The culinary herb cilantro, Coriandrum sativum, is very polarizing; some people love it and others hate it. A genetic component is suspected to be at play. A survey of 12087 American adults of European ancestry asked whether they like or dislike the taste of cilantro. A total of 3181 said that they dislike the taste of cilantro.

(a) Estimate with 95% confidence (and interpret) the proportion of all American adults of European an- cestry who dislike the taste of cilantro
(b) Estimate with 95% confidence (and interpret) a lower confidence bound for the proportion of all Amer- ican adults of European ancestry who dislike the taste of cilantro

Answer:

3/8

Step-by-step explanation:

Since we have given that

Measure of an angle = 135°

As we know that

Total angle formed by a circle = 360°

So, Fraction of a circle when the an angle 135° turn is given by

135/360=

Answer:

(0, –5)

Step-by-step explanation:

If you don't see this option there is another option, (0 , 11) is also correct.

Answer:

B=22.348 Inches per minutes

Step-by-step explanation:

A snail is traveling along a straight path. The snail’s velocity can be modeled by [tex]v(t)=1.4ln(1+t^2)[/tex] inches per minute for 0 ≤ t ≤ 15 minutes.

If the snail's velocity is [tex]v(t)=1.4ln(1+t^2)[/tex] per minute, its displacement for 0 ≤ t ≤ 15 minutes is given by the integral:

[tex]\int_{0}^{15}1.4ln(1+t^2)dt=76.04307[/tex]

The ant travels with a constant acceleration of 2 Inches per minute.

Therefore, the velocity of the ant will be:

[tex] \int 2 dt=2t+c,[/tex] inches per minutes, for some constant c.

For the interval, 12≤t≤15, the displacement of the ant is:

[tex] \int_{12}^{15}(2t+c)dt=t^2+ct|_{12}^{15}=81+3c[/tex]

Since the snails displacement and that of the ant are equal in 12≤t≤15.

81+3c=76.04307

3c=76.04307-81

3c=-4.95693

c=-1.65231

The velocity of the ant at t=12 is therefore:

2t+c=2(12)-1.65231=22.348 Inches per minutes

B=22.348 Inches per minutes

The value of B to the nearest whole number, given all of the factors enumerated above, is 22.4 inches/Min. (For the full answer, please see the attached.)

This simply refers to the pace or rate at which an object or a person changes their position in relation to a frame of reference. It is also a function of time.

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Answer:

A   -2 to -1

Step-by-step explanation:

Answer:

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

Step-by-step explanation:

Let p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected. Of these adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

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

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

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

Compute the critical value of z for 95% confidence level as follows:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

*Use a z-table.

Compute the 95% confidence interval for the population proportion as follows:

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

    [tex]=0.86\pm 1.96\times \sqrt{\frac{0.86(1-0.86)}{1006}}\\=0.86\pm 0.0214\\=(0.8386, 0.8814)\\\approx (0.839, 0.881)[/tex]

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

This confidence interval implies that:

We are 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

95% confident that the proportion of all adults who believe that the shows are either  made up" or "mostly distorted" is within 83.9% and 88.1%.

Given that,

An article included the following statement: "Few people believe there's much reality in reality TV:

Total of 86% said the shows are either 'totally made up' or 'mostly distorted.'" This statement was based on a survey of 1006 randomly selected adults

We have to determine,

Compute a bound on the error (based on 95% confidence) of estimation for the reported proportion of 0.86

According to the question,

Let, p = proportion of people who believe that the reality TV shows are either "totally made up" or "mostly distorted".

A random sample of n = 1006 adults are selected.

These adults 86% believes that the reality TV shows are either "totally made up" or "mostly distorted".

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

[tex]C.I. = p\ \pm z_\frac{\alpha}{2} \times \sqrt{\frac{p(1-p)}{n} }[/tex]

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

[tex]z_\frac{\alpha}{2} = z_\frac{0.05}{2} = z_0._0_2_5 = 1.96[/tex]

To calculate the critical value of z for 95% confidence level as follows:

Compute the 95% confidence interval for the population proportion as follows:

[tex]C.I. = p \pm z_\frac{\alpha}{2} \sqrt{\frac{p(1-p)}{n} } \\\\C.I. = 0.86\pm 1.96\sqrt{\frac{0.86(1-0.86)}{1006} }\\\\C.I. = 0.86 \pm 0.0214\\\\C.I. = (0.8366, 0.8814)\\\\C.I = (0.839, 0.88)[/tex]

The 95% confidence interval for the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is (0.839, 0.881).

Hence, 95% confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within 83.9% and 88.1%.

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Answer:

The probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.

Step-by-step explanation:

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 1,000.

If X follows Poisson (λ) and λ > 1,000 then the distribution of X can be approximated but he Normal distribution.

The mean of the approximated distribution of X is:

μ = λ

The standard deviation of the approximated distribution of X is:

σ = √λ

Thus, if λ > 1,000, then [tex]X\sim N(\mu=\lambda,\ \sigma^{2}=\lambda)[/tex].

Let X = number of asbestos particles in a sample of 1 squared centimetre of dust.

The random variable X follows a Poisson distribution with mean, μ = 1000.

Then the average number of asbestos particles in a sample of 10 squared centimetre of dust will be, [tex]\lambda = 10\times \mu=10\times 1000=10,000[/tex].

Compute the probability that 10 squared centimetres of dust contains more than 10150 particles as follows:

[tex]P(X>10150)=P(\frac{X-\mu}{\sigma}>\frac{10150-10000}{\sqrt{10000}})[/tex]

                       [tex]=P(Z>1.50)\\=1-P(Z<1.50)\\=1-0.93319\\=0.06681\\\approx0.067[/tex]

*Use a z-table for the probability.

Thus, the probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.

We use the Poisson distribution and normal approximation to calculate the probability of observing more than 10150 particles in 10 square centimeters of dust, given the average particle count per centimeter. The normal approximation is necessary due to the large area being considered.

The question pertains to the calculation of a probability using the Poisson distribution in conjunction with the normal approximation. Since the average number of asbestos particles in a square centimeter of dust is 1000, for 10 square centimeters, the expected number of particles will be 10 * 1000 = 10000. However, when considering a larger area, the number of events (in this case, asbestos particles) might not follow the Poisson distribution strictly, which necessitates the use of the normal approximation.

Normal approximation of a Poisson variable involves transforming the Poisson variable to a normally distributed variable. The mean (μ) of this normal distribution stays the same (μ = λ = 10000 for 10 square centimeters), and the standard deviation (σ) becomes the square root of the mean (σ = √λ = √10000 = 100).

To calculate the probability of getting more than 10150 particles (denoted as X) in 10 square centimeters of dust, we use the normal cumulative distribution function (normalcdf). We want P(X > 10150), which with the normalcdf function, becomes 1 - P(X <= 10150) = 1 - normalcdf(-∞, 10150, 10000, 100). The result of the calculation using a standard normal table or calculator gives the probability, rounded to three decimal places.

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Answer:

50 is more than 27

Step-by-step explanation:

27 comes before 50

Answer:

0.9%

Step-by-step explanation:

To determine the percentage of Earth's surface affected by a meteor strike, we calculate the surface area of Earth and the area of impact, then divide the area of impact by Earth's total surface area and multiply by 100 to get a percentage. Using the given Earth's radius of 3959 miles and the meteor's diameter of 750 miles, the affected area is approximately 0.224% of Earth's surface.

To calculate the percentage of the Earth's surface affected by a meteor strike, we can use the formula for the surface area of a sphere and the formula for the surface area of a circle, which represents the affected area.

The total surface area of the Earth (AEarth) is given by 4πr2, where r is the radius of the Earth. The radius of the Earth is 3959 miles.

The surface area affected by the meteor (AMeteor) can be approximated by the surface area of a circle, πd2/4, where d is the diameter of the meteor. The diameter of the meteor is 750 miles. Therefore, the affected area is approximately π * (7502)/4.

Now we can calculate both the total surface area of the Earth and the affected area:

To find the percentage of the surface affected, we divide the affected area by the total surface area and then multiply by 100:

Percentage = (AMeteor / AEarth) * 100

Percentage = (441,963 / 197,000,000) * 100 = 0.224% of the Earth's surface is affected by the meteor strike.

Answer:

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Step-by-step explanation:

We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                               P.Q. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\bar X[/tex] = sample mean meat consumption = 3 pounds

             s = population standard deviation = 1.4 pounds

            n = sample of people = 2092

            [tex]\mu[/tex] = population mean consumption of meat

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                  of significance are -1.96 & 1.96}  

P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]

                 = [ [tex]3-1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] , [tex]3+1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] ]

                 = [2.9 pounds , 3.1 pounds]

Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Answer: 5/13

Step-by-step explanation:

I got it correct

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

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

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

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

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

In this problem, we have that:

[tex]\mu = 79.7, \sigma = 8.4[/tex]

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

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

[tex]1.127 = \frac{X - 79.7}{8.4}[/tex]

[tex]X - 79.7 = 8.4*1.127[/tex]

[tex]X = 89[/tex]

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

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

[tex]0.15 = \frac{X - 79.7}{8.4}[/tex]

[tex]X - 79.7 = 8.4*0.15[/tex]

[tex]X = 81[/tex]

The numerical limits for a B grade is between 81 and 89.

Answer:

a) - 0.54

b) No, we expect Andy to lose money each time he plays this game. This game is not a good way to make money.

Step-by-step explanation:

We are told the game costs $2 to play.

In a card game, we have the following:

Total number of cards = 52

Number of number cards (2-10)=36

Number of face cards = 12

Number of ace = 3

Number of ace of clubs = 1

The probability and profits of the game is calculated below.

Probability he gets a number card 2-10 =[tex] \frac{36}{52} = 0.69 [/tex]

Profit = - $2

Probability he gets a face card:

[tex] \frac{12}{52} = 0.23 [/tex]

Profit = $3 - $2 = $1

Probability he gets an ace:

[tex] \frac{3}{52} = 0.06 [/tex]

Profit = $5 - $2 =$3

Probability he gets an ace of clubs:

[tex] \frac{1}{52} = 0.02[/tex]

Profit = $25 - $2 =$23

Andy's expected profit per game wil be given as:

E = probability * profit

= [(0.69 * -2)+(0.23 * 1)+(0.06 * 3)+(0.02 * 23)]

= -0.54

b) No, we expect Andy to lose money each time he plays this game. This game is not a good way to make money.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

Andy expected  profit per game is -0.54.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

Given that,

The game costs $2 to play.

He draws a card from a deck. If he gets a number card (2-10), he wins nothing.

For any face card ( jack, queen or king), he wins $3. For any ace, he wins $5, and he wins $25 if he draws the ace of clubs.

We have to determine,

Andy's expected profit per game is:

Would you recommend this game to Andy as a good way to make money.

According to the question,

The game costs $2 to play.

Total number of cards = 52

Number of number cards (2 - 10) =36

Number of face cards = 12

Number of ace = 3

Number of ace of clubs = 1

The probability and profits of the game is calculated below.

[tex]Probability \ he \ gets \ a\ number\ card \ 2-10 = \dfrac{36}{52} = 0.69[/tex]

Profit = $2

[tex]Probability\ he \ gets\ a \face\ card= \dfrac{12}{52} = 0.23[/tex]

Profit = $3 - $2 = $1

[tex]Probability\ he \ gets\ an \ace = \dfrac{3}{32 } = 0.06[/tex]

Profit = $5 - $2 =$3

[tex]Probability \ he \ gets \ an\ ace\ of \ clubs: = \dfrac{1}{52} = 0.02[/tex]

Profit = $25 - $2 =$23

Andy's expected profit per game will be given as:

[tex]E = \ Probability \times \ profit\\\\E = [(0.69 \times -2)+(0.23 \times 1)+(0.06 \times 3)+(0.02 \times 23)]\\\\E = -0.54[/tex]

The profit of Andy is -0.54.

Andy's expected profit per game (-0.542) is negative, which means he doesn't make any profit per game.

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To solve the equation -2/7x=-1/3 for x, you would divide both sides by -2/7. This translates to multiplication of -1/3 by the reciprocal of -2/7 which is -7/2. By performing this multiplication, x equals to 7/6.

The equation at question here is -2/7x=-1/3. To solve for x, we need to isolate x on one side of the equation. To do this, we can divide both sides of the equation by -2/7, which is the coefficient of x in the equation. Performing this operation (also known as the division property of equality), we get:

When we divide by a fraction, it's the same as multiplying by its reciprocal, so:

Performing this multiplication gives x = 7/6.

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Answer:    Step-by-step explanation:

Start by multiplying the ones: 6 times 28 is 168. Next, multiply the tens: 30 times 28 is 840. Add the products to get 1,008.

23.27

Use a calculator

Answer:

[tex]217.636-1.96\frac{20}{\sqrt{11}}=205.817[/tex]    

[tex]217.636+1.96\frac{20}{\sqrt{11}}=229.455[/tex]    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

Step-by-step explanation:

Assuming the following data: Weight 150 169 170 196 200 218 219 262 269 270 271

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

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

[tex]\sigma= \sqrt{400}= 20[/tex] represent the population standard deviation

n=11 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean we can use the following formula:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

The mean calculated for this case is [tex]\bar X=217.636[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]217.636-1.96\frac{20}{\sqrt{11}}=205.817[/tex]    

[tex]217.636+1.96\frac{20}{\sqrt{11}}=229.455[/tex]    

So on this case the 95% confidence interval would be given by (205.82;229.46)    

Answer:

The calculated chi -square  value =5.7356 Is less than12.01 at 0.1level of significance .

The null hypothesis is accepted.

The past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.

Step-by-step explanation:

Step:-(1)

Given data 28 25 29 25 32 36 27 24

Sample size 'n' = 8

mean of the sample (x⁻) = ∑x /n = [tex]\frac{28+ 25+ 29 +25 +32 +36 + 27 + 24}{8}[/tex]

mean of the sample (x⁻) =  28.25

x                           x- x⁻                         (x-x⁻ )²

28              28 - 28.25  = -0.25       0.0625    

25             25 -28.25   = -3.25       10.56

29             29-28.25    = 0.75        0.5625

25             25-28.25    = -3.25      10.56

32             32-28.25    =   3.75      14.06

36            36-28.25    =   7.75      60.06

27            27-28.25    = -1.25      1.5625

24            24-28.25    = -4.25      18.06

                                          ∑(x-x⁻)² =   115.4875    

Step:-(2)

The sample standard deviation

           S² = ∑(x-x⁻)²/n-1 = [tex]\frac{115.4875 }{7}[/tex]  = 16.49

   By using χ² distribution

                           χ²  = [tex]\frac{ns^2 }{variance}[/tex]

By above test can be applied only if the population from which sample is drawn normal.

Given data For the past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.

Population variance σ² = 23

Step:-(3)

Null hypothesis :H₀:σ² = 23

Alternative hypothesis :H₁:≠23

[tex]X^{2} = \frac{8 (16.49)}{23}=5.7356[/tex]

The calculated chi -square  value =5.7356

The degrees of freedom γ=n-1 =8-1=7

The tabulated value of chi -square = 12.01 at 0.1level of significance (check table)

The calculated chi -square  value =5.7356 Is less than12.01 at 0.1level of significance .

The null hypothesis is accepted.

Conclusion:-

The null hypothesis is accepted.

The past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.

Step-by-step explanation:

cosB=4/5

B=36.86°

Hope it helps u

Final answer:

There are 512 different combinations possible for an 8-number combination lock, as the same number can be reused and each of the three positions has 8 choices.

Explanation:

To calculate the number of different combinations possible for a combination lock with a dial containing 8 numbers (0 through 7), we can use the fundamental counting principle. Since the same number can be reused consecutively, there are no restrictions on the choices for each of the three numbers in the sequence. Therefore, for each position of the sequence (the first, second, and third number), there are 8 possibilities.

The total number of combinations is the product of the number of choices for each position. Thus, the number of combinations for the lock is:

8 (choices for the first number) × 8 (choices for the second number) × 8 (choices for the third number)

Doing the calculation, we get:

8 × 8 × 8 = 512

Therefore, there are 512 different combinations possible for the lock.

Answer:

Test: [tex]\mu[/tex] < 20 with normal distⁿ

Hypothesis test:

H0:  [tex]\mu \ge[/tex] 20                         (Null Hypothesis, H₀ : [tex]\mu[/tex] = 20 is also correct)

H1: [tex]\mu[/tex] < 20                             (Alternative Hypothesis, also called H₁)

This is lower tailed test.

Since sample is large, sample standard deviation can be taken as an  approximation of population standard deviation.

x = 19.3

[tex]\sigma[/tex] = 11.1

n = 444

significance level, [tex]\alpha[/tex] = 0.05 (If no value is given, we take level of 0.05)

Test statistic [tex]z* = \frac{x-\mu}{\sigma/\sqrt{n}}[/tex]

                         [tex]= \frac{19.3-20}{11.1/\sqrt{444}}[/tex]

                       = - 1.33

The attached files contains additional information

Answer:

y+2 = 2(x+2)

Step-by-step explanation:

given point (a,b) and slope m, point slope form of a line is:

y-b = m(x-a)

Answer:

For this case we want to test that the true proportion of americans have experienced difficulty in making mortgage payments is higher than 0.28 (alternative hypothesis) and the sytem of hypotheis are:

NUll hypothesis: [tex]p \leq 0.28[/tex]

Alternative hypotheis: [tex] p >0.28[/tex]

Step-by-step explanation:

Previous concepts

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

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

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

Solution to the problem

For this case we want to test that the true proportion of americans have experienced difficulty in making mortgage payments is higher than 0.28 (alternative hypothesis) and the sytem of hypotheis are:

NUll hypothesis: [tex]p \leq 0.28[/tex]

Alternative hypotheis: [tex] p >0.28[/tex]

Answer:

It takes approximately 13.737 hours for the 200 degree water to cool down to 125 degrees.

Step-by-step explanation:

Recall Newton's Law of heating and cooling for an object of initial temperature [tex]T_0[/tex], in an ambient temperature [tex]T_a[/tex]:

[tex]T(t)=T_a+(T_0-T_a)\,e^{-kt}[/tex]

where t is the time elapsed.

We know the ambient temperature and the initial temperature of the object, but we don't know the value of the constant "k" that describes the cooling process. We can obtain such value (k) by using the information that the 200 degrees water cooled 10 degrees in 1.25 hours.

In such case we have:

[tex]T(t)=T_a+(T_0-T_a)\,e^{-kt}\\200-10=75+(200-75)\,e^{-k(1.25)}\\190=75+125\,e^{-k(1.25)}\\190-75=125\,e^{-k(1.25)}\\115=125\,e^{-k(1.25)}\\\frac{115}{125}= e^{-k(1.25)}\\0.92=e^{-k(1.25)}\\ln(0.92)=-k\,(1.25)\\k=\frac{ln(0.92)}{-1.25} \\k=0.0667[/tex]

Therefore, we have now the complete expression for the cooling process:

[tex]T(t)=75+125\,e^{-0.0667\,t}[/tex]

To find the time it takes to cool the 200 degree water down to 125 degrees, we use:

[tex]125=75+125\,e^{-0.0667\,t}\\125-75=125\,e^{-0.0667\,t}\\50=125\,e^{-0.0667\,t}\\\frac{50}{125} =e^{-0.0667\,t}\\0.4=e^{-0.0667\,t}\\ln(0.4)=-0.0667\,\,t\\t=\frac{ln(0.4)}{-0.0667} \\t=13.737\,\,hours[/tex]

Answer:

[tex]3.786-2.262\frac{0.187}{\sqrt{10}}=3.652[/tex]    

[tex]3.786+2.262\frac{0.187}{\sqrt{10}}=3.920[/tex]    

So on this case the 95% confidence interval would be given by (3.652;3.920)    

Step-by-step explanation:

Previous concepts

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

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

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

[tex]\bar X[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=3.786[/tex]

The sample deviation calculated [tex]s=0.187[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]

Now we have everything in order to replace into formula (1):

[tex]3.786-2.262\frac{0.187}{\sqrt{10}}=3.652[/tex]    

[tex]3.786+2.262\frac{0.187}{\sqrt{10}}=3.920[/tex]    

So on this case the 95% confidence interval would be given by (3.652;3.920)    

To estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline, calculate the sample mean and the 95% confidence interval.

To estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline, we can calculate the sample mean and the 95% confidence interval.

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Answer:

Step-by-step explanation:

Systolic. Calculate the mean

118+128+158+96+156+122+116+136+126+120+

=1276÷10 = 127.6.

To calculate the median arrange the values from the lowest to the highest.

96,116,118,120,122,126,128,136,156,158.

122+126+2=124.

Disastolic

80+76+74+52+90+88+58+64+72+82= 736÷10=73.6

To calculate the median arrange the values from the lowest to the highest.

52,58,64,72,74,76,80,82,88,90

Median= 74+76=150÷2=75.

Comparison

The Systolic and diastolic blood pressure measures different things, so comparing them is of no use.

It will be good if the relationship between the blood pressure can be investigated because the data are in pairs.

Compute the mean and median for both the systolic and diastolic measurements. Compare the results to identify skewness in the data. The mean and median do not represent data variability, so also consider using the Standard Deviation or Interquartile Range for a broad view.

First, let's begin by calculating the mean or average of both systolic and diastolic data sets. For this, you'll add up all the measurements and divide by the number of measurements.

For the median, you'll arrange the measurements in ascending order. If the total number is odd, the median is the middle number. If it is even, the median will be the average of the two middle numbers.

Comparing the two data sets would involve looking at the calculated mean and median values and determine whether they're close in value or not. If they are close, it indicates a lack of skewness in the data. If they're not, there would appear to be skewness, with the mean being influenced by particularly high or low values.

The mean and median are measures of central tendency that provide some insight into the data, however, they do not show the spread or variability of the data. Therefore, using additional statistics like the Standard Deviation or Interquartile Range might be good options to understand dispersion in the data set.

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Answer:

$6,200   i hope this helps!   :)

Step-by-step explanation:

the 4% interest of the $5000   4% of 5000 = 200

multiply by 6 for the 6 years   200 * 6 = 1200

add the 1200 to the 5000   5000 + 1200 = 6,200

Answer:

(1) True / False: (a) To reduce the margin of error in a stratified sample, we must design strata so that the units within each stratum are as similar as possible.

R = True, since to have a more exact margin you have to eliminate the errors, and as we do it by running error by error, in order to obtain a point on the graph where the products are within the positive margin and you can obtain a graph less error than the beginning.

(b) To reduce the margin of error in a group sample, we must design groups so that the units within each group are as similar as possible.

R = if this type of procedure is reliable because it helps you to correct the errors of the procedure, but to be able to do it you have to have very well accounted for, each piece of information, whether negative or positive, of the study being carried out and from there to create data that help reduce the margin of error.

Answer:

The 95% confidence interval is between 26.5 ng/ml and 40.3 ng/ml

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{1-0.95}{2} = 0.025[/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]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error 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.

[tex]M = 1.96*\frac{19.6}{\sqrt{31}} = 6.9[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 33.4 - 6.9 = 26.5 ng/ml

The upper end of the interval is the sample mean added to M. So it is 6.4 + 33.4 + 6.9 = 40.3 ng/ml

The 95% confidence interval is between 26.5 ng/ml and 40.3 ng/ml

The 95% confidence interval for the mean measurement of the biomarker osteocalcin (OC) is approximately 26.64 ng/ml to 40.16 ng/ml.

To calculate the 95% confidence interval for the mean measurement of the biomarker osteocalcin (OC), we can use the formula:

CI = Mean ± Z * (SD / sqrt(n))

Where CI is the confidence interval, Mean is the sample mean, SD is the known standard deviation, Z is the Z-score corresponding to the desired confidence level, and n is the sample size.

In this case, the sample mean is 33.4 ng/ml, the known standard deviation is 19.6 ng/ml, and the sample size is 31. The Z-score for a 95% confidence level is approximately 1.96. Plugging these values into the formula:

CI = 33.4 ± 1.96 * (19.6 / sqrt(31))

Simplifying the formula:

CI = 33.4 ± 6.76

Therefore, the 95% confidence interval for the mean measurement of osteocalcin is approximately 26.64 ng/ml to 40.16 ng/ml.

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Answer:

99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

Step-by-step explanation:

We are given that the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for Wire electrical-discharge machining (WEDM) : 21, 16, 29, 35, 42, 24, 24, 25

Firstly, the pivotal quantity for 99% confidence interval for the population variance is given by;

                          P.Q. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex]  ~ [tex]\chi^{2}__n_-_1[/tex]

where,  [tex]s^{2}[/tex] = sample variance = [tex]\frac{\sum (X-\bar X)^{2} }{n-1}[/tex]  = 67.43

             n = sample of observations = 8

            [tex]\sigma^{2}[/tex] = population variance

Here for constructing 99% confidence interval we have used chi-square test statistics.

So, 99% confidence interval for the population variance, [tex]\sigma^{2}[/tex] is ;

P(0.9893 < [tex]\chi^{2}_7[/tex] < 20.28) = 0.99  {As the critical value of chi-square at 7

                                          degree of freedom are 0.9893 & 20.28}  

P(0.9893 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 20.28) = 0.99

P( [tex]\frac{0.9893 }{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{20.28 }{(n-1)s^{2} }[/tex] ) = 0.99

P( [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ) = 0.99

99% confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2} }{20.28 }[/tex] , [tex]\frac{(n-1)s^{2} }{0.9893 }[/tex] ]

                                                   = [ [tex]\frac{7\times 67.43 }{20.28 }[/tex] , [tex]\frac{7\times 67.43 }{0.9893 }[/tex] ]

                                                   = [23.275 , 477.115]

99% confidence interval for [tex]\sigma[/tex]  = [ [tex]\sqrt{23.275}[/tex] , [tex]\sqrt{477.115}[/tex] ]

                                                  = [4.824 , 21.843]

Therefore, 99% CI for the variance [tex]\sigma^{2}[/tex] , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

Answer:

[tex] 7p + 5 \le 50 [/tex]

Step-by-step explanation:

The number of pizzas is p.

One pizza costs $7.

p number of pizzas cost 7p.

The total cost is the cost of the pizzas plus the $5 tip.

The total cost is 7p + 5

He can spend $50 or less, so the total cost must be less than or equal to $50.

Answer: [tex] 7p + 5 \le 50 [/tex]