval for the proportion, which is [0.8808, 0.9192]. So we can be 95% confident that the true proportion of respondents who say the Internet has been a good thing for them is between 0.8808 and 0.9192.

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Answer:

Collection of numerical information is called Statistics

Step-by-step explanation:

From the Oxford dictionary, the definition of statistics (science) is: "the practice or science of collecting and analyzing numerical data in large quantities, especially to infer proportions in a whole from those in a representative sample".

While the definition of statistic (data) is: "a fact or piece of data obtained from a study of a large quantity of numerical data".

Statistics is a science, not a method, therefore answer can´t be a).

Not all sample data is necessarily a statistic, therefore answer can´t be b).

All statistics are numerical information.

Not all statistics are population data, therefore answer can´t be d).

Cleaned data is information filtered with a specific criterion, but not necessarily used in a statistic.

The most probable answer is c.

Answer:

Square the expressions to see the difference.

Step-by-step explanation:

[tex]$ \sqrt{(m + n)} $[/tex].

Squaring this we have: [tex]$ (\sqrt{m + n})^2 = m + n $[/tex]

Now, [tex]$ \sqrt{m} + \sqrt{n} $[/tex]

Squaring this we get: [tex]$ (\sqrt{m})^2 + (\sqrt{n})^2 = m + n + 2 \sqrt{mn} $[/tex]

For the two expressions to be equal, we should have

[tex]$ m + n = m + n +2\sqrt{mn} $[/tex] ⇔ [tex]$ \sqrt{mn} = 0 $[/tex].

This is possible iff mn = 0. i.e, m = 0 or n = 0.

Otherwise, they are not equal.

When m = 5 and n = 4.

[tex]$ \sqrt{5 + 4} = \sqrt{9} = 3 $[/tex]

[tex]$ \sqrt{5} + \sqrt{4} = \sqrt{5} + 2 $[/tex]

First is an integer. Second is an irrational number.

Clearly, they are not equal.

Answer:

P-value = 0.0261

We conclude that the machine is under-filling the bags.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 433 gram

Sample mean, [tex]\bar{x}[/tex] = 427 grams

Sample size, n = 26

Alpha, α = 0.05

Sample standard deviation, σ = 15 grams

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 433\text{ grams}\\H_A: \mu < 433\text{ grams}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{427 - 433}{\frac{15}{\sqrt{26}} } = -2.0396[/tex]

Now, we calculate the p-value using the standard table.

P-value = 0.0261

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

We conclude that the machine is under-filling the bags.

Answer:

a) Null hypothesis:[tex]p\leq 0.2[/tex]  

Alternative hypothesis:[tex]p > 0.2[/tex]  

b) [tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/tex]  

[tex]p_v =P(Z>1.25)=0.106[/tex]  

Step-by-step explanation:

1) Data given and notation

n=400 represent the random sample taken

X=90 represent the number of questions with B as the correct answer

[tex]\hat p=\frac{90}{400}=0.225[/tex] estimated proportion of arrests that were not prosecuted

[tex]p_o=0.2[/tex] is the value that we want to test, since we assume that each question present 5 options and just one is correct, 1/5 =0.2 if all five options were equally likely

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:  

Null hypothesis:[tex]p\leq 0.2[/tex]  

Alternative hypothesis:[tex]p > 0.2[/tex]  

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

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

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

3) Calculate the statistic  

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

[tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/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 significance level provided [tex]\alpha[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(Z>1.25)=0.106[/tex]  

If we compare the p value obtained and 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of B correct answers is not significantly higher than 0.2.  

Answer:

Yes, one of the properties of determinants is that they are real numbers (including zero) not matrix. This if the entries of the matrix are real. Determinants can be both positive or negative numbers.

Step-by-step explanation:

The determinant of a matrix is a number

What is a Matrix?

In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object

The determinant of inverse matrix can never be zero.

The determinant of an identity matrix is always 1.

The determinant of a zero matrix is always 0.

Given data ,

Let the matrix be represented as

[tex]A = \left[\begin{array}{ccc}a_{11} &a_{12} \\ a_{21} &a_{22} \\\end{array}\right][/tex]

Let the determinant of the matrix be d

Now , the determinant of the matrix A is calculated as

d = ( a₁₁ ) x ( a₂₂ ) - ( a₁₂ ) x ( a₂₁ )

So , the determinant will be a single number and will not be another matrix

The value of d will be scaling factor for the transformation of a matrix

Hence , the determinant of a matrix is a number

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The square root of 81 repeating is 9.05, (Rounded)

-It goes on forever.

Since it goes on forever, it's not a whole number or a natural number, both kind of numbers never keep going on forever.

Which leaves us with Irrational and Real Numbers

Therefore, 1 and 2 is your answer.

Best of Luck to you.

If you have any questions, or need more info, feel free to comment below.

Happy New Year!

Answer:

8/27 ≈ 29.6%

Step-by-step explanation:

Two of the six faces are B, which means four of the six faces are not B.

The probability of rolling not B three times is:

P = (4/6)^3

P = (2/3)^3

P = 8/27

P ≈ 29.6%

Answer:

8/27 = 0.296 = 29.6%

Step-by-step explanation:

Given that the die faces are as follows:

A A A   B B    C

i.e :

P( rolls A) = 3/6

P (rolls B) = 2/6

P (rolls C) = 1/6

for any single roll,

P (rolls not B) = P(rolls A) + P(rolls C)

P (rolls not B) = 3/6 + 1/6 = 4/6 = 2/3

for 3 consecutive rolls

P ( B does not appear) = P(rolls not B) * P(rolls not B) * P(rolls not B)

= P(rolls not B) ³

= (2/3)³

= 8/27 = 0.296 = 29.6%

Answer:

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.  

And rounded to 1 decimal place would be 33.3

Step-by-step explanation:

1) Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the scores for the ACT of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(21.5,5.2)[/tex]  

Where [tex]\mu=21.5[/tex] and [tex]\sigma=5.2[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.01[/tex]   (a)

[tex]P(X<a)=0.99[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(Z>2.33)=0.01

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99[/tex]  

[tex]P(Z<\frac{a-\mu}{\sigma})=0.99[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]Z=2.33<\frac{a-21.2}{5.2}[/tex]

And if we solve for a we got

[tex]a=21.2 +2.33*5.2=33.316[/tex]

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.  

And rounded to 1 decimal place would be 33.3