Mar 10, 2012 - Markups and Markdowns Word Problems - Independent Practice Worksheet. $6640. $3.201 ... 2) Fred buys a video game disk for $4. There was a discount of 20%.What is the sales price? 20% of 1 pay 8090 ... 5) Timmy wants to buy.a scooter and the price was $50. When ... at a simple interest rate of 54%.
Kelly plan to fence in her yard. The fabulous fence company charges $3.25 per foot of fencing and $15.57 an hour for labor. If Kelly needs 350 feet of fencing and the installers work a total of 6 hour installing the fence , how
much will she owe the fabulous fence company.
Answer:
Kelly will owe $1320.92 to the fabulous fence company.
Step-by-step explanation:
There is a cost related to the number of hours and a cost per feet. So the total cost is:
[tex]T = C_{h} + C_{f}[/tex]
In which [tex]C_{h}[/tex] is the cost related to the number of hours and [tex]C_{f}[/tex] is the cost related to the number of feet.
Cost per hour
Each hour costs $15.57.
They work for 6 hours total. So
[tex]C_{h} = 15.57*6 = 93.42[/tex]
Cost per feet
Each feet costs $3.25.
Kelly needs 350 feet. So
[tex]C_{f} = 350*3.25 = 1137.5[/tex]
The total cost is:
[tex]T = C_{h} + C_{f} = 93.42 + 1137.5 = 1230.92[/tex]
Kelly will owe $1320.92 to the fabulous fence company.
The age distribution of students at a community college is given below. Age (years) Number of students (f) Under 21 2890 21-24 2190 25-28 1276 29-32 651 33-36 274 37-40 117 Over 40 185 A student from the community college is selected at random. The events A and B are defined as follows. A = event the student is at most 32 B = event the student is at least 37 Are the events A and B disjoint? No Yes
Answer:
Are the events A and B disjoint? Yes
Step-by-step explanation:
Disjoint events are those events that cannot occur at the same time, i.e. for events X and Y to be disjoint, [tex]P(X\cap Y)=0[/tex].
The event A is defined as the number of students whose age is at most 32.
And event B is defined as the number of students whose age is at least 37.
The events A and B are disjoint events.
The sample space for event A consists of all the students of age group (under 21), (21 - 24), (25 - 28) and (29 - 32). Whereas the sample space for event B consists of all the students of age group (33 - 36), (37 - 40) and (Over 40).The sample space for the intersection of these two events is:
Sample space of (A ∩ B) = 0
As there are no common terms in both the sample.
Hence proved, events A and B are disjoint.
A 5-card hand is dealt from a well-shuffled deck of 52 playing cards. What is the probability that the hand contains at least one card from each of the four suits?
Answer:
0.2637
Step-by-step explanation:
We see from the question that the 5-card hand contains all 4 suits as shown below;
Number of cards = 52
Number of suits = 4
For the favorable cases therefore, we will choose two cards from the suit in which two cards are drawn. Then we will proceed to choose one card from each of the other suits.
4 suits will divide into 52 cards to give = (52 / 4) = 13 cards
Hence, the required probability;
[tex]= {\frac{4 *13c_2*13c_1*13c_1*13c_1}{52c_5}}\\= {\frac{2197}{8330}}\\= 0.2637[/tex]
The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the other 5 have selected desktops. Suppose that four computers are randomly selected.
(a) How many different ways are there to select four of the eight computers to be set up?
(b) What is the probability that exactly three of the selected computers are desktops?
(c) What is the probability that at least three desktops are selected?
Answer:
(a) There are 70 different ways set up 4 computers out of 8.
(b) The probability that exactly three of the selected computers are desktops is 0.305.
(c) The probability that at least three of the selected computers are desktops is 0.401.
Step-by-step explanation:
Of the 9 new computers 4 are laptops and 5 are desktop.
Let X = a laptop is selected and Y = a desktop is selected.
The probability of selecting a laptop is = [tex]P(Laptop) = p_{X} = \frac{4}{9}[/tex]
The probability of selecting a desktop is = [tex]P(Desktop) = p_{Y} = \frac{5}{9}[/tex]
Then both X and Y follows Binomial distribution.
[tex]X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})[/tex]
The probability function of a binomial distribution is:
[tex]P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}[/tex]
(a)
Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.
It is denotes as: [tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]
In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.
The number of ways to set up 4 computers of 8 is:
[tex]{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70[/tex]
Thus, there are 70 different ways set up 4 computers out of 8.
(b)
It is provided that 4 computers are randomly selected.
Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:
[tex]P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305[/tex]
Thus, the probability that exactly three of the selected computers are desktops is 0.305.
(c)
Compute the probability that of the 4 computers selected at least 3 are desktops as follows:
[tex]P(Y\geq 3)=1-P(Y<3)\\=1-[P(Y=0)+P(Y=1)+P(Y=2)]\\=1-[({4\choose 0}\times(\frac{5}{9} )^{0}\times (1-\frac{5}{9} )^{4-0}+({4\choose 1}\times(\frac{5}{9} )^{1}\times (1-\frac{5}{9} )^{4-1}+({4\choose 2}\times(\frac{5}{9} )^{2}\times (1-\frac{5}{9} )^{4-2}]\\=1-0.59918\\=0.40082\\\approx0.401[/tex]
Thus, the probability that at least three of the selected computers are desktops is 0.401.
An article reported on a school district's magnet schools program. Ofthe 1967 qualified applicants, 985 were accepted, 327 were waitlisted, and 655 were turned away for lack of space.a. The relative frequency of accepted qualified students (to three places after the decimal) is:________ b. The proportion of waitlisted students (to three places after the decimal) is:________ c. The percentage of students turned away from lack of space (to one place after the decimal) is:______
Answer:
a) 0.501
b) 0.166
c) 0.3
Step-by-step explanation:
We have the following information:
1967 qualified applicants
985 accepted
327 waitlisted
655 turned away for lack of space
a. The relative frequency of accepted qualified students (to three places after the decimal) is:________
This is the number of accepted qualified students divided by the number of qualified students.
So
985/1967 = 0.501
b. The proportion of waitlisted students (to three places after the decimal) is:________
This is the number of waitlisted students divided by the number of qualified students.
So
327/1967 = 0.166
c. The percentage of students turned away from lack of space (to one place after the decimal) is:______
This is the number of students turned away by lack of space divided by the number of qualified students.
So
655/1967 = 0.3
The relative frequency of accepted students is approximately 0.501, the proportion of waitlisted students is 0.166, and the percentage of students turned away for lack of space is roughly 33.3%.
Explanation:To find the relative frequency, proportion, and percentage from the given data for the school district's magnet schools program, we will use simple mathematical computations. We have a total of 1967 qualified applicants, where 985 were accepted, 327 were waitlisted, and 655 were turned away due to lack of space.
The relative frequency of accepted qualified students is calculated as the number of accepted students divided by the total number of applicants. So it's 985 / 1967 = 0.501 (to three places after the decimal).The proportion of waitlisted students is calculated as the number of waitlisted students divided by the total number of applicants. So it's 327 / 1967 = 0.166 (to three places after the decimal).The percentage of students turned away for lack of space is calculated as the number of students turned away divided by the total number of applicants, and then multiplied by 100 to convert it to a percentage. So it's (655 / 1967) * 100 ≈ 33.3% (to one place after the decimal).Courtney is picking out material for her new quilt. At the fabric store, there are 9 solids, 7 striped prints, and 5 floral prints that she can choose from. If she needs 2 solids, 4 floral prints, and 4 striped fabrics for her quilt, how many different ways can she choose the materials?
Answer:
N = 6300 ways
She can choose the materials 6300 ways
Step-by-step explanation:
In this case order of selection is not important, so we use combination.
For solids,
She needs 2 out of 9 available solids = 9C2
For striped prints
She needs 4 out of 7 available = 7C4
For floral prints
She needs 4 out of 5 available = 5C4
The total number of ways she can choose the materials is;
N = 9C2 × 7C4 × 5C4
N = 9/(7!2!) × 7/(4!3!) × 5/(4!1!)
N = 6300 ways
To find the number of different ways Courtney can choose the materials for her quilt, we can use the concept of combinations. The total number of ways she can choose the materials is the product of the number of choices for each type of fabric.
Explanation:To find the number of different ways Courtney can choose the materials for her quilt, we can use the concept of combinations. The total number of ways she can choose the materials is given by the product of the number of choices for each type of fabric. So, the answer is:
Total number of ways = number of ways to choose solids * number of ways to choose floral prints * number of ways to choose striped fabrics
Given that she needs 2 solids, 4 floral prints, and 4 striped fabrics, we can calculate:
Number of ways to choose solids = combinations(9, 2) = 36Number of ways to choose floral prints = combinations(5, 4) = 5Number of ways to choose striped fabrics = combinations(7, 4) = 35Substituting these values into the formula:
Total number of ways = 36 * 5 * 35 = 6300
So, there are 6300 different ways Courtney can choose the materials for her quilt.
Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z2 = 49. Then find parametric equations for this curve.
To sketch the curve of intersection, we substitute the equation of the parabolic cylinder into the equation of the ellipsoid. We use the discriminant to determine the nature of the curve and find its parametric equations.
Explanation:To sketch the curve of intersection of the parabolic cylinder and the top half of the ellipsoid, we can substitute the equation of the parabolic cylinder into the equation of the ellipsoid and then solve for the remaining variable. By doing this, we obtain a quadratic equation.
We can then use the discriminant to determine the nature of the solutions, which will help us identify if the curve is a parabola or an ellipse. Based on the discriminant, we can find the parametric equations for the curve and determine its shape.
For example, if the quadratic equation has two distinct real solutions, then the curve is an ellipse, but if it has one repeated real solution, the curve is a parabola.
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You go to Applebee’s and spend $98.42 on your meal. How much was the bill before 6% sales tax
Answer: the bill before 6% sales tax is $92.85
Step-by-step explanation:
Let x represent the bill before the 6% sales tax.
It means that you paid 6% tax on x and the amount of tax paid would be
6/100 × x = 0.06 × x = 0.06x
Total amount that you paid for the meal including the 6% tax would be
x + 0.06x = 1.06x
If you spent $98.42 on the meal after the 6% tax, it means that
1.06x = 98.42
Dividing the left hand side and the right hand side of the equation by 1.06, it becomes
1.06x/1.06 = 98.42/1.06
x = $92.85
The mean waiting time at the drive-through of a fast-food restaurant from the time the food is ordered to when it is received is 85 seconds. A manager devises a new system that he believes will decrease the wait time. He implements the new system and measures the wait time for 10 randomly sampled orders. They are provided below:
109 67 58 76 65 80 96 86 71 72
Assume the population is normally distributed.
(a) Calculate the mean and standard deviation of the wait times for the 10 orders.
(b) Construct a 99% confidence interval for the mean waiting time of the new system.
Answer:
a) And if we replace we got: [tex]\bar X= 78[/tex]
[tex] s = 15.391[/tex]
b) [tex]78-3.25\frac{15.391}{\sqrt{10}}=62.182[/tex]
[tex]78-3.25\frac{15.391}{\sqrt{10}}=93.818[/tex]
So on this case the 99% confidence interval would be given by (62.182;93.818)
Step-by-step explanation:
Dataset given: 109 67 58 76 65 80 96 86 71 72
Part a
For this case we can calculate the sample mean with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And if we replace we got: [tex]\bar X= 78[/tex]
And the deviation is given by:
[tex] s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And if we replace we got [tex] s = 15.391[/tex]
Part b
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 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 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that [tex]t_{\alpha/2}=3.25[/tex]
Now we have everything in order to replace into formula (1):
[tex]78-3.25\frac{15.391}{\sqrt{10}}=62.182[/tex]
[tex]78-3.25\frac{15.391}{\sqrt{10}}=93.818[/tex]
So on this case the 99% confidence interval would be given by (62.182;93.818)
The circumference of a circle is 5picm.
What is the area of the circle?
A.) 6.25 pi cm2
B.) 2.5 pi cm2
C.) 25 pi cm2
D.) 10 pi cm2
During the registration at the State University every semester, students in the college of business must have their courses approved by the college adviser. It takes the adviser an average of 2 minutes to approve each schedule, and students arrive at the adviser’s office at the rate of 28 per hour.17.How long does a student spend waiting on average for the adviser?A) 13 minutesB) 14 minutesC) 28 minutesD) 30 minutesE) none of the above
Answer:
Correct answer is option C i.e 28 minutes
Step-by-step explanation:
Number of students arriving at adviser's office per hour = x = 28
Number of students get approved = [tex]\frac{1}{2min}[/tex] = 30/hour
∴ y = 30
Number of students on average on waiting =Lq
Lq = [tex]\frac{x^{2} }{y(y-x)}[/tex]
= [tex]\frac{28^{2} }{30(30-28)}[/tex]
= 13.07
Average time student has to spend in
Waiting = Wq = [tex]\frac{x}{y(y-x)}[/tex]
= [tex]\frac{28}{30(30-28)}[/tex]
= 0.466 hours
= 28 minutes
A new streaming service charges 5 dollars per month for students, and 10 dollars per month for everyone else. This month, the service had 55 users, and collected 425 dollars. Set up a system of linear equations, and find the number of students using the service this month.
Answer:
Number of student = 25
Step-by-step explanation:
Let x be the number of student and y be the others
A new streaming service charges 5 dollars per month for students, and 10 dollars per month for everyone else
[tex]x+y=55\\y=55-x[/tex]
[tex]5x+10y=425[/tex]
replace y with 55-x
[tex]5x+10y=425\\5x+10(55-x)=425\\5x+550-10x=425\\-5x+550= 425[/tex]
Subtract 550 from both sides
[tex]-5x+550= 425\\-5x= -125\\x=25[/tex]
[tex]y=55-x\\y=55-25\\y=30\\[/tex]
Number of student = 25
Answer: 25 students used the service this month.
Step-by-step explanation:
Let x represent the number of students that used the streaming service this month.
Let y represent the number of people apart from students that used the streaming service this month.
This month, the service had 55 users. It means that
x + y = 55
The new streaming service charges 5 dollars per month for students, and 10 dollars per month for everyone else. They collected a total of 425 dollars. It means that
5x + 10y = 425 - - - - - - -1
Substituting x = 55 - y into equation 1, it becomes
5(55 - y) + 10y = 425
275 - 5y + 10y = 425
- 5y + 10y = 425 - 275
5y = 150
y = 150/5 = 30
x = 55 - y = 55 - 30
x = 25
The sum of 5 times a number and
minus −2, plus 7 times a number
Answer:
12x + 2
Step-by-step explanation:
Let the number be represented by x.
Then five times the number = 5*x
Seven times the number = 7*x
Sum of 5 times the number minus -2 = [tex]\[5*x - (-2)\][/tex] = [tex]\[5x +2\][/tex]
Adding seven times the number to this expression yields, [tex]\[5x+2+7x\][/tex]
[tex]\[= (5+7)x+2\][/tex]
[tex]\[= 12x+2\][/tex]
So the simplified expression corresponds to 12x + 2.
The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 2014. Being from Pennsylvania at that time, you believed automobile insurance was cheaper there and decided to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,425 with a standard deviation ofs = $160.(a) Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium.H0: μ ≥ 1,503Ha: μ < 1,503H0: μ ≤ 1,503Ha: μ > 1,503 H0: μ > 1,503Ha: μ ≤ 1,503H0: μ < 1,503Ha: μ ≥ 1,503H0: μ = 1,503Ha: μ ≠ 1,503(b) What is a point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean? (Use the mean annual premium in Pennsylvania minus the national mean.)
Final answer:
The hypothesis test to determine the mean annual premium in Pennsylvania compared to the national mean annual premium is H0: μ ≥ 1,503 and Ha: μ < 1,503. The point estimate of the difference between the mean annual premiums is -$78.
Explanation:
(a) To determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium, we need to develop a hypothesis test. The null hypothesis (H0) states that the mean annual premium in Pennsylvania is greater than or equal to the national mean annual premium. The alternative hypothesis (Ha) states that the mean annual premium in Pennsylvania is less than the national mean annual premium. Therefore, the correct answer is:
H0: μ ≥ 1,503
Ha: μ < 1,503
(b) The point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean is calculated by subtracting the national mean annual premium ($1,503) from the mean annual premium in Pennsylvania ($1,425). Therefore, the point estimate is $1,425 - $1,503 = -$78.
There are two traffic lights on Darlene's route from home to work. Let E denote the event that Darlene must stop at the first light, and define the event F in a similar manner for the second light. Suppose that P(E) = 0.2, P(F) = 0.3, and P(E ∩ F) = 0.13.
Answer:
The question is incomplete, below is the complete question,"There are two traffic lights on Darlene's route from home to work. Let E denote the event that Darlene must stop at the first light, and define the event F in a similar manner for the second light. Suppose that P(E) = 0.2, P(F) = 0.3, and P(E ∩ F) = 0.13.
a) What is the probability that the individual needn't stop at either light?
b) What is the probability that the individual must stop at exactly one of the two lights? c) What is the probability that the individual must stop just at the first light?"
Answer:
A. 0.63
B. 0.24
C. 0.07
Step-by-step explanation:
Data given,
P(E) = 0.2, P(F) = 0.3, and P(E ∩ F) = 0.13.
From the question, we can conclude that the event are dependent, hence
a. P(needn't stop at either light) = 1 - P(Need to stop at either light)
P(EUF)' =1-P(EUF)
P(EUF)' =1- (P(E)+P(F) -P(E ∩ F))
P(EUF)' =1-(0.2+0.3-0.13)
P(EUF)' =1-0.37
P(EUF)' =0.63
b. P(must stop at exactly one of the two lights) = P(must stop at either light) - P(must stop at both lights)
P(must stop at exactly one of the two lights) = P(E u F) - P(En F)
but P(E u F)=0.37,
P(En F)=0.13,
P(must stop at exactly one of the two lights) = 0.37 - 0.13 = 0.24
c. P(must stop at just the first light) = P(must stop at either light) - P(must stop at the second light)
P(must stop at just the first light) = P(E u F)-P(F)
P(must stop at just the first light) = 0.37 - 0.3 = 0.07
The question deals with the topic of Probability in Mathematics. It presents the probabilities of two events, denoted as E and F, which are stopping at the first and second traffic lights, respectively. The question also provides the concurrent occurrence of both events.
Explanation:The mathematics topic this question deals with is Probability. In the scenario given, E represents the event that Darlene must stop at the first traffic light and F represents the event that she needs to stop at the second traffic light. The probabilities of these events are given as P(E)=0.2 and P(F)=0.3, respectively. Additionally, we're given that the probability of both events happening (denoted P(E ∩ F)) is 0.13.
In order to analyze the situation, we can leverage the rule of joint probability, which states that the probability of two independent events both happening is the product of their individual probabilities. However, in this case the events E and F are not independent (since the probability of the intersection P(E ∩ F) is not equal to the product of probabilities P(E)*P(F)) so we know that the occurrence of E does influence the occurrence of F, and vice versa.
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Solve 4x2 - x + 5 = 0.
Answer:
x
=
1
+
i
√
79
8
,
1
−
i
√
79
8
Step-by-step explanation:
Use the square roots property to solve the quadratic equation (y+150)2=50.
We can take the square root of both sides, adding a plus/minus sign of the right hand side:
[tex]\sqrt{(y+150)^2}=\pm\sqrt{50}\iff y+150 = \pm\sqrt{50}[/tex]
Then, we subtract 150 from both sides:
[tex]y=\pm\sqrt{50}-150[/tex]
So, the two solutions are
[tex]y_1 = \sqrt{50}-150,\quad y_2 = -\sqrt{50}-150[/tex]
Joe, Megan, and Santana are salespeople. Their sales manager has 21 accounts and must assign seven accounts to each of them. In how many ways can this be done?
Answer:
116,280 ways
Step-by-step explanation:
The number of ways of assigning the accounts to each of the salesperson is computed by combination
Number of ways = n combination r = n!/(n-r)!r!
n = 21, r = 7
Number of ways = 21 combination 7 = 21!/(21-7)!7! = 21!/14!7! = 116,280 ways
A copyeditor thinks the standard deviation for the number of pages in a romance novel is six. A sample of 25 novels has a standard deviation of nine pages. At , is this higher than the editor hypothesized?
Answer:
No, the standard deviation for number of pages in a romance novel is six only.
Step-by-step explanation:
First we state our Null Hypothesis, [tex]H_o[/tex] : [tex]\sigma[/tex] = 6
and Alternate Hypothesis, [tex]H_1[/tex] : [tex]\sigma[/tex] > 6
We have taken these hypothesis because we have to check whether our population standard deviation is higher than what editor hypothesized of 6 pages in a romance novel.
Now given sample standard deviation, s = 9 and sample size, n = 25
To test this we use Test Statistics = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] follows chi-square with (n-1) degree of freedom [[tex]\chi ^{2}_n__-1[/tex]]
Test Statistics = [tex]\frac{(25-1)9^{2} }{6^{2} }[/tex] follows [tex]\chi ^{2}_2_4[/tex] = 54
and since the level of significance is not stated in question so we assume it to be 5%.
Now Using chi-square table we observe at 5% level of significance the [tex]\chi ^{2}_2_4[/tex] will give value of 36.42 which means if our test statistics will fall below 36.42 we will reject null hypothesis.
Since our Test statistics is more than the critical value i.e.(54>36.42) so we have sufficient evidence to accept null hypothesis and conclude that our population standard deviation is not more than 6 pages which the editor hypothesized.
Lillian earns $44 in 4 hours. At this rate, how many dollars will she earn
in 30 hours?
1 of 38 QUESTIONS
$440
$300
O $330
$110
SUBMIT
Answer:
(44/4)*30 = $330
Step-by-step explanation:
divide by four and multiply by 30
Find all the second order partial derivatives of g (x comma y )equalsx Superscript 4 Baseline y plus 5 sine (y )plus 4 y cosine (x ).
Answer:
Step-by-step explanation:
Check attachment for solution
Five players agree to divide a cake fairly using the last diminisher method. The players play in the following order: Anne first, Betty second, Cindy third, Doris fourth, and Ellen last. In round 1, there are no diminishers In round 2, Doris is the only diminisher In round 3, Cindy and Ellen are the only diminishers Which player gets her fair share at the end of:
Using the Last Diminisher method, in the first round, Anne gets her fair share because no one diminishes. In the second round, Doris is the only one who diminishes, thus gets her fair share. In the third round, despite Cindy and Ellen both diminishing, Ellen gets her fair share because she is later in turn order.
Explanation:The Last Diminisher method is a fair division protocol used when a divisible good, like a cake in this example, needs to be divided amongst several players. This method removes discrepancies by having each player in turn reduce the piece until they don't want to diminish it further, and then giving that piece to the last to diminish.
In this case, Anne, Betty, Cindy, Doris, and Ellen are dividing the cake and playing in that order. In the first round, no one diminishes, so Anne gets her fair share of the cake. In the second round, Doris is the only one who diminishes, so she gets her fair share at the end of this round. In the third round, the last to diminish are Cindy and Ellen, but since Ellen is later in order, Ellen is the one who gets her fair share at the end of the round.
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In response to a survey question about the number of hours daily spent watching TV, the responses by the eight subjects who identified themselves as Hindu were 2 , 2 , 1 , 3 , 1 , 0 , 4 , 1
a. Find a point estimate of the population mean for Hindus.
--------------(Round to two decimal places as needed)
b. The margin of error at the 95% confidence level for this point estimate is 0.89. Explain what this represents.
The margin of error indicates we can be__%confident that the sample mean falls within __ of the _____(population mean/ standard error/ sample mean)
Answer:
a) [tex] \bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75[/tex]
b) The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean
Step-by-step explanation:
Part a
The best point of estimate for the population mean is the sample mean given by:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
Since is an unbiased estimator [tex] E(\bar X) = \mu[/tex]
Data given: 2 , 2 , 1 , 3 , 1 , 0 , 4 , 1
So for this case the sample mean would be:
[tex] \bar X = \frac{2+2+1+3+1+0+4+1}{8}= 1.75[/tex]
Part b
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
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)
The margin of error is given by this formula:
[tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)
And for this case we know that ME =0.89 with a confidence of 95%
So then the limits for our confidence level are:
[tex] Lower= \bar X -ME= 1.75- 0.89=0.86[/tex]
[tex] Upperr= \bar X +ME= 1.75+0.89=2.64[/tex]
So then the best answer for this case would be:
The margin of error indicates we can be 95%confident that the sample mean falls within 0.89 of the population mean
A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting: (a) at least one nun? (b) exactly 2 nuns? (c) exactly 1 hei? (d) at most 2 gimels?
So, the probabilities are: (a) 37/64, (b) 3/64, (c) 27/64, (d) 57/64
In each spin, there are four possible outcomes (nun, gimel, hei, shin), and each outcome is equally likely.
(a) Probability of getting at least one nun:
The probability of getting no nuns in a single spin is 3/4. So, the probability of getting no nuns in three spins is [tex](3/4)^3[/tex]. Therefore, the probability of getting at least one nun is 1 - [tex](3/4)^3[/tex].
Probability of getting at least one nun:
1 - [tex](3/4)^3[/tex] = [tex]1-\frac{27}{64}=\frac{37}{64}[/tex] = 0.58
(b) Probability of getting exactly 2 nuns:
The probability of getting a nun in a single spin is 1/4. So, the probability of getting exactly 1 hei in three spins is [tex]\left(\begin{array}{l}\frac{3}{1} \\\end{array}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2[/tex].
= [tex]3 \times \frac{1}{16} \times \frac{3}{4}=\frac{3}{64}[/tex] = 0.05
(c) Probability of getting exactly 1 hei:
The probability of getting a hei in a single spin is 1/4. So, the probability of getting exactly 1 hei in three spins is [tex]\left(\begin{array}{l}\frac{3}{1} \\\end{array}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2[/tex]
= [tex]3 \times \frac{1}{4} \times \frac{9}{16}=\frac{27}{64}[/tex] = 0.42
(d) Probability of getting at most 2 gimels:
The probability of getting 0 gimels is [tex](3/4)^3[/tex]. The probability of getting 1 gimel is [tex]\left(\begin{array}{l}\frac{3}{1} \\\end{array}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2[/tex]. The probability of getting 2 gimels is [tex]\left(\begin{array}{l}\frac{3}{2} \\\end{array}\right)\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)[/tex].
Add these probabilities to get the total probability.
[tex]\left(\frac{3}{4}\right)^3+\left(\begin{array}{l}\frac{3}{1} \\\end{array}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2+\left(\begin{array}{l}\frac{3}{2} \\\end{array}\right)\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)[/tex]
[tex]=\frac{27}{64}+\frac{27}{64}+\frac{3}{64}=\frac{57}{64}[/tex] = 0.9
Calculating probabilities of specific outcomes when spinning a dreidel multiple times.
Dreidel Probability Calculations:
(a) Probability of getting at least one nun: 1 - Probability of getting no nuns = 1 - [tex](3/4)^3[/tex].(b) Probability of getting exactly 2 nuns: Combination of outcomes with exactly 2 nuns / Total possible outcomes = (3 choose 2) x [tex](1/4)^2[/tex] x (3/4).(c) Probability of getting exactly 1 hei: Combination of outcomes with exactly 1 hei / Total possible outcomes = 3 x (1/4) x [tex](3/4)^2[/tex].(d) Probability of getting at most 2 gimels: Sum of probabilities of getting 0, 1, or 2 gimels.The top three corn producers in the world-country A, country B, and country C-grew a total of about 676 million metric tons (MT) of con in 2014 The country A produced 60 million MT more than the combined production of country B and country C Country B produced 122 million MT more than country C. Find the number of metric tons of com produced by each country The country Aproduced□minon MT ofcorn, the country B produced The country A producedmilsion MT of corn, the country B produced mition MT of corn, and the country C produced million MT of corn mati on MT of corn, and the country Cproduced□ma on MT of corn
Answer:
x = 368 production of country A (millions of (MT)
y = 215 production of country B (millions of (MT)
z = 93 production of country C (millions of (MT)
Step-by-step explanation:
Let call production as follows
Country A production x millions on MT
Country B production y millions on MT
Country C production z millions on MT
Then according to problem statement
x + y + z = 676 (1)
x = 60 + y + z
y = 122 + z
That system ( 3 equations and three unknown varables ) could be solved by any of the available procedures.
By subtitution we get
x = 60 + 122 + z + z ⇒ x = 182 + 2*z
And
182 + 2*z + 122 + z + z = 676
Solving for z
304 + 4*z = 676 ⇒ 4*z = 676 - 304 ⇒ 4*z = 372
z = 372/4 ⇒ z = 93 millions of (MT)
And
y = 122 + z ⇒ y = 122 + 93 ⇒ y = 215 millions of (MT)
x = 182 + 2*z ⇒ x = 182 + 2 ( 93) ⇒ x = 182 + 186
x = 368 millions of (MT)
We can cheked in equation 1
x = 368
y = 215
z = 93
Give a total of 676 millions of (MT)
A department store surveyed 428 shoppers, and the following information was obtained: 216 shoppers made a purchase, and 294 were satisfied with the service they received. If 47 of those who made a purchase were not satisfied with the service, how many shoppers did the following?a. made a purchase and were satisfied with the service
b. made a purchase or were satisfied with the serice
c. were satisfied with the service but did not mak a purchase
d. were not satisfied and did not make a purchase
The answer are (a) 169 (b) 341 (c) 125 (d) 87
What is a Venn diagram?A Venn diagram is an illustration that uses circles to show the commonalities and differences between things or groups of things.
Given that, A department store surveyed 428 shoppers, and the following information was obtained: 216 shoppers made a purchase, and 294 were satisfied with the service they received. If 47 of those who made a purchase were not satisfied with the service,
Refer to the Venn diagram attached.
The total number of shoppers surveyed is, N = 428.
Number of shoppers who made a purchase, n (P) = 216
Number of shoppers who were satisfied with the service they received,
n (S) = 294
Number of shoppers who made a purchase but were not satisfied with the service, n(S' ∩ P) = 47
(a) The number of shoppers who made a purchase and were satisfied with the service = n(S ∩ P)
n(S ∩ P) = n(P)-n(S'∩P)
= 216 - 47 = 169
(b) The numbers of shoppers who made a purchase or were satisfied with the service = n (P ∪ S)
n (P ∪ S) = n(P)+n(S)-n(S∩P)
= 216+294-169
= 341
(c) The numbers of shoppers who were satisfied with the service but did not make a purchase = n(S∩P')
= n(S)-n(S∩P)
= 241-169
= 125
(d) The number of shoppers who were not satisfied and did not make a purchase = n(S'∩P')
= N-n (S ∪ P)
= 428-341
= 87
Hence, the answer are (a) 169 (b) 341 (c) 125 (d) 87
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a. 169 shoppers made a purchase and were satisfied with the service.
b. 341 shoppers made a purchase or were satisfied with the service.
c. 125 shoppers were satisfied with the service but did not make a purchase.
d. 381 shoppers were not satisfied and did not make a purchase.
Let's break down the information given:
Total shoppers surveyed = 428
Shoppers who made a purchase = 216
Shoppers satisfied with the service = 294
Shoppers who made a purchase and were not satisfied = 47
We are asked to find:
a. Shoppers who made a purchase and were satisfied with the service.
To find this, we subtract the shoppers who made a purchase and were not satisfied from the total shoppers who made a purchase:
216 − 47 = 169
b. Shoppers who made a purchase or were satisfied with the service.
To find this, we add the shoppers who made a purchase and the shoppers who were satisfied, but we need to be careful not to count the overlap twice (those who made a purchase and were satisfied):
216+294−169=341
c. Shoppers who were satisfied with the service but did not make a purchase.
To find this, we subtract the shoppers who made a purchase and were satisfied from the total shoppers who were satisfied:
294−169=125
d. Shoppers who were not satisfied and did not make a purchase.
To find this, we subtract the shoppers who made a purchase and were not satisfied from the total shoppers surveyed:
428−47=381
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A distribution of measurements is relatively mound-shaped with a mean of 60 and a standard deviation of 14. Use this information to find the proportion of measurements in the given interval. between 46 and 74
Approximately 68.26% of the measurements fall between 46 and 74 in this distribution.
To find the proportion of measurements between 46 and 74 in a normal distribution with a mean (μ) of 60 and a standard deviation (σ) of 14, we can use the standard normal distribution (z-score) and the cumulative distribution function (CDF).
First, we need to convert the interval endpoints to z-scores using the formula:
z = (x - μ) / σ
Where x is the value in the interval, μ is the mean, and σ is the standard deviation.
For x = 46:
z₁ = (46 - 60) / 14
z₁ = -1
For x = 74:
z₂ = (74 - 60) / 14
z₂ = 1
Using the Excel functions:
=NORM.S.DIST(-1) and =NORM.S.DIST(1)
The probabilities are 0.1587 and 0.8413 respectively.
Now, we want the proportion of measurements between z₁ and z₂, which is:
Proportion = 0.8413 - 0.1587
≈ 0.6826
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Final answer:
Using the Empirical Rule for a normal distribution, approximately 68% of the measurements would fall between 46 and 74, as this range lies within one standard deviation above and below the mean of 60 in a distribution with a standard deviation of 14.
Explanation:
To find the proportion of measurements between 46 and 74 in a distribution with a mean of 60 and a standard deviation of 14, we can use the Empirical Rule, assuming the distribution is normal (bell-shaped). This rule states that approximately 68% of the data lies within one standard deviation of the mean, 95% within two, and more than 99% within three.
In this case, 46 is one standard deviation below the mean (60 - 14), and 74 is one standard deviation above the mean (60 + 14). So, we would expect approximately 68% of the measurements to lie between 46 and 74.
This is because the data is likely to be distributed symmetrically around the mean in a normal distribution, and the range given includes measurements falling within one standard deviation from the mean.
Suppose that out of 20% of all packages from Amazon are delivered by UPS, 12% of the packages that are delivered by UPS weighs 2 lbs or more. Also, 8% of the packages that are not delivered by UPS weighs less than 2 lbs.
a. What is the probability that a package is delivered by UPS if it weighs 2 lbs or more?
b. What is the probability that a package is not delivered by UPS if it weighs 2 lbs or more?
Answer:
(a) Probability that a package is delivered by UPS if it weighs 2 lbs or more = 0.0316.
(b) Probability that a package is not delivered by UPS if it weighs 2 lbs or more = 0.9684 .
Step-by-step explanation:
We are given that 20% of all packages from Amazon are delivered by UPS, from which 12% of the packages that are delivered by UPS weighs 2 lbs or more and 8% of the packages that are not delivered by UPS weighs less than 2 lbs.
Firstly Let A = Package from Amazon is delivered by UPS.
B = Packages that are delivered by UPS weighs 2 lbs or more.
So, P(A) = 0.2 and P(A') = {Probability that package is not delivered by UPS}
P(A') = 1 - 0.2 = 0.8
P(B/A) = 0.12 {means Probability that package weight 2 lbs or more given it
is delivered by UPS}
P(B'/A') = 0.08 [means Probability that package weight less than 2 lbs given
it is not delivered by UPS}
Since, P(B/A) = [tex]\frac{P(A\bigcap B)}{P(A)}[/tex] , [tex]P(A\bigcap B)[/tex] = P(B/A) * P(A) = 0.12 * 0.2 = 0.024 .
Also P(B) { Probability that package weight 2 lbs or more} is given by;
Probability that package weight 2 lbs or more and it delivered by UPS.Probability that package weight 2 lbs or more and is not delivered by UPS.So, P(B) = [tex]P(B\bigcap A) + P(B\bigcap A')[/tex] = P(B/A) * P(A) + P(B/A') * P(A')
= 0.12 * 0.2 + 0.92 * 0.8 { Here P(B/A') = 1 - P(B'/A') = 1 - 0.08 = 0.92}
= 0.76
(a) Probability that a package is delivered by UPS if it weighs 2 lbs or more is given by P(A/B);
P(A/B) = [tex]\frac{P(A\bigcap B)}{P(B)}[/tex] = [tex]\frac{0.024}{0.76}[/tex] = 0.0316
(b) Probability that a package is not delivered by UPS if it weighs 2 lbs or more = 1 - P(A/B) = 1 - 0.0316 = 0.9684 .
Find tea. Write your answer in simplest radical form
Answer:
2√6 ft
Step-by-step explanation:
Tan Ф = opposite/ adjacent
tan 60 = t / 2√2 ft
tan 60 = √3
t = (tan 60 )(2√2 ft)
t = (√3)(2√2 ft) = 2√6 ft
Given two dependent random samples with the following results: Population 1 58 76 77 70 62 76 67 76 Population 2 64 69 83 60 66 84 60 81 Can it be concluded, from this data, that there is a significant difference between the two population means? Let d=(Population 1 entry)−(Population 2 entry). Use a significance level of α=0.01 for the test. Assume that both populations are normally distributed.
Answer:
[tex]z=\frac{\bar d -0}{\frac{\sigma_d}{\sqrt{n}}}=\frac{-0.625 -0}{\frac{6.818}{\sqrt{8}}}=-0.259[/tex]
[tex]p_v =2*P(z<-0.259) =0.796[/tex]
So the p value is higher than the significance level given [tex]\alpha=0.01[/tex], then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is equal to 0. So we can conclude that we don't have significant differences between the two populations.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
Let's put some notation :
x=values popoulation 2 , y = values population 1
x: 64 69 83 60 66 84 60 81
y: 58 76 77 70 62 76 67 76
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -6,7,-6,10,-4,-8, 7, -5
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-5}{8}=-0.625[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]\sigma_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n} =6.818[/tex]
The 4 step is calculate the statistic given by :
[tex]z=\frac{\bar d -0}{\frac{\sigma_d}{\sqrt{n}}}=\frac{-0.625 -0}{\frac{6.818}{\sqrt{8}}}=-0.259[/tex]
Now we can calculate the p value, since we have a two tailed test the p value is given by:
[tex]p_v =2*P(z<-0.259) =0.796[/tex]
So the p value is higher than the significance level given [tex]\alpha=0.01[/tex], then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is equal to 0. So we can conclude that we don't have significant differences between the two populations.