Answer:
for sample = xbar
population = μ
Step-by-step explanation:
The arithmetic mean for sample can be represented by xbar and it can be calculated as
xbar=∑xi/n
Where xi represents data values and n represents number of data values in a sample.
The arithmetic mean for population can be represented by μ and it can be calculated as
μ=∑xi/N
Where xi represents data values and N represents number of data values in a population.
The symbol for the arithmetic mean when it's a sample statistic is 'x', and 'μ' when it's a population parameter. Sample mean (x) refers to the mean of a sample group, while population mean (μ) refers to the mean of an entire population.
Explanation:The symbol that is used for the arithmetic mean when it is a sample statistic is 'x' (pronounced as x-bar). This is used to denote the mean of a sample. On the other side, the symbol that is used when the arithmetic mean is a population parameter is 'μ' (pronounced as mu). This symbol is representative of the mean of an entire population.
Take for example a group of 50 students in a class who took a test. If you wanted to find the sample mean (x) of their scores, you'd add all their scores and divide by the total number of students (which is 50 in this case). However, if you were to calculate the population mean (μ) of the scores of all students in all high school classes in the nation, you'd add their scores and divide by the total number.
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Solve, graph, and give interval notation for the compound inequality:
7 (x + 2) −8 ≥ 13 AND 8x − 3 < 4x − 3
Answer:
The answer to your question is below
Step-by-step explanation:
Inequality 1
7(x + 2) - 8 ≥ 13
7x + 14 - 8 ≥ 13
7x + 6 ≥ 13
7x ≥ 13 - 6
7x ≥ 7
x ≥ 7/7
x ≥ 1
Inequality 2
8x - 3 < 4x - 3
8x - 4x < - 3 + 3
4x < 0
x < 0 / 4
x < 0
Interval notation (-∞ , 0) U [1, ∞)
See the graph below
A typical incoming telephone call to your catalog sales force results in a mean order of $28.63 with a standard deviation of $13.91. You may assume that orders are received independently of one another. a. Based only on this information, can you find the probability that a single incoming call will result in an order of more than $40? Why or why not? b. An operator is expected to handle 110 incoming calls tomorrow. Find the mean and standard devi- ation of the resulting total order. c. What is the approximate probability distribution of the total order to be received by the operator in part b tomorrow? How do you know? d. Find the (approximate) probability that the operator in part b will generate a total order of more than $3,300 tomorrow. e. Find the (approximate) probability that the operator in part b will generate an average order between $27 and $29 tomorrow.
Answer:
Step-by-step explanation:
Hello!
The study variable for this exercise is:
X: Price of an order of a sales catalog placed per telephone.
You don't have information about the distribution of this variable, but you know that the mean is μ= $28.63 and the standard deviation is σ= $13.91
a. You need to calculate the probability of a single incoming call resulting in an order of more than $40, symbolically P(X>$40). To be able to calculate this probability you need to know what the distribution of the variable is. Without knowing the form of the distribution, i.e. the probability density function, you cannot tell what is the asked probability.
b. If the operator is expected to handle 110 calls in a day (tomorrow) this means that you have a sample of n= 110 calls, and in each call, you are going to take the information of the order placed by the customer.
Note:
If X₁, X₂, ..., Xₙ be the n random variables that constitute a sample, then any function of type θ = î (X₁, X₂, ..., Xₙ) that depends solely on the n variables and does not contain any parameters known, it is called the estimator of the parameter.
When the function i (.) It is applied to the set of the n numerical values of the respective random variables, a numerical value is generated, called parameter estimate θ.
This follows the concepts:
1) The function i (.) It is a function of random variables, so it is also a random variable, that is to say, that every estimator is a random variable.
2) From the above, it follows that Î has its probability distribution and therefore mathematical hope, E (î), and variance, V (î).
And:
The central limit theorem states that if a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
This means that if you have the study variable X with a certain distribution, then it's the sample mean X[bar], that is also an aleatory variable, will have the same distribution as it's origin variable. On the other hand, if the distribution of the said variable is unknown, so will be the distribution of its sample mean, but if the sample is large enough, then you can apply the central limit theorem and approximate the distribution of the sample mean to normal, symbolically:
X~?(μ;δ²) and n ≥ 30 then X[bar]≈N(μ;δ²/n )
The mean of said approximation is the same as the mean of the variable of origin.
μ= $28.63
And the standard deviation will be the same as the original variable but affected by the sample size:
δ/√n = $13.91/110= $0.126 ≅ $0.13
c. X[bar]≈N(μ;δ²/n )
d. Using the aproximation you can calculate the asked probabilities with the standard normal:
P(X[bar] > $3,300) = P(Z > [tex]\frac{3.300- 28.63}{13.91/\sqrt{110} }[/tex]) = P(Z > -19.098)= 0
e.
P(27 < X[bat] < 29) = P(X<29) - P(X<27) = P(Z<[tex]\frac{29-28.63}{0.13}[/tex]) - P(Z<[tex]\frac{27-28.63}{0.13}[/tex])
P(Z<0.278) - P(Z<-1.229)= 0.609 - 0.110= 0.499
I hope it helps!
A short quiz has two true-false questions and one multiple-choice question with four possible answers. A student guesses at each question. Assuming the choices are all equally likely and the questions are independent of each other, the following is the probability distribution of the number of answers guessed correctly. What is the Probability of getting less than all three right
Final answer:
This probability question in Mathematics aims to calculate the chances of guessing answers correctly on quizzes or exams with true-false and multiple-choice questions.
The probability of getting less than all three right on the quiz with true-false and multiple-choice questions can be calculated by summing the probabilities of getting 0, 1, or 2 correct answers.
Explanation:
Probability of Getting Less Than All Three Right:
For the quiz described, the probability distribution of the number of correct answers is as follows:
0 correct: 1/8
1 correct: 3/8
2 correct: 3/8
3 correct: 1/8
To find the probability of getting less than all three right, you would add the probabilities of getting 0, 1, or 2 correct, which is 1/8 + 3/8 + 3/8 = 7/8.
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. How long would it take to accumulate 5 centimeters of clay?
Answer:
It will take 50,000 years to accumulate 5 centimeters of clay.
Step-by-step explanation:
The relationship between millimeter and centimer is that:
1ml = 0.1cm
So
How many ml are 5 cm?
1 ml - 0.1cm
x ml - 5cm
[tex]0.1x = 5[/tex]
[tex]x = \frac{5}{0.1}[/tex]
[tex]x = 50[/tex] ml
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. How long would it take to accumulate 5 centimeters of clay?
5cm is 50 ml.
1 ml per 1000 years.
So
1 ml - 1000 years
50 ml - x years
[tex]x = 50*1000[/tex]
[tex]x = 50,000[/tex]
It will take 50,000 years to accumulate 5 centimeters of clay.
Answer:
It will take 50000 years to accumulate 5 centimeters of clay.
Step-by-step explanation:
Clay on the deep seafloor accumulates at a rate of about 1 millimeter per 1,000 years. To determine the amount of time it will take to accumulate 5 centimeters of clay, we would convert 5 centimeters to millimeters.
1 centimeter = 10 millimeters
5 centimeters = 5 × 10 = 50 millimeters
Therefore,
If 1 millimeter = 1000 years,
Then, 50 millimeters = 50 × 1000 =
50000 years.
The price-demand equation for gasoline is 0.1 x + 4 p = 85 where p is the price per gallon in dollars and x is the daily demand measured in millions of gallons.a. What price should be charged if the demand is 40 million gallons?b. If the price increases by $0.4 by how much does the demand decrease?
Answer:
a. The price that should be charged if the demand is 40 million gallons is $20.25.
b. The demand decreases by 16 millions of gallons.
Step-by-step explanation:
We know that the price-demand equation for gasoline is given by
[tex]0.1 x + 4 p = 85[/tex]
where
p is the price per gallon in dollars and
x is the daily demand measured in millions of gallons.
a. To find what price should be charged if the demand is 40 million gallons you must
Solve for p,
[tex]0.1x\cdot \:10+4p\cdot \:10=85\cdot \:10\\x+40p=850\\40p=850-x\\p=\frac{850-x}{40}[/tex]
We know that the demand is 40 million gallons (x = 40). So,
[tex]p=\frac{850-40}{40}=\frac{81}{4}=20.25[/tex]
b. To find how much does the demand decrease when the price increases by $0.4 you must
Solve for x,
[tex]0.1x\cdot \:10+4p\cdot \:10=85\cdot \:10\\x+40p=850\\x=850-40p[/tex]
We know that the price increases by $0.4. So,
[tex]-40\left(0.4\right)=-16[/tex]
The demand decreases by 16 millions of gallons.
When the demand is 40 million gallons, the price per gallon should be $20.25. The impact of a $0.4 price increase on the demand can be calculated by substitifying p in the equation, solving for x, and subtracting the original x value.
Explanation:The subject of this question is algebra, specifically dealing with the use of equations representing real-world scenarios. In this case, the equation represents price-demand dynamics for gasoline.
a. To find the price that should be charged when the demand is 40 million gallons, substitute x with 40 in the equation, which gives 0.1 * 40 + 4p = 85. By simplifying this, we get 4 + 4p = 85. Further solving for p, we get 4p = 81, therefore p = 81 / 4, which is $20.25 per gallon.
b. When the price increases by $0.4, substitute p with p + 0.4 in the equation. This gives 0.1x + 4(p + 0.4) = 85. Solving this for x, and then subtracting the original x value, gives us the decrease in demand due to the increase in price.
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The table shows the functions representing the height and base of a triangle for different values of x The area of the triangle when x= 2 is 14. Which equation can be used to represent the area of the triangle, A(x)?
Answer: option 2 is the correct answer.
Step-by-step explanation:
When x = 2, the area is 14.
It means that height = 2² + 3 = 7
It means that base = 2² = 4
Area = 1/2 × 7 × 4 = 14
Therefore, it is a right angle triangle, the formula for determining the area of the triangle is expressed as
Area = 1/2bh
Where
b represents the base of the triangle.
h represents the height of the triangle.
Since height is f(x) = x² + 3 and base is g(x) = 2x,
The equation that can be used to represent the area is
A(x) = 0.5(f.g)(x)
Use the graph to fill in the blank with the correct number. f(−2) = ________ X, Y graph. Plotted points negative 3, 0, negative 2, 2, 0, 1, and 1, negative 2. Numerical Answers Expected! Answer for Blank 1:
The given points
[tex](-3, 0)[/tex]
[tex](-2, 2)[/tex]
[tex](0, 1)[/tex]
[tex](1, -2)[/tex]
Imply that
[tex]f(-3)=0[/tex]
[tex]f(-2)=2[/tex]
[tex]f(0)=1[/tex]
[tex]f(1)=-2[/tex]
Answer:
f(2) = -1.
Step-by-step explanation:
In a data set with mean of 12 and standard deviation of 4, at least what percent of data falls between 4 and 20?
Answer:
At least 95% of data falls between 4 and 20.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 12
Standard deviation = 4
At least what percent of data falls between 4 and 20?
4 = 12 - 2*4
So 4 is two standard deviations below the mean
20 = 12 + 2*4
So 20 is two standard deviations above the mean
By the Empirical Rule, at least 95% of data falls between 4 and 20.
A certain museum has five visitors in two minutes on average. Let a Poisson random variable denote the number of visitors per minute to this museum. Find the variance of (write up to first decimal place).
Answer:
The variance is 2.5.
Step-by-step explanation:
Let X = number of visitors in a museum.
The random variable X has an average of 5 visitors per 2 minutes.
Then in 1 minute the average number of visitors is, [tex]\frac{5}{2} =2.5[/tex]
The random variable X follows a Poisson distribution with parameter λ = 2.5.
The variance of a Poisson distribution is:
[tex]Variance=\lambda[/tex]
The variance of this distribution is:
[tex]V(X)=\lambda=2.5[/tex]
Thus, the variance is 2.5.
A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?
a. 0.7510 and 0.8290
b. 0.6592 and 0.7044
c. 0.6659 and 0.6941
d. 0.6795 and 0.6805
Answer:
b. 0.6592 and 0.7044
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. This means that [tex]n = 1100, p = \frac{750}{1100} = 0.6818[/tex]
90% confidence interval
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 - 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.6592[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 + 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.7044[/tex]
So the correct answer is:
b. 0.6592 and 0.7044
By the fourth quarter of 2015, U.S. households had accumulated $12.5 trillion in housing equity, which represents about 14 percent of their net worth. What proportion of U.S. households own their home
Answer:
two-thirds
Step-by-step explanation:
In 2015, there was a campaign for the accumulation of households in the United States of America. Most of the citizens tried their best possible to acquire property in terms of buildings and other facilities. In the last quarter of the year, approximately two-thirds of the home in the United States of America were owned by households.
(Pitman 3.4.9) Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up first on the nth toss. If we play this game repeatedly, how much money do you expect to win or lose per game over the long run?
Answer:
Let's make a couple of assumptions to clarify the situation. First, the coin flipping is fair, that is, each flip is independent of all the others and for each flip, the probabilities of heads and tails are both 1/2. Second, you have enough money to pay me no matter how many tails are flipped before the first head.
Under those assumptions, the expected amount of money I will win in infinite.
In decision theory, utility is often used to make decisions rather than money. If my utility is proportional to expected monitory payoff, I should pay whatever I can scrape up, my total assets. For some reason, economists often assume utility functions have deminishing returns, and eventually flatten out.
In that case, the expected amount of utility payoff will be lower than the maximum utility. What does that mean for this game? It means it won't matter to me whether I get some large quantity of money like a trillion dollars, or any larger quantity of money, like a quadrillion dollars. All my needs are met by a trillion dollars. That's 240 dollars. So I certainly shouldn't pay more than $40 to play the game. As the utility function starts to flatten out earlier, perhaps $30 would come out to be a fair payment.
A whistle is made of a square tube with a notch cut in its edge, into which a baffle is brazed. Determine the dimensions d and θ for the baffle. Take b = 6.5 cm.
Answer:
d = 7.51 cm
θ = 60°
Step-by-step explanation:
The baffle used in the notch of a whistle is a triangular baffle (Isosceles triangle OAB).
For the isosceles triangle, the sides with equal dimension are OA and OB which is represented with d
d = the vertical component of the side of a triangle
It is given by
6.5cm = d sin60
d = 6.5/sin60
d = 7.505553499465134
d = 7.51 cm -------- Approximated
To calculate angle θ
Angle on a straight line is = 180
So, 60 + 60 + θ = 180
θ= 180 - 120
θ = 60°
(See attachments below)
Suppose we know that the functions r and s are are everywhere differentiable and that u(3)=0. Suppose we also know that for 1 ≤ x ≤ 3, the area between the x-axis and the non negative functions h(x)=u(x)dv/dx is 15, and that on the same interval, the area between the x-axis and the non negative function k(x) = v(x)du/dx is 20. Determine u(1)v(1).
Integrating by parts, we have
[tex]\displaystyle\int_1^3\underbrace{u(x)\dfrac{\mathrm dv}{\mathrm dx}}_{h(x)}\,\mathrm dx=u(3)v(3)-u(1)v(1)-\int_1^3\underbrace{v(x)\dfrac{\mathrm du}{\mathrm dx}}_{k(x)}\,\mathrm dx[/tex]
We're given [tex]u(3)=0[/tex] and [tex]\int_1^3h(x)\,\mathrm dx=15[/tex] and [tex]\int_1^3k(x)\,\mathrm dx=20[/tex]. So we have
[tex]15=-u(1)v(1)-20\implies\boxed{u(1)v(1)=-35}[/tex]
This histogram shows the times, in minutes, required for 25 rats in a animal behavior experiment to successfully navigate a maze. What percentage of the rats navigated the maze in less than 5.5 minutes? 34% 60% 68% 70% 84%
Answer:
The question is lacking the image of the histogram, but the file attachment to this answer contains the complete question and histogram image.
The percentage of the rats that navigated the maze in less than 5.5 minutes is 84%
Step-by-step explanation:
First of all let us compute the total frequency which represents the total number of rats in the experiment.
from the information given in the question, we are told that the total number of rats involved in the experiment are 25 rats and this makes up the total frequency.
To calculate the percentage of rats that navigated the maze in less than 5.5 minutes, you will first of all need to identify the 5.5 minute point on the histogram and add all the frequencies below that point. This gives the total number of rat that navigated in less than 5.5 minutes. These frequencies from 0 to <5.5 are; 3, 8, 6, and 4. And the total is given as the sum of these frequencies shown below:
Total number of rats that navigated in less than 5.5 minutes = 3 + 8 + 6 + 4 = 21. Hence, a total of 21 rats navigated the maze in less than 5.5 minutes.
Now to find the percentage of the number of rats that navigated in less than 5.5 minutes, we have to find out what percentage of 25 (total number of rats) is 21 (number of rats that navigated in less than 5.5 minutes). This is calculated thus:
[tex]\frac{21}{25}[/tex] × 100 = 0.84 × 100 = 84 %.
Therefore 84% of the total number of rats navigated the maze in less than 5.5 minutes in the experiment.
To find the percentage of rats that navigated the maze in less than 5.5 minutes, locate the bar on the histogram that represents that interval and calculate the percentage based on the number of rats in that interval compared to the total number of rats.
Explanation:The histogram shows the times required for 25 rats to navigate a maze. To find the percentage of rats that navigated the maze in less than 5.5 minutes, we need to look at the data on the histogram. The histogram is divided into various time intervals, and the bars represent the number of rats that fall into each interval. You need to locate the bar that corresponds to the interval less than 5.5 minutes and calculate the percentage of rats represented by that bar.
Let's say the bar for the interval less than 5.5 minutes represents 10 rats. To calculate the percentage, we divide the number of rats in that interval (10) by the total number of rats (25) and multiply by 100:
(10/25) x 100 = 40%
Therefore, 40% of the rats navigated the maze in less than 5.5 minutes.
A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 0.5 meters, its length is 5 meters, and its top is 2 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use g=9.8 m/s2.)
Final answer:
The work needed to pump the gasoline out of the underground tank is approximately 51.7 kJ, calculated using the density of gasoline, the volume of the cylindrical tank, and the gravitational energy required to lift the gasoline 2 meters up to ground level.
Explanation:
The total amount of work needed to pump the gasoline out of the tank can be determined using the concepts of physics specifically mechanical work and fluid dynamics. We know the density of gasoline is 673 kg/m3, the gravitational acceleration is 9.8 m/s2, the cylinder's radius is 0.5 meters, its length is 5 meters, and the top of the cylinder is 2 meters below ground level. The total volume of the cylinder is given by the formula for the volume of a cylinder V = πr2h, where r is the radius and h is the length of the cylinder. In this case, V = π(0.5)2(5) ≈ 3.927 m3. The total mass m of the gasoline can be calculated by multiplying the density of gasoline by the volume, m = density × volume = 673 kg/m3 × 3.927 m3 ≈ 2643.871 kg.
Since the gas tank is underground, the work done to lift the gasoline to ground level is W = mgh, where m is the mass of the gasoline, g is acceleration due to gravity, and h is the height the gasoline is lifted. We must lift the gasoline 2 meters to reach ground level, so the work done is W = 2643.871 kg × 9.8 m/s2 × 2 m ≈ 51738.7856 J or 51.7 kJ (since 1 J = 1 kg·m2/s2). Thus, the work required to pump the gasoline out of the tank would be approximately 51.7 kJ.
There are five sales associates at Mid-Motors Ford. The five associates and the number of cars they sold last week are: Sales Associate Cars Sold Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6
a. How many different samples of size 2 are possible?
b. List all possible samples of size 2, and compute the mean of each sample.
c. Compare the mean of the sampling distribution of the sample mean with that of the
population.
Answer:
a) There are 10 different samples of size 2.
b) See the explanation section
c) See the explanation section
Step-by-step explanation:
a) We need to select a sample of size 2 from the given population of size 5. We use combination to get the number of difference sample.
[tex]\{ {{5} \atop {2}} \} = \frac{5!}{2!(5-2)!} \\= \frac{5!}{2!3!} \\= \frac{120}{2*6} \\= \frac{120}{12} \\=10[/tex]
b) Possible sample of size 2:
Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6
Peter Hankish and Connie Stallter ( Mean = (8 + 6)/2 = 14/2 = 7)Peter Hankish and Juan Lopez (Mean = (8 + 4)/2 = 12/2 = 6)Peter Hankish and Ted Barnes (Mean = (8 + 10)/2 = 18/2 = 9)Peter Hankish and Peggy Chu (Mean = (8 + 6)/2 = 14/2 = 7)Connie Stallter and Juan Lopez (Mean = (6 + 4)/2 = 10/2 = 5)Connie Stallter and Ted Barnes (Mean = (6 + 10)/2 = 16/2 = 8)Connie Stallter and PeggyChu (Mean = (6 + 6)/2 = 12/2 = 6)Juan Lopez and Ted Barnes (Mean = (4 + 10)/2 = 14/2 = 7)Juan Lopez and Peggy Chu (Mean = (4 + 6)/2 = 10/2 = 5)Ted Barnes and Peggy Chu (Mean = (10 + 6)/2 = 16/2 = 8)c) The mean of the population is:
[tex]mean = \frac{(8+6+4+10+6)}{5} \\= \frac{34}{5} \\= 6.8[/tex]
Comparing the mean of the population and the sample; we can say that most of the 2-size sample have their mean higher than that of the population sample. And the variation with the mean is not much. Some sample have their mean greater than population mean, while some sample have their mean greater than the population mean.
This question is based on the statistics. Therefore, the answers of all the questions are explained below.
Given:
There are five sales associates at Mid-Motors Ford. Sales Associate Cars Sold Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6.
(a) We have to find different samples of size 2 are possible.
Thus, we have to select sample of size 2 from given population of size 5.
So, by using combination,
[tex]5_{c_2} = \dfrac{5!}{2! (5-2)!} =\dfrac{120}{12} = 10[/tex]
Thus, 10 different samples of size 2 are possible.
(b) We have to find list all possible samples of size 2, and compute the mean of each sample.
Peter Hankish 8 ,Connie Stallter 6, Juan Lopez 4, Ted Barnes 10, Peggy Chu 6.
Peter Hankish and Connie Stallter ( Mean = [tex]\dfrac{8+6}{2} = 7[/tex] Peter Hankish and Juan Lopez (Mean =[tex]\dfrac{8+4}{2} = 6[/tex] Peter Hankish and Ted Barnes (Mean = [tex]\dfrac{8+10}{2} = 9[/tex] Peter Hankish and Peggy Chu (Mean = [tex]\dfrac{8+6}{2} = 7[/tex] Connie Stallter and Juan Lopez (Mean = [tex]\dfrac{4+6}{2} =5[/tex] Connie Stallter and Ted Barnes (Mean = [tex]\dfrac{10+6}{2} = 8[/tex] Connie Stallter and PeggyChu (Mean = [tex]\dfrac{6+6}{2} = 6[/tex] Juan Lopez and Ted Barnes (Mean = [tex]\dfrac{4+10}{2} = 7[/tex] Juan Lopez and Peggy Chu (Mean = [tex]\dfrac{4+6}{2} = 5[/tex] Ted Barnes and Peggy Chu (Mean = [tex]\dfrac{10+6}{2} = 8[/tex]
(c) The mean of the population is:
[tex]Mean = \dfrac{8+6+4+10+6}{5}\\\\Mean = \dfrac{34}{5}\\\\Mean = 6.8[/tex]
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You have a right circular cone of height 1530 mm and volume 2.2 x 104 in3 . Calculate the base diameter of the cone.
Answer:
37.35 in
Step-by-step explanation:
The volume of a cone is given by the formula ...
V = (π/3)r²h
where r is the radius of the base and h is the height. We want to find the diameter of the base, so we can rewrite this in terms of diameter and solve for d. Please note that the height is given in millimeters, not inches, so a conversion is necessary.
V = (π/3)(d/2)²h
12V/(πh) = d²
d = 2√(3V/(πh)) = 2√(3(2.2×10^4 in^3)/(π·1530 mm/(25.4 mm/in))
= 2√(1.6764×10^6/(π·1.53×10^3) in^2)
d ≈ 37.35 in
The base diameter of the cone is about 37.35 inches.
The diameter of the base of the cone is found to be approximately 18.28802 inches after substituting the given values into the volume formula for a cone and solving for radius, and then doubling to get diameter.
Explanation:The volume V of a right circular cone is given by the formula V = 1/3πr²h, where r represents the radius of the cone's base and h is the cone's height. The diameter of the base of a cone is double the radius, so we will be solving for diameter instead of radius.
We were given V = 2.2 x 104 in³ and h = 1530 mm. Firstly it is important to note that 1 mm is equal to approximately 0.0393701 in, so h becomes approximately 60.2362205 in. Now we substitute these values into the equation and solve for r as follows: 2.2 x 104 = 1/3πr²*60.2362205. Solving for r we get r = approximately 9.14401 in.
We obtain the diameter by doubling the radius, so the diameter d = 2r = 18.28802 in.
Learn more about Cone diameter calculation here:https://brainly.com/question/34207679
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An 18-meter-tall cylindrical tank with a 4-meter radius holds water and is half full. Find the work (in mega-joules) needed to pump all of the water to the top of the tank. (The mass density of water is 1000 kg/m3. Let g = 9.8 m/s2.
Final answer:
To find the work needed to pump all of the water to the top of the tank, we need to calculate the potential energy at the half-full level and the top of the tank. By using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height, we can calculate the potential energy for each level and find the difference. The work needed is approximately 41 million mega-joules.
Explanation:
To calculate the work needed to pump all the water to the top of the tank, we need to find the potential energy of the water at the half-full level and the potential energy of the water at the top of the tank. The potential energy of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
Given that the tank is half full, the height of the water column is 18/2 = 9 meters. The radius of the tank is 4 meters, so the volume of the water is πr^2h = π(4^2)(9) = 144π cubic meters. The mass of the water is density × volume = 1000 × 144π = 144,000π kg.
The potential energy at the half-full level is PE = mgh = 144,000π × 9 × 9.8 = 12,931,200π joules. The potential energy at the top of the tank is PE = mgh = 144,000π × 18 × 9.8 = 25,862,400π joules. Therefore, the work needed to pump all the water to the top of the tank is the difference in potential energy = 25,862,400π - 12,931,200π = 12,931,200π joules. To convert to mega-joules, divide by 1,000,000, so the work needed is approximately 41 million mega-joules.
Scura makes sun block and their annual revenues depend on how much they sell. Let x be the quantity of 5 oz. bottles of sun block that they make and sell each year measured in 1000 's of bottles. Thus if x=10 then they make and sell 10000 bottles of sun block each year. If x=25 then they make and sell 25000 bottles of sun block each year.
a. If x=50 how many bottles of sun block does Scura make and sell?
b. What is x equal to if Scura produces and sells 45000 bottles of sunblock?
Answer:
a) 50,000 bottles
b) x = 45
Step-by-step explanation:
We are given the following in the question:
The annual revenue of sun block depends on how much they sell.
Let x be the quantity of 5 oz. bottles of sun block that they make and sell each year measured in 1000 's of bottles.
For x = 10,
10,000 bottles were made and sell each year.
For x = 25,
25,000 bottles of sun block were made and sell each year.
a) x = 50
[tex]\text{Number of bottles} = 50\times 1000 = 50,000[/tex]
Thus, 50,000 bottles of sun block does Scura make and sell.
b) Scura produces and sells 45000 bottles of sunblock
We have to find the value of x
[tex]x = \displaystyle\frac{\text{Number of bottles}}{1000} = \frac{45000}{1000} = 45[/tex]
Thus, x = 45 if Scura produces and sells 45000 bottles of sunblock.
For Scura's Sunblock Sales, if x=50, they sell 50,000 bottles, and when Scura sells 45,000 bottles of sunblock, x equals 45.
Explanation:Answer to Scura's Sunblock Salesa. If x=50, then according to the relationship given where x represents thousands of bottles, Scura makes and sells 50,000 bottles of sun block.
b. To determine what x is equal to when Scura produces and sells 45,000 bottles of sunblock, we take the total number of bottles and divide by 1000, since x is measured in 1000s. So, x=45 when Scura produces and sells 45,000 bottles of sunblock.
is 0.963 close to 3/4
Answer:
I would say No
Step-by-step explanation:
3/4 = .75
like 3 quarters
.75 about .963 no its not its about .2 away
There's no such concept as "close" in mathematics. Or at least, you have to specify when you consider two numbers to be "close".
All we can say is that, since 3/4=0.75, the two numbers are
[tex]0.963-0.75=0.213[/tex]
units apart. Is this small enough to consider them as "close"? Is this big enough to consider them not to be "close"?
You should clarify more what you mean so that a definitive answer can be given.
Write an equation in slope-intercept form of the line perpendicular to y = - 1 5 x + 1 4 that passes through the point (3, 4).
The equation of the line is [tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
Explanation:
The equation is [tex]y=-15x+14[/tex] and passes through the point (3,4)
To find the equation of the line in slope intercept form, first we shall find the slope.
This equation is of the slope-intercept form [tex]y=m x+b[/tex], we shall find the value of slope.
Thus, slope m = -15
Since, the line is perpendicular, the negative slope is given by [tex]\frac{-1}{m}[/tex]
Thus, the new slope is [tex]m=\frac{1}{15}[/tex]
Now, we shall find the equation of the line perpendicular to the slope [tex]\frac{1}{15}[/tex] is
[tex]y-y_{1}=\frac{1}{15} \left(x-x_{1}\right)[/tex]
Let us substitute the points (3,4), we have,
[tex]y-4=\frac{1}{15} \left(x-3\right)[/tex]
Muliplying the term within the bracket, we get,
[tex]y-4=\frac{1}{15}x-\frac{1}{5}[/tex]
Adding both sides of the equation by 4, we get,
[tex]y=\frac{1}{15}x-\frac{1}{5}+4[/tex]
Adding the like terms, we have,
[tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
Thus, the equation in slope intercept form of the line is [tex]y=\frac{1}{15} x+\frac{19}{5}[/tex]
A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 104.5 , 96.2 , 102.2 , 98.3 , and 101.8 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. A. 104.5 mg/dL absolute error = 2.5 mg / dL relative error = 0.025 B. 96.2 mg/dL absolute error = −5.8 mg / dL relative error = 0.057 C. 102.2 mg/dL absolute error = 0.2 mg / dL relative error = 0.020 D. 98.3 mg/dL absolute error = −3.7 mg / dL relative error = 0.036 E. 101.8 mg/dL absolute error = −0.2 mg / dL relative error =
Answer:
The Absolute Error is the difference between the actual and measured value.
[tex]Absolute \:error = |Actual \:value - Measured \:value|[/tex]
The Relative Error is the Absolute Error divided by the actual measurement.
[tex]Relative \:error = \frac{Absolute \:error}{Actual \:value}[/tex]
We know that the actual value is 102.0 mg/dL.
To find the absolute error and relative error for each measurement made by the glucose monitor you must use the above definitions.
a) For a concentration of 104.5 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102-104.5\right|\\Absolute \:error =\left|-2.5\right|\\Absolute \:error =2.5[/tex]
[tex]Relative \:error = \frac{2.5}{102.0}=0.0245[/tex]
b) For a concentration of 96.2 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-96.2\right|\\Absolute \:error =\left|5.8\right|\\Absolute \:error =5.8[/tex]
[tex]Relative \:error = \frac{5.8}{102.0}=0.0569[/tex]
c) For a concentration of 102.2 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-102.2\right|\\Absolute \:error =\left|-0.2\right|\\Absolute \:error =0.2[/tex]
[tex]Relative \:error = \frac{0.2}{102.0}=0.00196[/tex]
d) For a concentration of 98.3 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-98.3\right|\\Absolute \:error =\left|3.7\right|\\Absolute \:error =3.7[/tex]
[tex]Relative \:error = \frac{3.7}{102.0}=0.0363[/tex]
e) For a concentration of 101.8 mg/dL the absolute error and relative error are
[tex]Absolute \:error = \left|102.0-101.8\right|\\Absolute \:error =\left|0.2\right|\\Absolute \:error =0.2[/tex]
[tex]Relative \:error = \frac{0.2}{102.0}=0.00196[/tex]
Let R be the event that a randomly chosen person lives in the city of Raleigh. Let O be the event that a randomly chosen person is over 50 years old. Place the correct event in each response box below to show: Given that the person lives in Raleigh, the probability that a randomly chosen person is over 50 years old.
Answer:
P(O|R)
Step-by-step explanation:
The conditional probability notation of two events A and B can be written as either P(A|B) or P(B|A).
The '|' sign is read as 'given'. So, P(A|B) is read as the probability of event A given event B which implies that it is the probability that event A will occur given that event B has already occurred.
In the question,
Event R = Person lives in the city of Raleigh
Event O = Person is over 50 years old
The statement says, 'given that the person lives in Raleigh' which means that event R has already occurred and we need to find the probability of event O (the randomly chosen person is over 50 years old).
Hence, this statement can be given in conditional probability notation as
P(O|R)
The question pertains to conditional probability (P(O|R)) and requires specific data to calculate the probability that a randomly chosen person is over 50 years old given they live in Raleigh.
Explanation:The student is asking about the concept of conditional probability, which in this case refers to the probability of a randomly chosen person being over 50 years old, given that they live in Raleigh. To express this mathematically, we say P(O|R), which means the probability of event O occurring given that event R has occurred.
To find this probability, one would typically use information from a study or a survey giving population counts or percentages of those over 50 within the city of Raleigh population. Without specific data, one cannot provide a precise probability value.
However, the concept is important in probability theory and widely applied across different domains.
The following data represent the social ambivalence scores for 15 people as measured by a psychological test. (The higher the score, the stronger the ambivalence.) 8 12 11 15 14 10 8 3 8 7 21 12 9 19 11 (a) Guess the value of s using the range approximation. s ≈ (b) Calculate x for the 15 social ambivalence scores. Calculate s for the 15 social ambivalence scores. (c) What fraction of the scores actually lie in the interval x ± 2s? (Round your answer to two decimal places.).
Answer:
a) 4.5
b) x = 11.2, s = 4.65
c) 93.33%
Step-by-step explanation:
We are given he following data in the question:
8, 12, 11, 15, 14, 10, 8, 3, 8, 7, 21, 12, 9, 19, 11
a) Estimation of standard deviation using range
Sorted data: 3, 7, 8, 8, 8, 9, 10, 11, 11, 12, 12, 14, 15, 19, 21
Range = Maximum - Minimum = 21 - 3 = 18
Range rule thumb:
It states that the range is 4 times the standard deviation for a given data.[tex]s = \dfrac{\text{Range}}{4} = \dfrac{18}{4} = 4.5[/tex]
b) Mean and standard deviation
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{168}{15} = 11.2[/tex]
Sum of squares of differences = 302.4
[tex]S.D = \sqrt{\dfrac{302.4}{14}} = 4.65[/tex]
c) fraction of the scores actually lie in the interval x ± 2s
[tex]x \pm 2s = 11.2 \pm 2(4.65) = (1.9,20.5)[/tex]
Since 14 out of 15 entries lie in this range, we can calculate the percentage as,
[tex]\dfrac{14}{15}\times 100\% = 93.33\%[/tex]
Suppose that a recent poll of American households about car ownership found that for households with a car, 39% owned a sedan, 33% owned a van, and 7% owned a sports car. Suppose that three households are selected randomly and with replacement. What is the probability that at least one of the three randomly selected households own a sports car
Answer:
The probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.
Step-by-step explanation:
Let X = number of household owns a sports car.
The probability of X is, P (X) = p = 0.07.
Then the random variable X follows a Binomial distribution with n = 3 and p = 0.07.
The probability function of a binomial distribution is:
[tex]P(X=x) = {n\choose x}p^{x}[1-p]^{n-x}\\[/tex]
Compute the probability that of the 3 households randomly selected at least 1 owns a sports car:
[tex]P(X\geq 1)=1-P(X<1)\\=1-P(X=0)\\=1- {3\choose 0}(0.07)^{0}[1-0.07]^{3-0}\\=1-0.8044\\=0.1956[/tex]
Thus, the probability that of the 3 households randomly selected at least 1 owns a sports car is 0.1956.
A certain college graduate borrows 7277 dollars to buy a car. The lender charges interest at an annual rate of 11%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year.
1. Determine the payment rate that is required to pay off the loan in 5 years.
2. Also determine how much interest is paid during the 5-year period?
Answer:
a. $1773.82
b. $1592.1
Step-by-step explanation:
1. If he pays k dollar in the first year, then the amount that he owned without interest is
7277 - k
The amount that he owned including interest of 11% in the 2nd year is
(7277 - k)*1.11 or 7277*1.11 - 1.11k
After 2nd year and paying k then the amount he owned (without interest is)
(7277 - k)*1.11 - k
With interest
[(7277 - k)*1.11 - k]1.11 or [tex]7277*1.11^2 - 1.11^2k - 1.11k[/tex]
So after 5 years
[tex]7277*1.11^5 - (1.11^5 + 1.11^4 + 1.11^3 + 1.11^2 +1.11)k[/tex]
[tex]12262.17 - 6.91 k[/tex]
Since he's dept-free after 5 year then
[tex]12262.17 - 6.91 k = 0[/tex]
[tex]k = 12262.17 / 6.91 = 1773.82[/tex] dollar
2. The total amount he would have to pay over 5 years is 5k = 5*1773.82 = 8869.1
So the interest we has to pay over 5 years is the total subtracted by the principal, which is 8869.1 - 7277 = 1592.1 dollar
Driving under the influence of alcohol (DUI) is a serious offense. The following data give the ages of a random sample of 50 drivers arrested while driving under the influence of alcohol. This distribution is based on the age distribution of DUI arrests given in the Statistical Abstract of the United States46 16 41 26 22 33 30 22 36 3463 21 26 18 27 24 31 38 26 5531 47 27 43 35 22 64 40 58 2049 37 53 25 29 32 23 49 39 4024 56 30 51 21 45 27 34 47 35(b) Make a frequency table using seven classes.Class Limits... Class Boundaries...Midpoint...Frequency...RelativeFrequency...CumulativeFrequency
Answer:
Explanation below.
Step-by-step explanation:
For this case we have the following dataset:
46, 16, 41, 26, 22 ,33, 30, 22 ,36, 34,
63, 21, 26, 18, 27, 24, 31, 38, 26, 55,
31, 47, 27, 43 ,35, 22 ,64,40, 58, 20,
49, 37, 53, 25, 29, 32, 23, 49, 39, 40,
24, 56, 30, 51, 21, 45, 27, 34, 47, 35
So we have 50 values. The first step on this case would be order the dataset on increasing way and we got:
16, 18, 20, 21, 21, 22, 22, 22, 23, 24,
24, 25, 26, 26, 26, 27, 27, 27, 29, 30
30, 31, 31, 32, 33, 34, 34, 35, 35, 36,
37, 38, 39, 40, 40, 41, 43, 45, 46, 47,
47, 49, 49, 51, 53, 55, 56, 58, 63, 64
We can find the range for this dataset like this:
[tex] Range = Max-Min = 64-16 =48[/tex]
Then since we need 7 classes we can find the length for each class doing this:
[tex] W = \frac{48}{7}=6.86[/tex]
And now we can define the classes like this and counting how many observations lies on each interval we got the frequency:
Class Frequency Midpoint RF CF
________________________________________________
[16-22.86) 8 19.43 (8/50)=0.16 0.16
[22.86-29.71) 11 26.29 (11/50)=0.22 0.38
[29.71-36.57) 11 33.14 (11/50)=0.22 0.6
[36.57-43.43) 7 40.0 (7/50)=0.14 0.74
[43.43-50.29) 6 46.86 (6/50)=0.12 0.86
[50.29-57.14) 4 53.72 (4/50)=0.08 0.94
[57.14-64] 3 60.57 (3/50)=0.06 1.0
________________________________________________
Total 50 1.00
RF= Relative frequency. CF= Cumulative frequency
The relative frequency was calculated as the individual frequency for a class divided by the total of observations (50)
The mid point is the average between the limits of the class.
And the cumulative frequency is calculated adding the relative frequencies for each class.
Find g prime left parenthesis x right parenthesis for the given function. Then find g prime left parenthesis negative 3 right parenthesis, g prime left parenthesis 0 right parenthesis, and g prime left parenthesis 2 right parenthesis. g left parenthesis x right parenthesis equals StartRoot 4 x EndRoot
Answer: For x = 0, -3, our expression is undefined and for x = 2, we have 0.707.
Step-by-step explanation: From the question, we have
g(x) = \sqrt{4x}
Simplifying the right-hand side, we have:
g(x) = 2x^{1/2}
Differentiating with respect to $x$ using the second principle, we have,
g'(x) = 2 * \frac{1}{2} * x^{\frac{1}{2} - 1}
= x^{-1/2}
So from the indical laws, g'(x) =x^\frac{-1}{2} = 1/\sqrt{x}
For values of g'(x) when x = -3, we have
g(x) = 1/\sqrt{-3}
g(x) is undefined for values of x when x is -3 since the square root of a negative number is not defined. However, using complex solution we have
g(x) = 1/\sqrt{-3}
But \sqrt{-1} = i; then \sqrt{-3} = \sqrt(-1 * 3)
This is same as \sqrt(-1) * \sqrt(3)
And then we have 1.732i
For x = 2, we have
g’(2) = 1/\sqrt (x)
= 1/\sqrt(2) = 0.707
For x = 0, we have
g’(0) = 1/\sqrt (0)
= 1/0
Here again for x = 0, our expression is undefined.
The half-life of Sodium-24 is 15 hours. If you start with 600 grams of sodium-24 how much would be left after 4 days? Which of the following equations could you use to solve for the amount of grams left after 4 days?
Step-by-step explanation:
A = A₀ ½^(t / T)
where A is the amount left, A₀ is the original amount, t is time, and T is the half life.
4 days is 96 hours, so the amount left is:
A = 600 ½^(96 / 15)
To find how much Sodium-24 will be left after 4 days, we calculate the number of half-lives in 96 hours (which is 6.4) and use the formula remaining amount = initial amount × (1/2)n, resulting in approximately 8.79 grams remaining.
Explanation:The question asks how much Sodium-24 would be left after 4 days, given its half-life of 15 hours. Firstly, we convert 4 days into hours, which is 4 days × 24 hours/day = 96 hours. Next, we divide 96 hours by the half-life of Sodium-24, which is 15 hours, to find out how many half-lives have passed. The result is 96/15 = 6.4 half-lives.
Using the half-life decay formula, which is remaining amount = initial amount × (1/2)n, where 'n' is the number of half-lives, we can substitute the given values to find the number of grams left.
Remaining sodium-24 = 600 g × (1/2)6.4 ≈ 600 g × 0.01465 ≈ 8.79 g
Therefore, approximately 8.79 grams of Sodium-24 would remain after 4 days.