A jar contains 60 nickels and dimes worth $4.30. How many of each ki.pdf
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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.
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A high school football team with 40 players includes 16 play-ers w.pdf
A high school football team with 40 players includes 16 play-
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Month
Investment
additonal investment
Balance at the start of the month
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200.00
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D5+E5+H4
(F5*0.5%)+((F5+(F5*0.5%))*0.5%)
F5+G5
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9.76
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992.89
D9+E9+H8
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F9+G9
Jan-17
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F9+G9
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F14+G14
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2,462
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0
0
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Jul-16
200
200.00
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202.01
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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
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0
973.28
9.76
983.04
D8+E8+H7
(F8*0.5%)+((F8+(F8*0.5%))*0.5%)
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Dec-16
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0
983.04
9.85
992.89
D9+E9+H8
(F9*0.5%)+((F9+(F9*0.5%))*0.5%)
F9+G9
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992.89
9.95
1,002.85
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A large group of teenagers is divided into two smaller groups of 125 each, and each group
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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
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A kindergarten class consists of 14 boys and 11 girls. If the teache.pdf
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Solution
A kindergarten class consists of 14 boys and 11 girls. If the teacher selects children from the
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We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
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A kindergarten class consists of 14 boys and 11 girls. If the teache.pdf
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.pdf
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.pdf
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.pdf
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.pdf
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 is a 3x3 matrix with two eigenvalues. Each eigenspace is one-dimen.pdf
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.pdf
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.pdf
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 .pdf
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.pdf
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 .pdf
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 .pdf
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..
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