A jar contains m red balls and n black balls. The document defines probabilities for drawing balls from the jar:
a) The probability the first ball drawn is red is m/(m+n).
b) The probability the second ball is black given the first was red is [m/(m+n)]*[n/(m+n-1)].
c) The probability the first ball is red and the second is black is [n/(m+n)]*[m/(m+n-1)].
A high school football team with 40 players includes 16 play-ers w.pdfabbasbhagat4
A high school football team with 40 players includes 16 play-
ers who played offense last year,17 who played defense,and
12 who were not on last year
Solution
no of players either offence or defence= 40 - 12 = 28
no of players played both offence and defemce = 17+16 -28 = 5.
A group of ten children want to play cards. They split into three g.pdfabbasbhagat4
A group of ten children want to play cards. They split into three groups, one of these groups has
four children in it, the other two have three each. Then each group sits around a table. Two
seatings are considered the same if everyone\'s left neighbor is the same. In how many ways can
this be done if the three tables are identical?
Solution
(4-1)! * (3-1)! * (3-1)! = 24.
A group of teenagers were monitored closely over a period of several.pdfabbasbhagat4
A group of teenagers were monitored closely over a period of several weeks to measure how
often they watched violent television programs. The researcher believes that the adolescent male
watch more violent television than their female counterparts. a.What is the dependent variable?
b.What is the independent variable? c.Dependent variable is what type of data (nominal, ordinal,
interval, ratio)? d.Independent variable is what type of data
Solution
a)Independent variable=gender b)Dependent variable=how often they watched
violent tv programs c)DV=interval d)IV=nominal d).
A group of friends decides to go on a trip next summer in July 2017..pdfabbasbhagat4
A group of friends decides to go on a trip next summer in July 2017. To have the money required
for the trip, each of them will need to invest $200 every month starting next month and
additional $50 every month after the next month. However, they will not invest any money in the
months of November, December, and January. Will they each have the needed $5150 for the trip
by the end of June 2017? How much more or less will they have? Assume the following:
a) The interest rate is 12% compounded semi-monthly.
b) The current month is June and the year is 2016.
c) Each of them receives $10/month starting next month, which they do not invest and put
aside for the use in the trip.
Solution
Month
Investment
additonal investment
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Jun-16
0
0
-
Jul-16
200
200.00
2.01
202.01
D4+E4
(F4*0.5%)+((F4+(F4*0.5%))*0.5%)
F4+G4
Aug-16
200
50
452.01
4.53
456.54
D5+E5+H4
(F5*0.5%)+((F5+(F5*0.5%))*0.5%)
F5+G5
Sep-16
200
50
706.54
7.08
713.62
D6+E6+H5
(F6*0.5%)+((F6+(F6*0.5%))*0.5%)
F6+G6
Oct-16
200
50
963.62
9.66
973.28
D7+E7+H6
(F7*0.5%)+((F7+(F7*0.5%))*0.5%)
F7+G7
Nov-16
0
0
973.28
9.76
983.04
D8+E8+H7
(F8*0.5%)+((F8+(F8*0.5%))*0.5%)
F8+G8
Dec-16
0
0
983.04
9.85
992.89
D9+E9+H8
(F9*0.5%)+((F9+(F9*0.5%))*0.5%)
F9+G9
Jan-17
0
0
992.89
9.95
1,002.85
D10+E10+H9
(F10*0.5%)+((F10+(F10*0.5%))*0.5%)
F10+G10
Feb-17
200
50
1,252.85
12.56
1,265.41
D11+E11+H10
(F11*0.5%)+((F11+(F11*0.5%))*0.5%)
F11+G11
Mar-17
200
50
1,515.41
15.19
1,530.60
D12+E12+H11
(F12*0.5%)+((F12+(F12*0.5%))*0.5%)
F12+G12
Apr-17
200
50
1,780.60
17.85
1,798.45
D13+E13+H12
(F13*0.5%)+((F13+(F13*0.5%))*0.5%)
F13+G13
May-17
200
50
2,048.45
20.54
2,068.98
D14+E14+H13
(F14*0.5%)+((F14+(F14*0.5%))*0.5%)
F14+G14
Jun-17
200
50
2,318.98
23.25
2,342.23
D15+E15+H14
(F15*0.5%)+((F15+(F15*0.5%))*0.5%)
F15+G15
balance of investment (A)
2,342
Amount kept aside (B)
120
Amount available C= (A+B)
2,462
Amount required (D)
5,150
Amount short (D) - (C )
2,688
*it is assumed that amount is invested at the start of the month
*Interst rate is annual hence semi monthly will be (12/12)/2=0.50%
Month
Investment
additonal investment
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Jun-16
0
0
-
Jul-16
200
200.00
2.01
202.01
D4+E4
(F4*0.5%)+((F4+(F4*0.5%))*0.5%)
F4+G4
Aug-16
200
50
452.01
4.53
456.54
D5+E5+H4
(F5*0.5%)+((F5+(F5*0.5%))*0.5%)
F5+G5
Sep-16
200
50
706.54
7.08
713.62
D6+E6+H5
(F6*0.5%)+((F6+(F6*0.5%))*0.5%)
F6+G6
Oct-16
200
50
963.62
9.66
973.28
D7+E7+H6
(F7*0.5%)+((F7+(F7*0.5%))*0.5%)
F7+G7
Nov-16
0
0
973.28
9.76
983.04
D8+E8+H7
(F8*0.5%)+((F8+(F8*0.5%))*0.5%)
F8+G8
Dec-16
0
0
983.04
9.85
992.89
D9+E9+H8
(F9*0.5%)+((F9+(F9*0.5%))*0.5%)
F9+G9
Jan-17
0
0
992.89
9.95
1,002.85
D10+E10+H9
(F10*0.5%)+((F10+(F10*0.5%).
A group of HIV positive subjects, all of whom are being treated medi.pdfabbasbhagat4
A group of HIV positive subjects, all of whom are being treated medically in a similar fashion,
are randomly assigned to one of three groups. Two groups are given two different diet
supplements while the remaining group is given an inert similar appearing supplement. Interest
lies in determining whether the three treatments will produce different average CD4 counts.
Which of the following tests was most likely used to assess this continuous variable?
Solution.
A group of 60 players is to be assigned to four soccer teams. The re.pdfabbasbhagat4
A group of 60 players is to be assigned to four soccer teams. The red, blue, green, and yellow
teams will each get 15 players. How many way can teams be formed? If players are assigned
randomly, what is the probability Jane and Mary will end up playing together on the blue team?
What is the probability Jane and Mary end up on different teams?
Solution
4!* 60C15 ways team can be formed.
A kindergarten class consists of 14 boys and 11 girls. If the teach.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling, what is the probability that the first child selected will be a girl?
Explain in detail.
Solution
no of ways he can select a child = 11+14=25 no of ways he can select a girl =11 so
probability=no of ways he can select a girl/no of ways he can select a child = 11/25 so ans is
11/25 or 0.44.
A large group of teenagers is divided into two smaller groups of 125.pdfabbasbhagat4
A large group of teenagers is divided into two smaller groups of 125 each, and each group
watches a different movie. One group watches a love story while the other group watches a
movie containing several scenes with violence. The teenagers are then asked questions to
uncover unconscious associations and determine their level of aggression.
I. What is the population?
II. What is the sample?
III. Is the study observational or experimental? Justify your answer.
IV. What are the variables?
V. For each of those variables, what level of measurement (nominal, ordinal, interval, or ratio)
was used to obtain data from these variables?
Solution
population is all teenagers sample is the 250 teenagers in the study If the teens
were randomly assigned to the different movies, then it\'s an experiment; if they decided which
movie they watched, then it\'s observational. The variables are movie watched (independent)
and level of aggression (dependent). movie is nominal, level of aggression is ordinal.
A high school football team with 40 players includes 16 play-ers w.pdfabbasbhagat4
A high school football team with 40 players includes 16 play-
ers who played offense last year,17 who played defense,and
12 who were not on last year
Solution
no of players either offence or defence= 40 - 12 = 28
no of players played both offence and defemce = 17+16 -28 = 5.
A group of ten children want to play cards. They split into three g.pdfabbasbhagat4
A group of ten children want to play cards. They split into three groups, one of these groups has
four children in it, the other two have three each. Then each group sits around a table. Two
seatings are considered the same if everyone\'s left neighbor is the same. In how many ways can
this be done if the three tables are identical?
Solution
(4-1)! * (3-1)! * (3-1)! = 24.
A group of teenagers were monitored closely over a period of several.pdfabbasbhagat4
A group of teenagers were monitored closely over a period of several weeks to measure how
often they watched violent television programs. The researcher believes that the adolescent male
watch more violent television than their female counterparts. a.What is the dependent variable?
b.What is the independent variable? c.Dependent variable is what type of data (nominal, ordinal,
interval, ratio)? d.Independent variable is what type of data
Solution
a)Independent variable=gender b)Dependent variable=how often they watched
violent tv programs c)DV=interval d)IV=nominal d).
A group of friends decides to go on a trip next summer in July 2017..pdfabbasbhagat4
A group of friends decides to go on a trip next summer in July 2017. To have the money required
for the trip, each of them will need to invest $200 every month starting next month and
additional $50 every month after the next month. However, they will not invest any money in the
months of November, December, and January. Will they each have the needed $5150 for the trip
by the end of June 2017? How much more or less will they have? Assume the following:
a) The interest rate is 12% compounded semi-monthly.
b) The current month is June and the year is 2016.
c) Each of them receives $10/month starting next month, which they do not invest and put
aside for the use in the trip.
Solution
Month
Investment
additonal investment
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Jun-16
0
0
-
Jul-16
200
200.00
2.01
202.01
D4+E4
(F4*0.5%)+((F4+(F4*0.5%))*0.5%)
F4+G4
Aug-16
200
50
452.01
4.53
456.54
D5+E5+H4
(F5*0.5%)+((F5+(F5*0.5%))*0.5%)
F5+G5
Sep-16
200
50
706.54
7.08
713.62
D6+E6+H5
(F6*0.5%)+((F6+(F6*0.5%))*0.5%)
F6+G6
Oct-16
200
50
963.62
9.66
973.28
D7+E7+H6
(F7*0.5%)+((F7+(F7*0.5%))*0.5%)
F7+G7
Nov-16
0
0
973.28
9.76
983.04
D8+E8+H7
(F8*0.5%)+((F8+(F8*0.5%))*0.5%)
F8+G8
Dec-16
0
0
983.04
9.85
992.89
D9+E9+H8
(F9*0.5%)+((F9+(F9*0.5%))*0.5%)
F9+G9
Jan-17
0
0
992.89
9.95
1,002.85
D10+E10+H9
(F10*0.5%)+((F10+(F10*0.5%))*0.5%)
F10+G10
Feb-17
200
50
1,252.85
12.56
1,265.41
D11+E11+H10
(F11*0.5%)+((F11+(F11*0.5%))*0.5%)
F11+G11
Mar-17
200
50
1,515.41
15.19
1,530.60
D12+E12+H11
(F12*0.5%)+((F12+(F12*0.5%))*0.5%)
F12+G12
Apr-17
200
50
1,780.60
17.85
1,798.45
D13+E13+H12
(F13*0.5%)+((F13+(F13*0.5%))*0.5%)
F13+G13
May-17
200
50
2,048.45
20.54
2,068.98
D14+E14+H13
(F14*0.5%)+((F14+(F14*0.5%))*0.5%)
F14+G14
Jun-17
200
50
2,318.98
23.25
2,342.23
D15+E15+H14
(F15*0.5%)+((F15+(F15*0.5%))*0.5%)
F15+G15
balance of investment (A)
2,342
Amount kept aside (B)
120
Amount available C= (A+B)
2,462
Amount required (D)
5,150
Amount short (D) - (C )
2,688
*it is assumed that amount is invested at the start of the month
*Interst rate is annual hence semi monthly will be (12/12)/2=0.50%
Month
Investment
additonal investment
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Balance at the start of the month
Interest @ 12%
Investment balance at the end of the month
Jun-16
0
0
-
Jul-16
200
200.00
2.01
202.01
D4+E4
(F4*0.5%)+((F4+(F4*0.5%))*0.5%)
F4+G4
Aug-16
200
50
452.01
4.53
456.54
D5+E5+H4
(F5*0.5%)+((F5+(F5*0.5%))*0.5%)
F5+G5
Sep-16
200
50
706.54
7.08
713.62
D6+E6+H5
(F6*0.5%)+((F6+(F6*0.5%))*0.5%)
F6+G6
Oct-16
200
50
963.62
9.66
973.28
D7+E7+H6
(F7*0.5%)+((F7+(F7*0.5%))*0.5%)
F7+G7
Nov-16
0
0
973.28
9.76
983.04
D8+E8+H7
(F8*0.5%)+((F8+(F8*0.5%))*0.5%)
F8+G8
Dec-16
0
0
983.04
9.85
992.89
D9+E9+H8
(F9*0.5%)+((F9+(F9*0.5%))*0.5%)
F9+G9
Jan-17
0
0
992.89
9.95
1,002.85
D10+E10+H9
(F10*0.5%)+((F10+(F10*0.5%).
A group of HIV positive subjects, all of whom are being treated medi.pdfabbasbhagat4
A group of HIV positive subjects, all of whom are being treated medically in a similar fashion,
are randomly assigned to one of three groups. Two groups are given two different diet
supplements while the remaining group is given an inert similar appearing supplement. Interest
lies in determining whether the three treatments will produce different average CD4 counts.
Which of the following tests was most likely used to assess this continuous variable?
Solution.
A group of 60 players is to be assigned to four soccer teams. The re.pdfabbasbhagat4
A group of 60 players is to be assigned to four soccer teams. The red, blue, green, and yellow
teams will each get 15 players. How many way can teams be formed? If players are assigned
randomly, what is the probability Jane and Mary will end up playing together on the blue team?
What is the probability Jane and Mary end up on different teams?
Solution
4!* 60C15 ways team can be formed.
A kindergarten class consists of 14 boys and 11 girls. If the teach.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling, what is the probability that the first child selected will be a girl?
Explain in detail.
Solution
no of ways he can select a child = 11+14=25 no of ways he can select a girl =11 so
probability=no of ways he can select a girl/no of ways he can select a child = 11/25 so ans is
11/25 or 0.44.
A large group of teenagers is divided into two smaller groups of 125.pdfabbasbhagat4
A large group of teenagers is divided into two smaller groups of 125 each, and each group
watches a different movie. One group watches a love story while the other group watches a
movie containing several scenes with violence. The teenagers are then asked questions to
uncover unconscious associations and determine their level of aggression.
I. What is the population?
II. What is the sample?
III. Is the study observational or experimental? Justify your answer.
IV. What are the variables?
V. For each of those variables, what level of measurement (nominal, ordinal, interval, or ratio)
was used to obtain data from these variables?
Solution
population is all teenagers sample is the 250 teenagers in the study If the teens
were randomly assigned to the different movies, then it\'s an experiment; if they decided which
movie they watched, then it\'s observational. The variables are movie watched (independent)
and level of aggression (dependent). movie is nominal, level of aggression is ordinal.
A kindergarten class consists of 14 boys and 11 girls. If the teache.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling,
a. What is the probability that the first child selected will be a girl?
b. If the teacher selects a random sample of n _ 3 children and the first two children are both
boys, what is the probability that the third child selected will be a girl?
Solution
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling,
we have 14 boys and 11 girls.
a) probability of selected is girl = 11 / (14+11) = 11/ 25
b) the first two children are both boys, what is the probability that the third child selected will be
a girl?
first two already boys then
we have 12boys and 11 girls
probability = 11 / ( 11+12) = 11/23.
A landscaper wishes to use several different types of plants. The ca.pdfabbasbhagat4
A landscaper wishes to use several different types of plants. The categories include color
(red, yellow, blue and green), type (flowers and shrubs), and height (tall, medium and
short).How many different combinations can he/she use if she selects one color, one type, and
one height?
A) 28 B) 24 C)18 D) 16
Solution
Choices to select one color = 4
and
Choices to select type = 2
and
Choices to select height = 3
Thus overall possibilities = 4x2x3 = 24
-----------------
(Please rate, thanks).
A group of n families, each with m members, are to be lined up for a.pdfabbasbhagat4
A group of n families, each with m members, are to be lined up for a photograph. In how many
ways can the nm people be arranged if members of a family must stay together?
Solution
first we have treat each family as one chunk. so we have n chunks to arrange. = n!
ways next for each family we can shuffle within family = m! ways for each family. = m! ^ n
ways putting them together. n! (m!) ^n ways.
A kindergarten class consists of 14 boys and 11 girls If the teacher.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls If the teacher selects a random sample of n
= 3 children and the first two children are both boys, what is the probability that the third child
selected will be a girl? Explain your response.
Solution
probability= (14c2 *11c1)/(25c3) =1001/2300 so ans is 1001/2300.
A group of employees of Unique Services will be surveyed about a new.pdfabbasbhagat4
A group of employees of Unique Services will be surveyed about a new pension plan. In-depth
interviews with each employee selected in the sample will be conducted. The employees are
classified as follows. Classification Event Number of Employees Supervisors A 120
Maintenance B 50 Production C 1460 Management D 302 Secretarial E 68 What
is the probability that the first person selected is classified as a maintenance employee?
0.200 0.500 0.025 1.000
Solution
0.025.
A jar contains 60 nickels and dimes worth $4.30. How many of each ki.pdfabbasbhagat4
A jar contains 60 nickels and dimes worth $4.30. How many of each kind of coin are in the jar?
Solution
0.05N + 0.10D = 4.30 and
N + D = 60
So lets solve!
N = 60 -D
So...
0.05(60-D) + 0.10D = 4.30
3 - 0.05D + 0.10D = 4.30
0.05D = 1.30
D = 26
Since we have 26 Dimes, we have 60-26 = 34 nickels.
A is a 3x3 matrix with two eigenvalues. Each eigenspace is one-dimen.pdfabbasbhagat4
A is a 3x3 matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A
diagonalizable?
Solution
FALSE If A is 3x3, its characteristic polynomial is of degree 3 and has 3 roots
counting with multiplicities. If there are only 2 eigenvalues, one of them is of multiplicity 2, and
the associated eigenspace should be 2-dimensional for A to be diagonalizable..
A group of college students were surveyed to learn how many times th.pdfabbasbhagat4
A group of college students were surveyed to learn how many times they had visited a dentist in
the previous year 12. the probability distribution for Y, the number of the visits, is given by the
following table,
Y(No. Visits) Probability
0 0.15
1 0.50
2 0.35
Total 1.0
Calculate the mean,, of the number of the visit!
Solution
To find the mean we will calculate the weighted mean of Y. mean of Y = E(Y) =
.15(0) + .50(1) + .35(2) = 1.2.
A hypothesis test produces t = 2.00 for a sample of n = 16 scores. W.pdfabbasbhagat4
A hypothesis test produces t = 2.00 for a sample of n = 16 scores. What is the value of r2?
Explain your response.
2/18
2/17
4/20
4/19
Solution
The answer is 4/19. Formula for r2=t2/t2+df. df is 16-1=15. Plugging in numbers you will get
4/4+15=4/19.
A hypothesis tests that two population means are equal. A sample of .pdfabbasbhagat4
A hypothesis tests that two population means are equal. A sample of 10 with a standard deviation
of 5 is selected from the first population and a sample of 15 with a standard deviation of 5 from
the second population. Testing the claim at the 0.01 level, what is the critical value? Assume
unequal standard deviations.
Solution
±2.576.
A horse race has 14 entries. Assuming that there are no ties, what i.pdfabbasbhagat4
A horse race has 14 entries. Assuming that there are no ties, what is the probability that the five
horses owned by one person finish first, second, third, fourth, and fifth
Solution
(5 / 13) * (4 / 12) * (3 / 11) * (2 / 10) * (1 / 9) = .0008.
a horse race has 12 entries. assuming that there are no ties , what .pdfabbasbhagat4
a horse race has 12 entries. assuming that there are no ties , what is the probabalilty that the five
horses owned by one person finish first, second,third, fourth, and fith?
please round to four decimal places as needed
Solution
1/12+1/11+1/10+1/9+1/8=0.510.
A group of 100 students were randomly assigned to receive vitamin C .pdfabbasbhagat4
A group of 100 students were randomly assigned to receive vitamin C (50 students) or a placebo
(50 students). The groups were followed for 2 weeks and the proportions with colds were
compared.
A. Independent samples
B. Paired samples
Solution
A. Independent samples group to group was compared. not individual to individual..
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
A kindergarten class consists of 14 boys and 11 girls. If the teache.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling,
a. What is the probability that the first child selected will be a girl?
b. If the teacher selects a random sample of n _ 3 children and the first two children are both
boys, what is the probability that the third child selected will be a girl?
Solution
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
class using random sampling,
we have 14 boys and 11 girls.
a) probability of selected is girl = 11 / (14+11) = 11/ 25
b) the first two children are both boys, what is the probability that the third child selected will be
a girl?
first two already boys then
we have 12boys and 11 girls
probability = 11 / ( 11+12) = 11/23.
A landscaper wishes to use several different types of plants. The ca.pdfabbasbhagat4
A landscaper wishes to use several different types of plants. The categories include color
(red, yellow, blue and green), type (flowers and shrubs), and height (tall, medium and
short).How many different combinations can he/she use if she selects one color, one type, and
one height?
A) 28 B) 24 C)18 D) 16
Solution
Choices to select one color = 4
and
Choices to select type = 2
and
Choices to select height = 3
Thus overall possibilities = 4x2x3 = 24
-----------------
(Please rate, thanks).
A group of n families, each with m members, are to be lined up for a.pdfabbasbhagat4
A group of n families, each with m members, are to be lined up for a photograph. In how many
ways can the nm people be arranged if members of a family must stay together?
Solution
first we have treat each family as one chunk. so we have n chunks to arrange. = n!
ways next for each family we can shuffle within family = m! ways for each family. = m! ^ n
ways putting them together. n! (m!) ^n ways.
A kindergarten class consists of 14 boys and 11 girls If the teacher.pdfabbasbhagat4
A kindergarten class consists of 14 boys and 11 girls If the teacher selects a random sample of n
= 3 children and the first two children are both boys, what is the probability that the third child
selected will be a girl? Explain your response.
Solution
probability= (14c2 *11c1)/(25c3) =1001/2300 so ans is 1001/2300.
A group of employees of Unique Services will be surveyed about a new.pdfabbasbhagat4
A group of employees of Unique Services will be surveyed about a new pension plan. In-depth
interviews with each employee selected in the sample will be conducted. The employees are
classified as follows. Classification Event Number of Employees Supervisors A 120
Maintenance B 50 Production C 1460 Management D 302 Secretarial E 68 What
is the probability that the first person selected is classified as a maintenance employee?
0.200 0.500 0.025 1.000
Solution
0.025.
A jar contains 60 nickels and dimes worth $4.30. How many of each ki.pdfabbasbhagat4
A jar contains 60 nickels and dimes worth $4.30. How many of each kind of coin are in the jar?
Solution
0.05N + 0.10D = 4.30 and
N + D = 60
So lets solve!
N = 60 -D
So...
0.05(60-D) + 0.10D = 4.30
3 - 0.05D + 0.10D = 4.30
0.05D = 1.30
D = 26
Since we have 26 Dimes, we have 60-26 = 34 nickels.
A is a 3x3 matrix with two eigenvalues. Each eigenspace is one-dimen.pdfabbasbhagat4
A is a 3x3 matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A
diagonalizable?
Solution
FALSE If A is 3x3, its characteristic polynomial is of degree 3 and has 3 roots
counting with multiplicities. If there are only 2 eigenvalues, one of them is of multiplicity 2, and
the associated eigenspace should be 2-dimensional for A to be diagonalizable..
A group of college students were surveyed to learn how many times th.pdfabbasbhagat4
A group of college students were surveyed to learn how many times they had visited a dentist in
the previous year 12. the probability distribution for Y, the number of the visits, is given by the
following table,
Y(No. Visits) Probability
0 0.15
1 0.50
2 0.35
Total 1.0
Calculate the mean,, of the number of the visit!
Solution
To find the mean we will calculate the weighted mean of Y. mean of Y = E(Y) =
.15(0) + .50(1) + .35(2) = 1.2.
A hypothesis test produces t = 2.00 for a sample of n = 16 scores. W.pdfabbasbhagat4
A hypothesis test produces t = 2.00 for a sample of n = 16 scores. What is the value of r2?
Explain your response.
2/18
2/17
4/20
4/19
Solution
The answer is 4/19. Formula for r2=t2/t2+df. df is 16-1=15. Plugging in numbers you will get
4/4+15=4/19.
A hypothesis tests that two population means are equal. A sample of .pdfabbasbhagat4
A hypothesis tests that two population means are equal. A sample of 10 with a standard deviation
of 5 is selected from the first population and a sample of 15 with a standard deviation of 5 from
the second population. Testing the claim at the 0.01 level, what is the critical value? Assume
unequal standard deviations.
Solution
±2.576.
A horse race has 14 entries. Assuming that there are no ties, what i.pdfabbasbhagat4
A horse race has 14 entries. Assuming that there are no ties, what is the probability that the five
horses owned by one person finish first, second, third, fourth, and fifth
Solution
(5 / 13) * (4 / 12) * (3 / 11) * (2 / 10) * (1 / 9) = .0008.
a horse race has 12 entries. assuming that there are no ties , what .pdfabbasbhagat4
a horse race has 12 entries. assuming that there are no ties , what is the probabalilty that the five
horses owned by one person finish first, second,third, fourth, and fith?
please round to four decimal places as needed
Solution
1/12+1/11+1/10+1/9+1/8=0.510.
A group of 100 students were randomly assigned to receive vitamin C .pdfabbasbhagat4
A group of 100 students were randomly assigned to receive vitamin C (50 students) or a placebo
(50 students). The groups were followed for 2 weeks and the proportions with colds were
compared.
A. Independent samples
B. Paired samples
Solution
A. Independent samples group to group was compared. not individual to individual..
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Basic phrases for greeting and assisting costumers
A jar has m red balls and n black balls. we draw two balls from the .pdf
1. A jar has m red balls and n black balls. we draw two balls from the jar. let B1 denote that the
color of the first ball, and B2 denote the color of the second. let b denote black and r denote red.
evaluate the following probabilities:
a) P(B1=r)
b) P(b2=b|B1=r)
c) P(B1=r,B2=b)
d) P(B1=b,B2=b)
e) P(B2=b)
f) P(B1=r|B2=b)
Solution
a) P(B1=r) = m/(m+n) b) P(B2=b|B1=r) = [m/(m+n)]*[n/(m+n-1)] c) P(B1=r,B2=b)
= [n/(m+n)]*[m/(m+n-1)] d) P(B1=b,B2=b) = [n/(m+n)]*[(n-1)/(m+n-1)] e) P(B2=b) =
[n/(m+n)]*[(n-1)/(m+n-1)] + [m/(m+n)]*[n/(m+n-1)] f) P(B1=r|B2=b) = [n/(m+n)]*[m/(m+n-1)]
![A jar has m red balls and n black balls. we draw two balls from the jar. let B1 denote that the
color of the first ball, and B2 denote the color of the second. let b denote black and r denote red.
evaluate the following probabilities:
a) P(B1=r)
b) P(b2=b|B1=r)
c) P(B1=r,B2=b)
d) P(B1=b,B2=b)
e) P(B2=b)
f) P(B1=r|B2=b)
Solution
a) P(B1=r) = m/(m+n) b) P(B2=b|B1=r) = [m/(m+n)]*[n/(m+n-1)] c) P(B1=r,B2=b)
= [n/(m+n)]*[m/(m+n-1)] d) P(B1=b,B2=b) = [n/(m+n)]*[(n-1)/(m+n-1)] e) P(B2=b) =
[n/(m+n)]*[(n-1)/(m+n-1)] + [m/(m+n)]*[n/(m+n-1)] f) P(B1=r|B2=b) = [n/(m+n)]*[m/(m+n-1)]](https://image.slidesharecdn.com/ajarhasmredballsandnblackballs-230402102633-82db0cbe/85/A-jar-has-m-red-balls-and-n-black-balls-we-draw-two-balls-from-the-pdf-1-320.jpg)
