A random sample of 157 recent donations at a certain blood bank reveals that 86 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of 0.01. State the appropriate null and alternative hypotheses.

The probability that the proportion of politicians who are lawyers differs from the total politician proportion by less than 5% is approximately 0.9793 (rounded to four decimal places).

1. **Calculate the standard deviation [tex](\( \sigma \))[/tex]:**

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

[tex]\[ \sigma = \sqrt{\frac{0.42(1-0.42)}{628}} \][/tex]

[tex]\[ \sigma \approx \sqrt{\frac{0.42 \times 0.58}{628}} \][/tex]

[tex]\[ \sigma \approx \sqrt{\frac{0.2436}{628}} \][/tex]

[tex]\[ \sigma \approx \sqrt{0.0003882} \][/tex]

[tex]\[ \sigma \approx 0.019705 \][/tex]

2. **Find the z-scores for [tex]\( p + 0.05 \)[/tex] and [tex]\( p - 0.05 \)[/tex]:**

[tex]\[ Z_{\text{upper}} = \frac{0.42 + 0.05 - 0.42}{0.019705} \][/tex]

[tex]\[ Z_{\text{upper}} = \frac{0.05}{0.019705} \][/tex]

[tex]\[ Z_{\text{upper}} \approx 2.53 \][/tex]

[tex]\[ Z_{\text{lower}} = \frac{0.42 - 0.05 - 0.42}{0.019705} \][/tex]

[tex]\[ Z_{\text{lower}} = \frac{-0.05}{0.019705} \][/tex]

[tex]\[ Z_{\text{lower}} \approx -2.53 \][/tex]

3. **Find the probability using the standard normal distribution table:**

[tex]\[ \text{Probability} = P(-2.53 < Z < 2.53) \][/tex]

By looking up the values in the standard normal distribution table or using a calculator, the probability is approximately 0.9793.

Answer:

Step-by-step explanation:

The proportion of students in a psychology experiment who could remember an eight-digit number correctly for t minutes was :

0.9 − 0.3 ln(t)

the proportion that remembered the number for 5 minutes, we would substitute t = 5 into expression, 0.9 − 0.3 ln(t). It becomes

0.9 − 0.3 ln5

= 0.9 - 0.3 × 1.60943791243

= 0.9 - 0.4828

= 0.4172

Approximately 0.4 to 1 decimal place

Answer:

Mean = [tex]\mu = 1140[/tex]

Standard deviation = [tex]\sigma = 80[/tex]

Find the probabilities for the average length of life of the selected batteries.

A)The average is between 1128 and 1140.

We are supposed to fidn P(1128<x<1140)

Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]

At x = 1128

[tex]Z=\frac{1128-1140}{80}[/tex]

[tex]Z=-0.15[/tex]

Refer the z table for p value

P(x<1128)=0.4404

At x = 1140

[tex]Z=\frac{1140-1140}{80}[/tex]

[tex]Z=0[/tex]

Refer the z table for p value

P(x<1140)=0.5

So,P(1128<x<1140)=P(x<1140)-P(x<1128)=0.5-0.4404=0.0596

Hence the probabilities for the average length of life of the selected batteries  is between 1128 and 1140 is 0.0596

B)The average is greater than 1152.

P(x>1152)

At x = 1128

[tex]Z=\frac{1152-1140}{80}[/tex]

[tex]Z=0.15[/tex]

Refer the z table for p value

P(x<1152)=0.5596

So,P(x>1152)=1-P(x<1152)=1-0.5596=0.4404

Hence the probabilities for the average length of life of the selected batteries is greater than 1152 is 0.4404

C) The average is less than 940.

P(x<940)

At x = 940

[tex]Z=\frac{940-1140}{80}[/tex]

[tex]Z=-2.5[/tex]

Refer the z table for p value

P(x<940)=0.5596

Hence the probabilities for the average length of life of the selected batteries is less than 940 is 0.5596

The question involves calculating the probability of the lifespan of car batteries based on given statistical data using the standard normal distribution.

The student is asking about calculating probabilities related to the life expectancy of car batteries using the principles of statistics. They provided information about the average lifespan of the batteries (1140 days), the standard deviation (80 days), and the sample size (400).

For part (a):

To find the probability that the average lifespan of the selected batteries is between 1128 and 1140 days, you would use the sampling distribution of the sample mean. The mean of the sampling distribution is the same as the population mean, 1140, and the standard deviation (often termed the standard error) of this distribution is the population standard deviation divided by the square root of the sample size: σ/√n or 80/√400 = 4 days. You would then use a standard normal distribution to find the probability that a normally distributed random variable falls between the z-scores corresponding to 1128 and 1140 days.

For part (b):

To determine the probability that the average is greater than 1152, again find the z-score for 1152 and use the standard normal distribution.

For part (c):

Asking for the probability of the average being less than 940 is a theoretical scenario far from the mean. Given the standard deviation, this would likely yield a probability close to 0.

Answer:

0.3446,0.2119

Step-by-step explanation:

Given that Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are approximately normally distributed with mean 160 and standard deviation 15.

If X represents Jill scores and Y Jack scores we have

X is N(170,20) and Y is N(160,15)

since x and y are independent we have the difference

X-Y is [tex]N(170-16,\sqrt{20^2+15^2} )\\=N(10, 25)[/tex]

a) Prob that Jack’s score is higher

= P(-x+y>0)

=[tex]P(Z<\frac{10}{25} )\\= P(Z<0.4)\\\\= =0.3446[/tex]

b) X+Y is Normal with (330, 25)

[tex]P(X+Y>350) = P(Z>\frac{350-330}{25} )\\=P(Z>0.8)\\\\= =0.2119[/tex]

To find the probability that Jack's score is higher than Jill's, calculate the z-scores and compare them. For the probability that the total of their scores is above 350, find the z-score for the sum and use a standard normal distribution table. The probability of Jack's score being higher is 50% and the probability of the sum being above 350 is 78.81%.

To approximate the probability that Jack's score is higher than Jill's, we can use the concept of the z-score. The z-score measures how many standard deviations a value is from the mean. For Jack's score, we calculate his z-score as (his score - his mean) / his standard deviation, which is (160 - 160) / 15 = 0. For Jill's score, the z-score is (170 - 170) / 20 = 0.

Since both z-scores are 0, we can conclude that the probability of Jack's score being higher than Jill's is 0.5, or 50%.

To find the probability that the total of their scores is above 350, we need to find the z-score for the sum. The sum of their scores is 160 + 170 = 330. The mean of the sum is 160 + 170 = 330, and the standard deviation of the sum is sqrt((15^2) + (20^2)) = sqrt(625) = 25.

Therefore, the z-score for the sum is (350 - 330) / 25 = 0.8. Using a standard normal distribution table or calculator, we can find that the probability of the sum being above 350 is approximately 0.7881, or 78.81%.

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To determine if the poor economy caused a change in the frequency of consuming restaurant meals by young millennials in 2012, a hypothesis test can be conducted by comparing the mean number of meals consumed in 2012 to the mean from 2011.

To determine if the poor economy caused a change in the frequency of consuming restaurant meals by young millennials in 2012, we can conduct a hypothesis test.

The null hypothesis, denoted as H0, states that there is no change in the mean number of restaurant meals per person for young millennials in 2012.

The alternative hypothesis, denoted as Ha, states that there is a change in the mean number of restaurant meals per person for young millennials in 2012.

We would use statistical data from 2012 to compare the mean number of restaurant meals consumed by young millennials to the mean from 2011 (192 meals per person) to determine if there is a significant change.

Final answer:

To determine if there has been a change in the mean number of restaurant meals consumed by young millennials in 2012 from the 192 meals reported in 2011, a two-tailed hypothesis test is required. The null hypothesis states no change (H0: μ = 192), while the alternative hypothesis suggests a change (H1: μ ≠ 192). A statistical test like a t-test or z-test, based on sample data, can be employed to assess this hypothesis with a commonly used significance level of 0.05.

Explanation:

Hypothesis Testing for Changes in Restaurant Meal Consumption

To investigate whether the economic downturn resulted in a change in the frequency of consuming restaurant meals by young millennials in 2012, compared to the mean of 192 meals reported in 2011, we should conduct a two-tailed hypothesis test. The null hypothesis (H0) states that there is no change in the annual mean number of restaurant meals consumed per person among young millennials. In formulaic terms, this is H0: μ = 192. The alternative hypothesis (H1) suggests that there has been a change, which can be phrased as H1: μ ≠ 192.

To conduct the test, we would need sample data from young millennials in 2012 regarding their restaurant meal consumption. Statistical software or methods such as a t-test or z-test would be used depending on the sample size and variance. The test will compare the sample mean to the known mean of 192, with a chosen level of significance, usually 0.05. If the p-value obtained from the test is less than the significance level, we reject the null hypothesis, indicating a significant change in the mean number of meals consumed.

To analyze trends related to meal consumption, factors such as technology use, employment, and economic factors (such as NEET rates) may also provide contextual insights on the behaviors of young millennials. While some young adults are increasingly engaged with online avenues for food purchase, a change in eating-out behavior may be attributable to these broader lifestyle and economic shifts.

Answer: 0.8031

Step-by-step explanation:

Binomial distribution:

For a binomial variable x,

The probability of getting success in x trials = [tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]

, where n = Total trials .

p= Probability of getting success in each trial.

For the given situation , we take x as the number of damaged apple .

Given : The proportion of damaged apples : p=0.15

n= 10

Then, the probability that his sample will contain at least one damaged apple.  :-

[tex]P(x\geq1)=1-P(x<1)\\\\=1-P(x=0)\\\\=1-^{10}C_{0}(0.15)^0(1-0.15)^{10}\\\\=1-(1)(0.85)^{10}\ \ [\because\ ^nC_0=1]\\=1-0.196874404341=0.803125595659\approx0.8031[/tex]

Hence, the required probability = 0.8031

Final answer:

To find the probability of at least one damaged apple in a sample of 10 from a shipment with a 15% damage rate, we calculate the complement (no damaged apples) and subtract it from 1, resulting in a probability of 0.8031.

Explanation:

The question asks us to use the binomial distribution to estimate the probability that a sample of 10 apples will contain at least one damaged apple, given that 15% of the apples are known to be damaged. To find the probability of at least one damaged apple, it's easiest to calculate the probability of the opposite event (no damaged apples) and subtract that from 1.

Using the binomial distribution formula, the probability of getting exactly k successes (damaged apples) in n trials (sampled apples) is given by P(X=k) = (nCk)*(p^k)*((1-p)^(n-k)), where p is the probability of success (0.15 in this case), and nCk is the binomial coefficient.

For k=0 (no damaged apples), the calculation becomes P(X=0) = (10C0)*(0.15^0)*((1-0.15)^(10-0)) = 1*(1)*(0.85^10) ≈ 0.1969. Therefore, the probability of finding at least one damaged apple is 1 - P(X=0) = 1 - 0.1969 = 0.8031.

To calculate the test statistic for a hypothesis test on population proportion, you need the size of the sample, sample proportion, level of significance, and proposed population proportion.

In order to calculate the test statistic for a hypothesis test on the proportion of people with a characteristic of interest, the following pieces of information are needed:

These pieces of information are necessary to calculate the test statistic, which is used to determine whether the sample data provides enough evidence to support or reject the hypothesis about the population proportion.

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The information needed to calculate the test statistic for a hypothesis test of a population proportion includes the size of the sample selected, the sample proportion calculated, and the proposed population proportion.

To calculate the test statistic for a hypothesis test about a population proportion, several pieces of information are required:

Note that while C. the level of significance used and F. the actual population proportion are important for different hypothesis testing steps such as defining the null and alternative hypotheses, calculating the p-value, or making the final decision for the hypothesis test, they do not directly factor into the calculation of the test statistic. Also, E. the characteristic of interest is not required to calculate the test statistic itself, but it is necessary to determine which is the appropriate test to apply and to structure the hypotheses correctly.

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The probability a student gets a Bachelor of Arts and Science degree is 0.18, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, and the probability a student gets only a Bachelor of Arts degree is 0.5.

The probability a student gets a Bachelor of Arts and Science degree can be found by multiplying the probabilities of getting a Bachelor of Arts degree and a Bachelor of Science degree.

So, P(Arts and Science) = P(Arts) * P(Science) = 0.6 * 0.3 = 0.18. Therefore, the probability a student gets a Bachelor of Arts and Science degree is 0.18, which is not listed as an option in the given choices.

The probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree can be found by adding the probabilities of getting each degree. So, P(Arts or Science) = P(Arts) + P(Science) = 0.6 + 0.3 = 0.9.

Therefore, the probability a student gets a Bachelor of Arts degree or a Bachelor of Science degree is 0.9, which is listed as option a) 0.9.

The probability a student gets only a Bachelor of Arts degree can be found by subtracting the probability of getting a Bachelor of Science degree and the probability of getting no degree from 1.

So, P(Only Arts) = 1 - P(Science) - P(no) = 1 - 0.3 - 0.2 = 0.5. Therefore, the probability a student gets only a Bachelor of Arts degree is 0.5, which is listed as option b) 0.5.

Answer:

We failed to accept null hypothesis

Step-by-step explanation:

x= 5.74

Standard deviation = [tex]0.26[/tex]

[tex]H_0:\mu = 5.8\\H_a:\mu < 5.8[/tex]

n = 65

We will use z test

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

Substitute the values :

[tex]z=\frac{5.74-5.8}{\frac{0.26}{\sqrt{65}}}[/tex]

[tex]z=-1.86[/tex]

Refer the z table for p value

p value = 0.0314

α = 0.10

p value < α

So, we failed to accept null hypothesis

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval:  (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{130}{20} = 6.5[/tex]

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

[tex]S.D = \sqrt{\frac{233.5}{19}} = 3.50[/tex]

b) Standard Error

[tex]= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826[/tex]

c) Point estimate for the mean time required for all college is given by the sample mean.

[tex]\bar{x} = 6.5[/tex]

d) 90% Confidence interval:  

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729[/tex]  

[tex]6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)[/tex]

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

Final answer:

The question requires calculating the sample mean, standard deviation, standard error, point estimate for the population mean, and constructing a confidence interval for the mean time to earn bachelor's degrees. The value of 4 years will need to be checked against the calculated confidence interval, and the reliability of these results may be scrutinized based on the distribution of the data.

Explanation:

The question involves calculating various statistical measures for a dataset representing the number of years college students took to earn bachelor's degrees. To address these parts:

To calculate the sample mean, you add up all the numbers and divide by the total count of numbers. The sample standard deviation measures the amount of variation or dispersion in a set of values. You use the formula for standard deviation for a sample.

The standard error (SE) is calculated by dividing the sample standard deviation by the square root of the sample size.

The point estimate for the mean time is the sample mean, as it provides the best estimate of the population mean based on the sample data.

To construct the confidence interval, you would typically use the sample mean ± (critical value from the t-distribution * standard error). For 90% confidence, you can find the critical t-value for 19 degrees of freedom (since sample size minus one equals degrees of freedom).

To answer whether the confidence interval contains the value of 4 years and discuss the reliability of this interval, you'll examine the calculated interval and consider factors like the presence of outliers and the shape of the distribution.

Answer: The lower bound of confidence interval would be 0.116.

Step-by-step explanation:

Since we have given that

p = 13.2%= 0.132

n = 1105

At 90%  confidence,

z = 1.645

So, Margin of error would be

[tex]z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.132\times 0.868}{1152}}}\\\\=0.0164[/tex]

So, the lower bound of the confidence interval would be

[tex]p-\text{margin of error}\\\\=0.132-0.0164\\\\=0.116[/tex]

Hence, the lower bound of confidence interval would be 0.116.

The confidence interval for percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

To estimate the percentage of U.S. adults who were victims of crime at the 90% confidence level, we can use the formula for a confidence interval:

[tex]CI = p +/- Z * \sqrt{((p * (1-p)) / n)[/tex]

where CI is the confidence interval, p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. In this case, p = 0.132, Z = 1.645 (corresponding to a 90% confidence level), and n = 1105. Plugging in these values, we can calculate the confidence interval as:

[tex]CI = 0.132 +/- 1.645 * \sqrt((0.132 * (1-0.132)) / 1105)[/tex]

Simplifying the expression gives us:

CI = 0.132 ± 0.018

Therefore, the confidence interval for the percentage of U.S. adults who were victims of crime at the 90% confidence level is (0.114, 0.15). The lower bound of the confidence interval is 0.114.

Answer: Our required probability is 0.32.

Step-by-step explanation:

Since we have given that

Number of doors to be selected in a manner  = 22

Number of sketches = 12

Number of oil paintings = 16

Total number of doors we have =12+16 =28

We need to find the probability that 10 sketches and 12 oil paintings are chosen to be displayed.

So, probability would be

[tex]\dfrac{^{12}C_{10}\times ^{16}C_{12}}{^{28}C_{22}}\\\\=\dfrac{66\times 1820}{376740}\\\\=\dfrac{120120}{376740}\\\\=0.3188\\\\\approx 0.32[/tex]

Hence, our required probability is 0.32.

Answer: t = 1.9287

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of pages should be sent by 2-day mail instead.

As per given we have,

[tex]H_0: \mu \leq7\\\\H_a:\mu>7[/tex]

Sample mean : [tex]\overline{x}=8.52[/tex]

Sample standard deviation : s=3.81

sample size : n= 24

Since , the sample size is less than 30 and populations standard deviation is unknown , so we use t-test.

The test statistic for population mean :-

[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

[tex]t=\dfrac{8.5-7}{\dfrac{3.81}{\sqrt{24}}}\\\\=\dfrac{1.5}{\dfrac{3.81}{4.8990}}\\\\=\dfrac{1.5}{0.777712993333}=1.92873208093\approx1.9287[/tex]

Hence, the test statistics : t = 1.9287

Answer:

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]

[tex]P(t_{68}<-1.06)=0.146[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        

Step-by-step explanation:

1) Data given and notation        

[tex]\bar X=18.94[/tex] represent the mean for the sample    

[tex]s=8.3[/tex] represent the standard deviation for the sample        

[tex]n=69[/tex] sample size        

[tex]\mu_o =20[/tex] represent the value that we want to test      

[tex]\alpha[/tex] represent the significance level for the hypothesis test.      

t would represent the statistic (variable of interest)        

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

2) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true average lateral recumbency time under these conditions is less than 20 min:      

Null hypothesis:[tex]\mu \geq 20[/tex]        

Alternative hypothesis:[tex]\mu < 20[/tex]        

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:        

[tex]t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)        

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic        

We can replace in formula (1) the info given like this:        

[tex]t=\frac{18.94-20}{\frac{8.3}{\sqrt{69}}}=-1.06[/tex]        

4) Calculate the P-value        

First we need to calculate the degrees of freedom

[tex]df=n-1=69-1=68[/tex]

The critical value for this case would be :

[tex]P(t_{68}<-1.06)=0.146[/tex]

5) Conclusion      

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL reject the null hypothesis, and the the actual sample average lateral recumbency is not significantly lower than 20.        

Answer:

1 because almost certain event

Step-by-step explanation:

whenever a fair die is rolled the number of getting a 6 is having probability 1/6. Each throw is independent of the other

Hence no of sixes would be binomial with p=1/6

when 48000 dice are rolled, using binomial would be a hectic task.

Hence we approximate to normal distribution

X - no of sixes in 48000 throws would be normal

with mean = np = 8000

Var =npq = 6666.667

Std dev = 81.650

Now it is easier to find out

[tex]P(7500<x<8500)\\=P(7499.5<x<8499.5)[/tex](using continuity correctin)

=[tex]P(|z|<6.1295)\\=1[/tex]

we get this probability almost equal to 1

d) Normal distribution is a good choice because when no of trials increase using binomial and combination formulae would not be easy

1km___100000cm

0.7km___X=7000cm

(0.7*100000)/1=7000

25cm___1 step

7000cm___X=280 steps

(7000*1)/25= 280

The quantity of oil in the first 180 minutes was 62.78 million gallons.

It is the reverse of differentiation.

Given

An oil tanker breaks apart and starts leaking. As time goes on, the rate at which the oil is leaking out will diminish.

The tanker breaks apart, the oil is leaking out at a rate of R(t).

[tex]\rm R(t) = \dfrac{0.7}{1 + t^2}[/tex], where t is time in minutes.

Convert the hours into minutes.

1 hours = 60 minutes

3 hours = 180 minutes

Then the quantity of oil in the first 180 minutes will be

V(t) = R(t)

Integrate the function.

[tex]\rm V(t) = \int_0^{180} R(t) dt\\\\V(t) = \int_0^{180} \dfrac{0.7}{1+t^2} dt\\\\V(t) = 0.7 \int_0^{180} \dfrac{1}{1 + t^2}dt\\\\V(t) = 0.7[tan^{-1}t]_0^{180}\\\\V(t) = 0.7 [tan^{-1}180 -tan^{-1}0]\\\\V(t) = 0.7 * 89.682\\\\V(t) = 62.78[/tex]

Thus, the quantity of oil in the first 180 minutes was 62.78 million gallons.

More about the integration link is given below.

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The number of million gallons of oil that will leak out in the first 3 hours after the shipwreck can be found by evaluating the integral of the rate function over the given interval.

To find the number of million gallons of oil that will leak out in the first 3 hours after the shipwreck, we need to find the integral of the rate function R(t) over the interval t = 0 to t = 3.

The integral of R(t) = (0.7)/(1+t^2) with respect to t is:

∫R(t) dt = 0.7 ∫(1+t^2)^-1 dt

Integrating this expression gives:

∫R(t) dt = 0.7 ln|1+t^2| + C

Evaluating this expression from t = 0 to t = 3, we get:

∫03R(t) dt = 0.7 ln|1+3^2| - 0.7 ln|1+0^2|

Simplifying further:

∫03R(t) dt = 0.7 ln(10) - 0.7 ln(1)

Since ln(1) = 0, we have:

∫03R(t) dt = 0.7 ln(10) - 0 = 0.7 ln(10)

Finally, converting the result to million gallons:

∫03R(t) dt = 0.7 ln(10) million gallons

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Answer:

The answer is $64.25 or you could say its B.

Hope this helps :)

Answer with Step-by-step explanation:

We are given that a demand function

[tex]x=D(p)=3100-20p[/tex]

Where x=The quantity of sodas sold

p=Per can price (in cents)

We  have to find the price p for which  the demand inelastic.

Differentiate the demand function w.r.t p

[tex]D'(P)=-20[/tex]

Price elasticity of demand=[tex]E(p)=\frac{-pD'(p)}{D(p)}[/tex]

Price elasticity of demand=[tex]E(p)=\frac{-p(-20)}{3100-20p}[/tex]

When demand  is inelastic then

E(p)<1

[tex]\frac{20p}{3100-20p}[/tex] <1

Multiply by (3100-20p) on both sides

[tex]20p<3100-20p[/tex]

Adding 20p on both side of inequality

[tex]40p<3100[/tex]

Divide by 40 on both sides

[tex]p<77.5[/tex]

When the value of price is less than 77.5 then the demand of elasticity is inelastic.

The demand function of a beverage company is inelastic when the percentage change in the quantity of demanded soda is less than 1% for a 1% change in price. If prices between points A and B change by 1%, and the quantity demanded changes by only 0.45% (less than 1%), the demand is inelastic.

In the given demand function, x = D(p) = 3100 - 20p, the elasticity of demand is considered inelastic if the absolute value of elasticity is less than 1. Elasticity of demand reflects how sensitive the demand for a good is to changes in its price. With a demand curve, we'd consider whether a 1% change in price changes the quantity of the good demanded.

Should the quantity demanded change by less than 1% for a 1% price change (for example, a 0.45% change), we have inelastic demand. This means, if prices between points A and B alter by 1%, the change in the quantity demanded will be only 0.45%. This percentage indicates an inelastic demand because the price change will result in a comparatively smaller percentage change in quantity demanded. Price elasticities of demand are usually negative, showcasing the negative relationship between price and demanded quantity (law of demand). However, while interpreting, the focus lays on the absolute value.

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Answer:

The probability that those first 3 employees are in the order of Jack, Bill, and John is [tex]\frac{1}{59280}[/tex]

Step-by-step explanation:

Consider the provided information.

Company A employs forty people. We need to find the probability that those first 3 employees are in the order of Jack, Bill, and John?

First find the total number of ways;

We have 40 employees out of which we need to select only 3.

This can be written as: [tex]40\times 39 \times 38=59280[/tex]

The number of ways of selecting jack, bill and john in the same order = 1

[tex]Probability=\frac{\text{Favorable outcomes}}{\text{Total number of outcomes}}[/tex]

[tex]Probability=\frac{1}{59280}[/tex]

Therefore, the probability that those first 3 employees are in the order of Jack, Bill, and John is [tex]\frac{1}{59280}[/tex].

Answer:

a sample of 29 network will have a voltage of 232V

Step-by-step explanation:

N = 66

variate = 232 V.

mean voltage = 231.7 V

standard deviation = 2.19 V.

The z-value given by = (variate -mean)/ standard deviation

= (232-231.7)/2.19 =

z = 0.3/2.19 = 0.137

From of normal distribution table, the z of 0.137 found between the area of z=0 and z=0.5 is given as 0.0557

Thus the area to the right of the z=0.0557 ordinate is 0.5000−0.0557=0.4443 = 44.43%

thus, this is the probability of Network voltage = 44.43%.

for sample of 66 network, it is likely that 66×0.4443 = 29.32, i.e.

a sample of 29 network will have a voltage of 232V

Answer:

(B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

Step-by-step explanation:

A confidence interval of 95% means that there is a 95% certainty that for a given sample, the population mean will be within the confidence interval estimated.

This is the same as saying that if he sampling procedure were repeated many times, 95% of the time the population mean would be contained in the resulting confidence interval.

Therefore, the answer is B)

The correct statement is (B) If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

A 95% confidence interval is a range of values that is likely to contain the population mean. This does not mean that there is a 95% probability that the population mean falls within the interval calculated from a single sample. Instead, it means that if we were to take many samples and calculate a confidence interval from each sample, approximately 95% of those intervals would contain the true population mean.

Let's evaluate the other options:

(A) 95% of the sample of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because the confidence interval refers to the population mean, not the distribution of individual sample values. It does not imply that 95% of the sample has values within the interval.

(C) If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139. - This statement is incorrect because it misinterprets the confidence interval. The confidence interval gives us a range where we expect the population mean to lie, not the sample means. While the sample means will vary, the confidence interval tells us about the variability of the estimate of the population mean, not the variability of the sample means themselves.

(D) 95% of the population of employees has a systolic blood pressure between 123 and 139. - This statement is incorrect because it generalizes the confidence interval to the entire population. The confidence interval does not tell us what proportion of the population falls within the interval; it only provides a range that is likely to contain the population mean.

Therefore, the correct interpretation of the 95% confidence interval is that if we were to take many samples from the population and calculate a 95% confidence interval for each, approximately 95% of those intervals would contain the true population mean systolic blood pressure. This is why option (B) is the valid statement.

Answer:

A"fishbone" diagram

Step-by-step explanation:

A"fishbone" diagram, may help to determine potential causes of an issue and organise concepts into valuable groups of brain storming. A depiction of the fishbone is a graphical way of looking at causes and effects.

It's a more organised solution than other brain storming tools that cause problems. The issue or effect is shown on the fish's head or mouth.

Answer:

We use z-test for this hypothesis.

[tex]z_{stat} = 2.23[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.88

Sample mean, [tex]\bar{x}[/tex] = 5.91

Sample size, n = 48

Alpha, α = 0.05

Population standard deviation, σ = 3.2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 4.88\\H_A: \mu > 4.88[/tex]

The null hypothesis states that the mean score of successful managers on a psychological test is 4.88 and the alternate hypothesis says that the mean score of successful managers on a psychological test is greater than 4.88.

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

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{5.91 - 4.88}{\frac{3.2}{\sqrt{48}} } = 2.23[/tex]

Answer:

(1.4328, 1.622)

Claim is supported by evidence.

Step-by-step explanation:

Given that a  recent study was conducted to determine the curing efficiency (time to harden) of dental composites (resins for the restoration of damaged teeth) using two different types of lights.

Let X be the halogen and y the Luxomax light

[tex]H_0: \bar x =\bar y\\H_a: \bar x > \bar Y[/tex]

(Two tailed test)

we are given data as:

Group   Group One     Group Two  

Mean   5.3500            3.9000

SD             0.7000     0.8000

SEM       0.2214     0.2530

N                10                 10      

The mean of Group One minus Group Two equals 1.4500

Std error for difference =  0.336

Test statistic t=4.3135

 df = 18

p value = 0.0004

Since p <0.01 at 1% level, reject H0

There is significant difference and hence the claim is valid.

There  is evidence to support this claim at 1% significance level

Margin of error =1.172

99% confidence interval = [tex](1.45-1.172, 1.45+1.172)\\\\=(1.4328, 1.622)[/tex]

To determine if the Halogen light produces a larger cure depth, a two-sample t-test with equal variances is used. The test statistic is calculated and compared with the critical value at α = 0.01 to test the claim. Additionally, a 99% confidence interval for the difference in mean cure depths is constructed.

To assess whether the Halogen light produces a larger cure depth after 40 seconds than LuxOMax lights, we perform a two-sample t-test assuming equal variances. The hypotheses are:

Using the given data, we calculate the test statistic:

μ1 = 5.35, μ2 = 3.90, n1 = n2 = 10, s1 = 0.7, s2 = 0.8.

The pooled standard deviation (Sp) is computed as: Sp = √[((n1-1)s1² + (n2-1)s2²) / (n1+n2-2)] = √[(9×0.7² + 9×0.8²) / 18].

The test statistic t is calculated by: t = (x1 - x2) / (Sp×√(1/n1 + 1/n2)).

Using α = 0.01, we compare the calculated t value to the critical t value from a t-distribution table. If t > t_critical, we reject H0, providing evidence to support the Halogen light's claim.

Next, we construct a 99% confidence interval (CI) for the difference in population mean cure depths:

CI = (x1 - x2) ± (t_critical×Sp×√(1/n1 + 1/n2)).

This interval estimates the range within which the true difference in mean cure depths between the two lights lies with 99% confidence.

This answer includes instructions for approximating a function using a third degree Taylor polynomial at a specific point, using Taylor's Inequality to estimate the accuracy of the approximation, and checking the results by graphing the function |Rn(x)|.

For part (a), to approximate the function f(x) = ln(1 + 2x) using a Taylor polynomial of degree 3 at a = 4, we first need to find the first 4 derivatives of the function evaluated at a=4. Substituting the values will give the Taylor polynomial - T3(x).

For part (b), Taylor's Inequality states that the remainder term Rn(x) is less than or equal to M|X-a|^(n+1) / (n+1)! where M is the max value of the |f^(n+1)(t)| where t lies in the interval between a and x. By substituting the values we can find |R3(x)|.

For the last part (c), by using graphing software, you can graph the function |Rn(x)| in the given interval to visually confirm your earlier result. These graphs will show the magnitude of the error.|

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Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

The value of the test statistic is 1.583.

To find the value of the test statistic, we need to calculate the z-score using the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, the sample mean is 6.9 ppm, the population mean is 6.5 ppm, the population variance is 1.21, and the sample size is 27. Calculate the test statistic:

z = (6.9 - 6.5) / (√(1.21 / 27)) = 0.4 / 0.252951 = 1.583

Rounding to three decimal places, the value of the test statistic is 1.583.

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Answer:

a) Around 2,052.583 Km.

b) Around 22 days.

c) Around 3.887 Km/h

Step-by-step explanation:

a)

In order to find and approximation of the distance traveled, we have to make some assumptions:

If these assumptions hold, then the distance d in Km can be calculated by using the haversine formula:

[tex]\large d=2R*arcsin\left(\sqrt{sin^2((\varphi_2-\varphi_1)/2)+cos(\varphi_1)cos(\varphi_2)sin^2((\lambda_2-\lambda_1)/2)}\right)[/tex]

where  

[tex]\large (\varphi_1, \lambda_1)[/tex] are the latitude and longitude of the starting point in radians.

[tex]\large (\varphi_2, \lambda_2)[/tex] are the latitude and longitude of the end point in radians.

R = radius of  Earth in kilometers.

Be careful to convert the angles into radians before computing the trigonometric functions. This can be done by cross-multiplication knowing that 180° ≅ 3.141592654 radians.

In the problem we have

[tex]\large \varphi_1[/tex] = 48° = 0.837758041 radians

[tex]\large \lambda_1[/tex] = 161° = 2.809980096 radians

[tex]\large \varphi_2[/tex] = 54° = 0.942477796 radians

[tex]\large \lambda_2[/tex] = 133° = 2.321287905 radians

so

[tex]\large (\varphi_2-\varphi_1)/2[/tex] = 3° = 0.052359878 radians

[tex]\large (\lambda_2-\lambda_1)/2[/tex] = -14° = -0.2443461 radians

Replacing in the formula for the distance:

[tex]\large d=2R*arcsin\left(\sqrt{sin^2(0.052359878)+cos(0.837758041)cos(0.942477796)sin^2(-0.2443461)}\right)=\\\\=0.32237835*R[/tex]

According to NASA, the radius of  Earth at the poles is around 6,356 Km and at the equator is 6,378 Km.

Since they traveled around the middle point between the equator and the North pole, a better estimate of the radius in this case would be the average (6,378+6,356)/2 = 6,367 Km

We have that an approximation to the distance traveled would be  

0.32237835*6,367 = 2,052.583 Km

b)

Assuming that the shoes where left and found the same year, there are 22 days from April 30th  to May 22nd , so they traveled for around 22 days (this may vary slightly depending on the exact time the shoes were left and found)

c)

They traveled 2,052.583 Km in 22 days.

22 days equals 22*24 = 528 hours.

By cross-multiplication

528 h _______  2,052.583 Km

1 h __________ x Km

x =  2,052.583/528 ≅ 3.887 Km

So they traveled at a rate of 3.887 Km/h

Answer:

76

Step-by-step explanation:

Final answer:

The worst exam score when compared to the other students is the one with the lowest z-score. In this case, with a z-score of -0.8, the 78 in Math is the worst score.

Explanation:

To determine which exam score was the worst compared to the other students, we need to calculate the number of standard deviations the student's score falls from the class mean. This calculation is known as a z-score. A z-score allows us to understand the position of a score within the context of a distribution.

For the Math exam:
Z = (Your score - Class mean) / Standard deviation
= (78 - 82) / 5
= -0.8.

For the History exam:
Z = (85 - 87) / 4
= -0.5.

For the English exam:
Z = (80 - 76) / 2.5
= 1.6.

The score with the lowest z-score is the worst because it is the farthest below the mean. Comparing the z-scores, the Math exam has the lowest z-score at -0.8, so the 78 in Math is the worst score when compared to the other students in your classes.