The amount by which we must raise the price in order to retain your existing dollar profit margin is $2.04; the right solution is (a).
What is an expression?The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.
Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.
There was a trucking strike that caused a prime ingredient in your menu item to shoot up by 17%. You were paying $12 per package.
The increases in the price of the ingredient = 17% of 12
Increase = 0.17 × 12
Increase = 2.04
Therefore, the amount by which we must raise the price in order to retain your existing dollar profit margin is $2.04; the right solution is (a).
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The price of the menu item should be increased by $2.04.
Let's begin by calculating the new cost per package after the price increase. If the original cost per package was $12 and it increased by 17%, we calculate the increase as follows:
Increase = 12 * 0.17 = 2.04
The new cost per package will be:
New cost = 12 + 2.04 = 14.04
To maintain the same dollar profit margin on your menu item, you need to increase the price by the same amount as the cost increase, which is $2.04.
(1) True/False: (a) To reduce the margin of error in a stratified sample, we should design strata so that the units within each stratum are as similar as possible. (b) To reduce the margin of error in a cluster sample, we should design clusters so that the units within each cluster are as similar as possible.
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.
Each point on the edge of a circle is equidistant from the center of the circle. The center of a circle is located at (6,3). Which point
on the y-axis could be on the edge of the circle if the distance from the center of the circle to the edge is 10 units?
Answer:
(0, –5)
Step-by-step explanation:
If you don't see this option there is another option, (0 , 11) is also correct.
An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores below the top 13%13% and above the bottom 56%56% C: Scores below the top 44%44% and above the bottom 21%21% D: Scores below the top 79%79% and above the bottom 9%9% F: Bottom 9%9% of scores Scores on the test are normally distributed with a mean of 79.779.7 and a standard deviation of 8.48.4. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.
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.
For many important processes that occur in the body, direct measurement of characteristics of the process is not possible. In many cases, however, we can measure a biomarker, a biochemical substance that is relatively easy to measure and is associated with the process of interest. Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and the building of new bone, called formation. A biomarker for bone formation measured was osteocalcin (OC), measured in the blood. The units are nanograms per milliliter (ng/ml). For the 31 subjects in the study the mean was 33.4 ng/ml. Assume that the standard deviation is known to be 19.6 ng/ml. Report the 95% confidence interval:
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.
Explanation: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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what is the product of -3/8 and -4/12
Answer:
0.125
Step-by-step explanation:
Answer:
1/8
Step-by-step explanation:
-3/8 * -4/12
We can simplify the second fraction
Divide the top and bottom by 4
4/12 = 1/3
-3/8 * -1/3
The threes in the top and bottom cancel
A negative times a negative cancel
1/8
A sample of blood pressure measurements is taken from a data set and those values (mm Hg) are listed below. The values are matched so that subjects each have systolic and diastolic measurements. Find the mean and median for each of the two samples and then compare the two sets of results. Are the measures of center the best statistics to use with these data? What else might be better?
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.
Explanation:
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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5. 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 Calculate a 99% CI for the variance σ 2 , and the standard deviation σ of the coating layer thickness distribution
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.
Joaquin has a $50 gift card from pizza palace . All pizzas at pizza palace cost $7 each . Joaquin wants to give the driver a $5 tip for delivering the pizza . Which inequality could be used to determine the possible number of pizzas,p, he could order with his gift card , including the tip?
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]
Which of these is an example of a non-random sample?
A.
At a school assembly, five students are randomly chosen to receive free admission to a theme park.
B.
Out of all the seventh grade students in a public school district, fifteen are chosen to win a trip to a vacation destination.
C.
Registered voters in Arizona are surveyed to determine if they have relatives in Florida.
D.
Airline passengers to Orlando, Florida, are asked about vacation plans.
Answer:
it is D :)
Step-by-step explanation:
trust me ;)
Final answer:
Option D, which is about asking airline passengers to Orlando about their vacation plans, is an example of a non-random sample, specifically convenience sampling.
Explanation:
An example of a non-random sample is option D: Airline passengers to Orlando, Florida, are asked about vacation plans. This is considered non-random sampling because the passengers already have something in common – they are traveling to a popular vacation destination, which likely influences their vacation plans. This method does not give every individual in the broader population an equal chance of being selected and is thus not a random sample. This type of sampling is referred to as convenience sampling, as it involves selecting individuals who are easily accessible rather than using a process that gives every individual an equal chance of being chosen.
A report on the U.S. economy indicates that 28% of Americans have experienced difficulty in making mortgage payments. A news organization randomly sampled 400 Americans from 10 cities judged to be especially economically depressed and found that 136 reported such difficulty. Does this indicate that the problem is more severe among these cities? What are the correct null and alternative hypotheses for testing such a claim?
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]
An ant arrives at the snail’s starting position at time t=12 minutes and follows the snail’s path. During the interval 12≤t≤15 minutes, the ant travels in the same direction as the snail with a constant acceleration of 2 inches per minute per minute. The ant catches up to the snail at time t=15 minutes. The ant’s velocity at time t=12 is B inches per minute. Find the value of B.
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.)
What is velocity?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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A single die is rolled twice. The set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), 4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}. 9. Find the probability of getting two numbers whose sum is 9. 10.Find the probability of getting two numbers whose sum is 4. 11.Find the probability of getting two numbers whose sum is less than 7. 12.Find the probability of getting two numbers whose sum is greater than 8 13.Find the probability of getting two numbers that are the same (doubles). 14.Find the probability of getting a sum of 7 given that one of the numbers is odd. 15.Find the probability of getting a sum of eight given that both numbers are even numbers. 16.Find the probability of getting two numbers with a sum of 14.
Answer:
(9)[tex]\frac{1}{12}[/tex] (10) [tex]\frac{1}{12}[/tex] (11)[tex]\frac{5}{12}[/tex] (12)[tex]\frac{1}{4}[/tex] (13)[tex]\frac{1}{6}[/tex] 14)[tex]\frac{5}{36}[/tex] (15)[tex]\frac{1}{12}[/tex] (16)0
Step-by-step explanation:
The sample Space of the single die rolled twice is presented below:
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),}.
n(S)=36
(9)Probability of getting two numbers whose sum is 9.
The possible outcomes are: (3, 6), (4, 5), (5, 4)
[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]
10) Probability of getting two numbers whose sum is 4.
The possible outcomes are: (1, 3),(2, 2),(3, 1),
[tex]P(\text{two numbers whose sum})=\frac{3}{36}=\frac{1}{12}[/tex]
11.)Find the probability of getting two numbers whose sum is less than 7.
The possible outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)
[tex]P(\text{two numbers whose sum is less than 7})=\frac{15}{36}=\frac{5}{12}[/tex]
12.Probability of getting two numbers whose sum is greater than 8
The possible outcomes are:(4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)
[tex]P(\text{two numbers whose sum is greater than 8})=\frac{9}{36}=\frac{1}{4}[/tex]
(13)Probability of getting two numbers that are the same (doubles).
The possible outcomes are:(1, 1)(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
[tex]P(\text{two numbers that are the same})=\frac{6}{36}=\frac{1}{6}[/tex]
14.Probability of getting a sum of 7 given that one of the numbers is odd.
The possible outcomes are: (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
[tex]P(\text{getting a sum of 7 given that one of the numbers is odd.})=\frac{5}{36}[/tex]
(15)Probability of getting a sum of eight given that both numbers are even numbers.
The possible outcomes are: (2, 6), (4, 4), (6, 2)
[tex]P(\text{getting a sum of eight given that both numbers are even numbers.})=\frac{3}{36}\\=\frac{1}{12}[/tex]
16.Probability of getting two numbers with a sum of 14.
[tex]P(\text{getting two numbers with a sum of 14.})=\frac{0}{36}=0[/tex]
The radius of the earth is 3959 miles. A meteor with a diameter of 750 miles strikes the earth in unpopulated area. What percentage of the earths surface is affected by the meteor strike
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.
Explanation: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:
AEarth = 4π * (39592) = about 197 million square milesAMeteor = π * (7502)/4 = about 441,963 square milesTo 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.
With gasoline prices increasing, drivers are more concerned with their cars’ gasoline consump- tion. 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. Now that his car is 5 years old, he would like to know whether the variability of gas mileage has changed. He recorded the gas mileage from his last eight fill-ups; these are listed here. Conduct a test at a 10% significance level to infer whether the variability has changed.
28 25 29 25 32 36 27 24
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.
2.36 Is it worth it?: Andy is always looking for ways to make money fast. Lately, he has been trying to make money by gambling. Here is the game he is considering playing: 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. Round answers to 2 decimal places. a) Andy's expected profit per game is: $ b) Would you recommend this game to Andy as a good way to make money? Explain. Yes, Andy could be lucky and might earn money in the long-run playing this game No, we expect Andy to lose money each time he plays this game
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.
No, 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.
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A chemist prepared ten 4.85 g quantities of aniline and purified it to acetanilide using fractional crystallization. The following dry yields were recorded.
3.83 3.82 3.90 3.87 3.92 3.34 3.64 3.99 3.70 3.85
Estimate the mean grams of acetanilide that can be recovered from an initial amount of 4.85 g of aniline. Use a 95% confidence interval. (Round your answers to three decimal places.)
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.
Explanation: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.
Calculate the sample mean by summing up all the dry yields and dividing by the number of samples (10).Calculate the standard deviation of the dry yields.Calculate the standard error of the mean by dividing the standard deviation by the square root of the sample size.Calculate the margin of error by multiplying the standard error by the appropriate t-value from the t-distribution table for a 95% confidence level.Finally, calculate the lower and upper bounds of the confidence interval by subtracting and adding the margin of error to the sample mean, respectively.Learn more about Mean and confidence interval here:https://brainly.com/question/32546207
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An article included the following statement: "Few people believe there's much reality in reality TV: a 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. Compute a bound on the error (based on 95% confidence) of estimation for the reported proportion of 0.86. (Round your answer to three decimal places.) Interpret the bound. (Round your answers to one decimal place.) We are % confident that the proportion of all adults who believe that the shows are either "totally made up" or "mostly distorted" is within % of the sample proportion of %.
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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To open a combination lock, you turn the dial to the right and stop at a number; then you turn it to the left and stop at a second number. Finally, you turn the dial back to the right and stop at a third number. If you used the correct sequence of numbers, the lock opens. If the dial of the lock contains 8 numbers, 0 through 7, determine the number of different combinations possible for the lock. Note: The same number can be reused consecutively
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.
an angle measures 135 degrees through what fraction does the angle turn
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=
A marketing research company desires to know the mean consumption of meat per week among people over age 30. 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. Construct the 95% confidence interval for the mean consumption of meat among people over age 30. Round your answers to one decimal place.
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].
256 divided by 11 equals what
23.27
Use a calculator
What’s the answer too 76.1*9.6
Answer:
[tex]76.1\times 9.6=730.56[/tex]
Step-by-step explanation:
In this case, we need to find the value of [tex]76.1\times 9.6[/tex]. The numbers are in decimal form. It is understood that the final answer will have decimal point after two places from right.
For an instance, multiply remove decimals from both numbers. Now simply multiply 761 and 96 such that we will get 73056.
Now, keep the decimal point after two places from right. So, we will get 730.56.
Final answer:
The answer to the multiplication of 76.1 and 9.6 is 730 when rounded to two significant figures in accordance with the rules of significant figures in multiplication.
Explanation:
The student asked for the product of 76.1 and 9.6. To compute this, you multiply the two numbers. Since 76.1 has three significant figures and 9.6 has two significant figures, our final answer should be reported with two significant figures, owing to the least number of significant figures in the given numbers.
Now, calculating the product:
76.1 times 9.6 = 730.56
When rounding to two significant figures, the answer is 730.
In calculations involving significant figures, it is important to report your final answer with the correct number of significant figures. In multiplication and division, the number of significant figures in the final answer should be the same as the least number of significant figures in any of the numbers being calculated.
what is the equation of the line that passes through the point (-2,-2)and has a slope of 2
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)
Alicia earned the following amounts pet sitting: $28, $36, $15, $43, $28. Explain how you would describe the data. Use at least 4 words from the Word List in your explanation.
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.
Based on this data, what is a reasonable estimate of the probability that Todd runs more than 15 km tomorrow?
Answer: 5/13
Step-by-step explanation:
I got it correct
Suppose that the number of asbestos particles in a sample of 1 squared centimeter of dust is a Poisson random variable with a mean of 1000. What is the probability that 10 squared centimeters of dust contains more than 10150 particles? Use normal approximation without continuity correction. Round your answer to 3 decimal places. Enter your answer in accordance to the question statement
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.
Explanation: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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ASAP, due soon Please help! I'll mark you brainliest if it's right! Fill in the blanks!
Answer:
(x + 10)^2 + (-186)
Step-by-step explanation:
x^2 + 20x - 86 =
Move the constant term to the right leaving a space between the x-term and the constant term.
= x^2 + 20x - 86
To complete the square, take half of the x-term coefficient and square it.
Half of 20 is 10. 10 squared is 100. This is the number that completes the square. Add it right after the x-term. Now you need to subtract the same amount at the end.
= x^2 + 20x + 100 - 86 - 100
Since 100 was added and subtracted, the expression has the same value. The first three terms are a perfect square trinomial, so we write it as the square of a binomial.
= (x + 10)^2 - 186
Since you have an addition sign, we write -186 as a sum:
= (x + 10)^2 + (-186)
Answer:
Step-by-step explanation:
+10 and -186 is your answer.
Suppose the weights of tight ends in a football league are normally distributed such that σ2=400. A sample of 11 tight ends was randomly selected, and the weights are given below. Calculate the 95% confidence interval for the mean weight of all tight ends in this league. Round your answers to two decimal places and use ascending order.
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)
Given the exponential growth equation y = 50(1.6)^x what is the initial value?
Answer:
50 is the initial value
Step-by-step explanation:
The equation is of the form
y = ab^x where a is the initial value and b is the growth/decay value
b>1 is growth and b<1 means decay
y = 50(1.6)^x
50 is the initial value and 1.6 means growth since it is greater than 1
1.6 -1 = .6 so it has a 60% growth rate
Consider testing Upper H 0 : mu equals 20H0: μ=20 against Upper H Subscript a Baseline : mu less than 20Ha: μ<20 where muμ is the mean number of latex gloves used per week by all hospitalemployees, based on the summary statistics nequals=444, x overbarxequals=19.3 and sequals=11.1 Complete parts a and b.
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
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