10.1: Appendix A - Solution to Select "Try it Now" Questions
1: Mathematics and Problem-Solving
- At least one Icelandic child did not learn English in school.
- C ⋁ ~S
-
A
B
~A
~A ⋀ B
T
T
F
F
T
F
F
F
F
T
T
T
F
F
T
F
-
A
B
~A
~A ⋀ B
~B
(~A ⋀ B) ⋁ ~B
T
T
F
F
F
F
T
F
F
F
T
T
F
T
T
T
F
T
F
F
T
F
T
T
- Choice b is correct because it is the contrapositive of the original statement.
- Choice b is equivalent to the negation; it keeps the first part the same and negates the second part.
- Choices a & b are false; c is true.
- Failing to meet just one of the three conditions is all it takes to be disqualified. A person is disqualified if they were not born in the US, or are not at least 35 years old, or have not lived in the US for at least 14 years. The key word here is “or” instead of “and”.
- Valid. Cats are a subset of creatures that are scared by vacuum cleaners. Max is in the set of cats, so he must also be in the set of creatures that are scared by vacuum cleaners.
- Invalid. The set of bicycles is a subset of the set of vehicles with two wheels; the Harley-Davidson is in the set of two-wheeled vehicles but not necessarily in the smaller circle.
- Invalid. Since no cows are purple, we know there is no overlap between the set of cows and the set of purple things. We know Fido is not in the cow set, but that is not enough to conclude that Fido is in the purple things set.
- Invalid. Lipstick on your collar is a subset of scenarios in which you are cheating, and cheating is a subset of the scenarios in which I will divorce you. Although it is wonderful that you don’t have lipstick on your collar, you could still be cheating on me, and I will divorce you. In fact, even if you aren’t cheating on me, I might divorce you for another reason. You’d better shape up.
-
Let
S
= have a shovel,
D
= dig a hole.
The first premise is equivalent to S → D . The second premise is D . The conclusion is S .
We are testing [( S → D ) ⋀ D ] → S
|
S |
D |
S→D |
(S→D) ⋀ D |
[(S→D) ⋀ D] → S |
|
T |
T |
T |
T |
T |
|
T |
F |
F |
F |
T |
|
F |
T |
T |
T |
F |
|
F |
F |
T |
F |
T |
This is not a tautology, so this is an invalid argument.
-
Let
b
= brushed teeth and
w
= toothbrush is wet.
Premise: b → w
Premise: ~ w
Conclusion: ~ b
This argument is valid by the Law of Contraposition. - This argument is valid by the Transitive Property, which can involve more than two premises, as long as they continue the chain reaction. The premises f → s , s → b , b → c , c → d , d → g , g → w , w → h , h → x can be reduced to f → x . (Because we had already used c and d , we decided to use w for cow and x for death.) If the old lady swallows the fly, she will eventually eat a horse and die.
-
Let
p
= wrote a paper and
s
= gave a speech.
Premise: p ⋁ s
Premise: ~ s
Conclusion: p
This argument is valid by Disjunctive Syllogism. Alison had to do one or the other; she didn’t choose the speech, so she must have chosen the paper. -
Let
f
= pulled fire alarm and
t
= got in big trouble.
Premise: f → t
Premise: t
Conclusion: f
This argument is invalid because it has the form of the Fallacy of the Converse. The young rascal may have gotten in trouble for any number of reasons besides pulling the fire alarm. -
Let
t
= tripped and
p
= got a penalty.
Premise: t → p
Premise: ~ t
Conclusion: ~ p
This argument is invalid because it has the form of the Fallacy of the Inverse. Alexei may have gotten a penalty for an infraction other than tripping. -
Let
p
= go to party,
t
= be tired, and
f
= see friends.
Premise: p → t
Premise: p → f
Conclusion: ~ f → ~ t
We could try to rewrite the second premise using the contrapositive to state ~ f → ~ p , but that does not allow us to form a syllogism. If I don’t see friends, then I didn’t go the party, but that is not sufficient to claim I won’t be tired tomorrow. Maybe I stayed up all night watching movies.
(Explanation continued on next page)A Venn diagram can help, if we set it up correctly. The “party” circle must be completely contained within the intersection of the other circles. We know that I am somewhere outside the “friends” circle, but we cannot determine whether I am in the “tired” circle. All we really know for sure is that I didn’t go to the party.
-
a. Circular
b. Correlation does not imply causation
c. Post hoc
d. Appeal to consequence
e. Straw man
2: Mathematics and Numbers
- The sale price is $799(0.70) = $559.30. After tax, the price is $559.30(1.092) = $610.76
-
2001-2002: Absolute change: $0.43 trillion. Relative change: 7.45%
2005-2006: Absolute change: $0.54 trillion. Relative change: 6.83%
2005-2006 saw a larger absolute increase, but a smaller relative increase.
- Without more information, it is hard to judge these arguments. This is compounded by the complexity of Medicare. As it turns out, the $716 billion is not a cut in current spending, but a cut in future increases in spending, largely reducing future growth in health care payments. In this case, at least the numerical claims in both statements could be considered at least partially true. Here is one source of more information if you’re interested: http://factcheck.org/2012/08/a-campaign-full-of-mediscare/
- \(18 \text {inches} \cdot \dfrac{\text { 1 foot }}{\text { 12 inches }} \cdot \dfrac{19.8 \text { pounds }}{1000 \text { feet }} \cdot \dfrac{16 \text { ounces }}{\text { lpound }} \approx$ 0.475 \text {ounces}\)
- The original sandbox has volume 64 \(ft^3\). The smaller sandbox has volume 24\(ft^3\). \(\dfrac{48 \text { bags }}{64 ft^{3}}=\dfrac{x \text { bags }}{24 ft^{3}} \) results in \(x = 18\) bags.
-
There is not enough information provided to answer the question, so we will have to make some assumptions, and look up some values.
Assumptions:
a) We own a car. Suppose it gets 24 miles to the gallon. We will only consider gas cost.
b) We will not need to rent a car in Spokane, but will need to get a taxi from the airport to the conference hotel downtown and back.
c) We can get someone to drop us off at the airport, so we don’t need to consider airport parking.
d) We will not consider whether we will lose money by having to take time off work to drive.
Values looked up (your values may be different)
a) Flight cost: $184
b) Taxi cost: $25 each way (estimate, according to hotel website)
c) Driving distance: 280 miles each way
d) Gas cost: $3.79 a gallon
Cost for flying: $184 flight cost + $50 in taxi fares = $234.
Cost for driving: 560 miles round trip will require 23.3 gallons of gas, costing $88.31.
Based on these assumptions, driving is cheaper. However, our assumption that we only include gas cost may not be a good one. Tax law allows you deduct $0.55 (in 2012) for each mile driven, a value that accounts for gas as well as a portion of the car cost, insurance, maintenance, etc. Based on this number, the cost of driving would be $319.
Taxes
1. While sales tax is a flat percentage rate, it is often considered a regressive tax for the same reasons as the gasoline tax.
3: Mathematics and Finance
-
\(I\)=$30 of interest
\(P_0\) = $500
\(r\)= unknown
\(t\)= 1 month
Using \(I=P_{0} r t\), we get \(30=500 \cdot r \cdot 1\). Solving, we get \(r\)=0.06, or 6%. Since the time was monthly, this is monthly interest. The annual rate would be 12 times this: 72% interest.
-
\(d\)=$5 The daily deposit
\(r\)=0.03 3% annual rate
\(k\)=365 since we’re doing daily deposits, we’ll compound daily
\(N\)=10 we want the amount after 10 years
\(P_{10}=\dfrac{5\left(\left(1+\frac{0.03}{365}\right)^{365 \times 10}-1\right)}{\frac{0.03}{365}}=\$ 21,282.07\)We would have deposited a total of \(\$5\cdot 365 \cdot 10=\$18,250\), so $3,032.07 is from interest
-
\(d\)= unknown
\(r\)=0.16 16% annual rate
\(k\)=12 since we’re making monthly payments
\(N\)=2 2 years to repay
\(P_0\)=3,000 we’re starting with a $3000 loan
\(3,000=\dfrac{d\left(1-\left(1+\frac{0.16}{12}\right)^{-2 \times 12}\right)}{\frac{0.16}{12}}\)Solving for \(d\) gives $146.89 as monthly payments.
In total, she will pay $3,525.36 to the store, meaning she will pay $525.36 in interest over the two years.
-
a. This is a savings annuity problem. He will have saved up $7,524.11.
b. This is a compound interest problem. She would need to deposit $22,386.46.
c. This is a loans problem. She can buy $4,609.33 of new equipment.
d. This is a savings annuity problem. You would need to save $200.46 each month.
4: Probability and Odds
- There are 60 possible readings, from 00 to 59. a.\(\dfrac{1}{60}\) b. \(\dfrac{16}{60}\) (counting 00 through 15)
- Since the second draw is made after replacing the first card, these events are independent. The probability of an ace on each draw is \(\dfrac{4}{52}=\dfrac{1}{13}\), so the probability of an Ace on both draws is \(\dfrac{1}{13} \cdot \dfrac{1}{13}=\dfrac{1}{169}\)
-
P(white sock and white tee) = \(\dfrac{6}{10} \cdot \dfrac{3}{7}=\dfrac{9}{35}\)
P(white sock or white tee) = \(\dfrac{6}{10}+\dfrac{3}{7}-\dfrac{9}{35}=\dfrac{27}{35}\)
- a. \(\dfrac{6}{10}+\dfrac{3}{7}-\dfrac{9}{35}=\dfrac{27}{35}\)
-
Out of 100,000 people, 500 would have the disease. Of those, all 500 would test positive. Of the 99,500 without the disease, 2,985 would falsely test positive and the other 96,515 would test negative.
\(P(\text { disease } \mid \text { positive })=\dfrac{500}{500+2985} \approx 14.3 \%\)
- \(8 \cdot 11 \cdot 5=440\) menu combinations
- There are 26 characters. a. \(26^{5}=11,881,376\) b. \({ }_{26} \mathrm{P}_{5}=26 \cdot 25 \cdot 24 \cdot 23 \cdot 22=7,893,600\)
- Order does not matter. \({ }_{29} C_{19}=20,030,010\) possible subcommittees
-
There are \(5^{10}=9,765,625\) different ways the exam can be answered. There are 10 possible locations for the one missed question, and in each of those locations there are 4 wrong answers, so there are 36 ways the test could be answered with one wrong answer.
\(P(9 \text { answers correct })=\dfrac{40}{5^{10}} \approx 0.0000041\) chance
- \(P(\text { three Aces and two Kings })=\dfrac{\left (_{4} C_{3}\right)\left (_{4} C_{2}\right)}{ _{52} C_{5}}=\dfrac{24}{2598960} \approx 0.0000092\)
- \(P(\text { shared birthday })=1-\dfrac {_{365}{P}_{10}}{365^{10}} \approx 0.117\)
- \((\$ 3,995) \cdot \dfrac{1}{2000}+(-\$ 5) \cdot \dfrac{1999}{2000} \approx-\$ 3.00\)
-
Suppose you roll the first die. The probability the second will be different is \(\dfrac{5}{6}\). The probability that the third roll is different than the previous two is \(\dfrac{4}{6}\), so the probability that the three dice are different is \(\dfrac{5}{6} \cdot \dfrac{4}{6}=\dfrac{20}{36}\). The probability that two dice will match is the complement, \(1-\dfrac{20}{36}=\dfrac{16}{36}\).
The expected value is: \(\text { (\$2) } \cdot \dfrac{16}{36}+(-\$ 1) \cdot \dfrac{20}{36}=\dfrac{12}{36} \approx \$ 0.33\) Yes, it is in your advantage to play. On average, you’d win $0.33 per play.
5: Data and Statistics
5.1: Basics of Statistics
- The sample is the 20 fish caught. The population is all fish in the lake. The sample may be somewhat unrepresentative of the population since not all fish may be large enough to catch the bait.
- This is a parameter, since the college would have access to data on all students (the population)
- a. Categorical. b. Quantitative c. Quantitative
-
a. Systematic
b. Stratified or Quota
c. Voluntary response
d. Simple random
e. Cluster
-
a. Response bias – historically, men are likely to over-report, and women are likely to under-report to this question.
b. Voluntary response bias – the sample is self-selected
c. Sampling bias – the sample may not be representative of the whole class
d. Lack of anonymity
e. Self-interest study
f. Loaded question
-
a. Observational study
b. Experiment; the treatment is the jumping jacks
c. Experiment; the treatments are coffee and tea
- The truth-telling group could be considered the control group, but really both groups are treatment groups here, since it is important for the lie detector to be able to correctly identify lies, and also not identify truth telling as lying. This study is blind, since the person running the test does not know what group each subject is in.
5.2: Describing Data
- While the pie chart accurately depicts the relative size of the people agreeing with each candidate, the chart is confusing, since usually percents on a pie chart represent the percentage of the pie the slice represents.
- Using a class intervals of size 55, we can group our data into six intervals:
|
Cost Interval |
Frequency |
|---|---|
|
$140-194 |
5 |
|
$195-249 |
3 |
|
$250-304 |
9 |
|
$305-359 |
12 |
|
$360-414 |
4 |
|
$415-469 |
3 |
- Adding the prices and dividing by 5 we get the mean price: $3.682
- First we put the data in order: $3.29, $3.59, $3.75, $3.79, $3.99. Since there are an odd number of data, the median will be the middle value, $3.75.
-
There are 23 ratings.
- The mean is \(\dfrac{1 \cdot 4+2 \cdot 8+3 \cdot 7+4 \cdot 3+5 \cdot 1}{23} \approx 2.5\)
- There are 23 data values, so the median will be the 12th data value. Ratings of 1 are the first 4 values, while a rating of 2 are the next 8 values, so the 12th value will be a rating of 2. The median is 2.
- The mode is the most frequent rating. The mode rating is 2.
- Earlier we found the mean of the data was $3.682
|
data value |
deviation: data value - mean |
deviation squared |
|---|---|---|
|
3.29 |
3.29 – 3.682 = -0.391 |
0.153664 |
|
3.59 |
3.59 – 3.682 = -0.092 |
0.008464 |
|
3.79 |
3.79 – 3.682 = 0.108 |
0.011664 |
|
3.75 |
3.75 – 3.682 = 0.068 |
0.004624 |
|
3.99 |
3.99 – 3.682 = 0.308 |
0.094864 |
This data is from a sample, so we will add the squared deviations, divide by 4, the number of data values minus 1, and compute the square root:
\[\sqrt{\dfrac{0.153664+0.008464+0.011664+0.004624+0.094864}{4} \approx 0.261} \nonumber \]
-
The data is already in order, so we don’t need to sort it first.
The minimum value is $140 and the maximum is $460.
There are 36 data values so \(n = 36\). \(n/2 = 18\), which is a whole number, so the median is the mean of the 18th and 19th data values, $305 and $310. The median is $307.50.
To find the first quartile, we calculate the locator, \(L = 0.25(36) = 9\). Since this is a whole number, we know Q1 is the mean of the 9th and 10th data values, $250 and $260. Q1 = $255.
To find the third quartile, we calculate the locator, \(L = 0.75(36) = 27\). Since this is a whole number, we know Q3 is the mean of the 27th and 28th data values, $345 and $350. Q3 = $347.50.
The 5 number summary of this data is: $140, $255, $307.50, $347.50, $460
- Boxplot of textbook costs
6: Growth and Decay
6.1: Growth Models
-
Letting \(n = 0\) correspond with 1976, then \(P_0 = 20,610\).
From 1976 to 2010 the number of stay-at-home fathers increased by
53,555 – 20,610 = 32,945
This happened over 34 years, giving a common different \(d\) of 32,945 / 34 = 969.
\(P_n = 20,610 + 969n\)
Predicting for 2020, we use \(n = 44\)
\(P_{44} = 20,610 + 969(44) = 63,246\) stay-at-home fathers in 2020.
-
Using \(n = 0\) corresponding with 2008,
\(P_{12} = (1+0.0134)^{12} (1.14) =\) about 1.337 billion people in 2020
-
Here we will measure
n
in months rather than years, with \(n = 0\) corresponding to the February when they went public. This gives \(P_0 = 45\) thousand. October is 8 months later, so \(P_8 = 60\).
\(\begin{array}{l}
P_{8}=(1+r)^{8} P_{0} \\
60=(1+r)^{8} 45 \\
\dfrac{60}{45}=(1+r)^{8} \\
\dfrac{\sqrt{60}}{45}=1+r \\
r=\sqrt[8]{\dfrac{60}{45}}-1=0.0366
\end{array} \) or 3.66%The general explicit equation is \(P_{n}=(1.0366)^{n} 45\)
Predicting 24 months (2 years) after they went public:
\(P_{24}=(1.0366)^{24} 45=106.63\) thousand users.
- \(1.14(1.0134)^{n}=1.2\). \(n = 3.853\), which is during 2011
-
\(P_{1}=P_{0}+0.70\left(1-\dfrac{P_{0}}{300}\right) P_{0}=20+0.70\left(1-\dfrac{20}{300}\right) 20=33\)
\(P_2 = 54\)
\(P_3 = 85\)
6.2: Logarithmic Scales in Natural Sciences
-
Letting \(n = 0\) correspond with 1976, then \(P_0 = 20,610\).
From 1976 to 2010 the number of stay-at-home fathers increased by
53,555 – 20,610 = 32,945
This happened over 34 years, giving a common different d of 32,945 / 34 = 969.
\(P_{n}=20,610+969 n\)
Predicting for 2020, we use \(n = 44\)
\(P_{44}=20,610+969(44)=63,246\) stay-at-home fathers in 2020.
-
Using \(n = 0\) corresponding with 2008,
\(P_{12}=(1+0.0134)^{12}(1.14)=\) about 1.337 billion people in 2020
-
Here we will measure n in months rather than years, with
n
= 0 corresponding to the February when they went public. This gives \(P_0= 45\) thousand. October is 8 months later, so \(P_8= 60\).
\(\begin{array}{l}
P_{8}=(1+r)^{8} P_{0} \\
60=(1+r)^{8} 45 \\
\dfrac{60}{45}=(1+r)^{8} \\
\sqrt[8]{\dfrac{60}{45}}=1+r \\
r=\sqrt[8]{\dfrac{60}{45}}-1=0.0366 \text { or } 3.66 \%
\end{array}\)The general explicit equation is \(P_{n}=(1.0366)^{n} 45\)
Predicting 24 months (2 years) after they went public:
\(P_{24}=(1.0366)^{24} 45=106.63\) thousand users.
- \(\log (1,000,000)=6\)
- \(\log (123) \approx 2.0899\)
- The difference in magnitudes was about 3.929.
7: Mathematics and the Arts
7.4: Fractals
8: Mathematics and Politics
8.3: Voting Theory
-
Using plurality method:
G gets 44+14+20 = 78 first-choice votes
M gets 70+22 = 92 first-choice votes
B gets 80+39 = 119 first-choice votes
Bunney (B) wins under plurality method.
-
Determining the Condorcet Winner:
G vs M: 44+14+20 = 78 prefer G, 70+22+80=172 prefer M: M preferred
G vs B: 44+14+20+70=148 prefer G, 22+80+39 = 141 prefer B: G preferred
M vs B: 44+70+22=136 prefer M, 14+80+39=133 prefer B: M preferred
M is the Condorcet winner, based on the information we have.
-
Using IRV:
G has the fewest first-choice votes, so is eliminated first. The 20 voters who did not list a second choice do not get transferred - they simply get eliminated
136
133
1st choice
M
B
2nd choice
B
M
McCarthy (M) now has a majority, and is declared the winner.
-
Using Borda Count:
We give 1 point for 3rd place, 2 points for 2nd place, and 3 points for 1st place.
44
14
20
70
22
80
39
1st choice
G
132 pt
G
42 pt
G
60 pt
M
210 pt
M
66 pt
B
240 pt
B
117 pt
2nd choice
M
88 pt
B
28 pt
G
140 pt
B
44 pt
M
160 pt
3rd choice
B
44 pt
M
14 pt
M 20 pt
B 20 pt
B
70 pt
G
22 pt
G
80 pt
M 39 pt
G 39 pt
G: 132+42+60+140+22+80+39 = 515 pts
M: 88+14+20+210+66+160+39 = 597 pts
B: 44+28+20+70+44+240+117 = 563 pts
McCarthy (M) would be the winner using Borda Count.
-
Using Copeland’s Method:
Looking back at our work from Try it Now #2, we see
G vs M: 44+14+20 = 78 prefer G, 70+22+80=172 prefer M: M preferred – 1 point
G vs B: 44+14+20+70=148 prefer G, 22+80+39 = 141 prefer B: G preferred – 1 point
M vs B: 44+70+22=136 prefer M, 14+80+39=133 prefer B: M preferred – 1 point
M earns 2 points; G earns 1 point. M wins under Copeland’s method.
-
Using Approval voting:
Seattle has 30+10+15+5 = 60 approval votes
Tacoma has 30+15+20+15+5 = 85 approval votes
Puyallup has 10+20+25+5 = 50 approval votes
Olympia has 15+15+5 = 35 approval votes
Tacoma wins under this approval voting
8.4: Weighted Voting
-
If we represent the players as \(M_{1}, M_{2}, M_{3}, M_{4}, U_{1}, U_{2}, U_{3}\), then we may be tempted to set up a system like [4: 1, 1, 1, 1, 1, 1, 1]. While this system would meet the first requirement that four members must support a proposal for it to pass, this does not satisfy the requirement that at least one member of the union must support it.
To accomplish that, we might try increasing the voting weight of the union members: [5: 1, 1, 1, 1, 2, 2, 2]. The quota was set at 5 so that the four management members alone would not be able to reach quota without one of the union members. Unfortunately, now the three union members can reach quota alone. To fix this, three management members need to have more weight than two union members.
After trying several other guesses, we land on the system [13: 3, 3, 3, 3, 4, 4, 4]. Here, the four management members have combined weight of 12, so cannot reach quota. Likewise, the three union members have combined weight of 12, so cannot reach quota alone. But, as required, any group of 4 members that includes at least one union member will reach the quota of 13. For example, three management members and one union member have combined weight of 3+3+3+4=13, and reach quota.
-
In the voting system \([q: 10, 5, 3]\), if the quota is 10, then player 1 is a dictator since they can reach quota without the support of the other players. This makes the other two players automatically dummies.
If the quota is 12, then player 1 is necessary to reach quota, so has veto power. Since at this point either player 2 or player 3 would allow player 1 to reach quota, neither player is a dummy, so they are regular players (not dictators, no veto power, and not a dummy).
If the quota is 16, then no two players alone can reach quota, so all three players have veto power.
-
The voting system tells us that the quota is 36, that Player 1 has 20 votes (or equivalently, has a weight of 20), Player 2 has 17 votes, Player 3 has 16 votes, and Player 4 has 3 votes.
A coalition is any group of one or more players. What we're looking for is winning coalitions - coalitions whose combined votes (weights) add to up to the quota or more. So the coalition \(\left\{P_{3}, P_{4}\right\}\) is not a winning coalition because the combined weight is 16+3=19, which is below the quota.
So we look at each possible combination of players and identify the winning ones:
{P1, P2} (weight: 37) {P1, P3} (weight: 36)
{P1, P2, P3} (weight: 53) {P1, P2, P4} (weight: 40)
{P1, P3, P4} (weight: 39) {P1, P2, P3, P4} (weight: 56)
{P2, P3, P4} (weight: 36)
Now, in each coalition, we need to identify which players are critical. A player is critical if the coalition would no longer reach quota without that person. So, in the coalition {P1, P2}, both players are necessary to reach quota, so both are critical. However in the coalition {P1, P2, P3}, we can see from the earlier two coalitions that either P2 or P3 could leave the coalition and it would still reach quota. But if P1 left, it would not reach quota, so P1 is the only player critical in this coalition. We evaluate the rest of the coalitions similarly, giving us this (underlining the critical players)
{ P1, P2 } { P1, P3}
{ P1 , P2, P3} { P1, P2 , P4}
{P1, P3 , P4} {P1, P2, P3, P4}
{ P2, P3, P4 }
Next we count how many times each player is critical:
P1: 5 times
P2: 3 times
P3: 3 times
P4: 1 time
In total, there were 5+3+3+1 = 12 times anyone was critical, so we take our counts and turn them into fractions, giving us our Banzhaf power:
P1: 5/12
P2: 3/12 = 1/4
P3: 3/12 = 1/4
P4: 1/12
[36: 20, 17, 15]
-
Listing all sequential coalitions and identifying the pivotal player:
< P1 , P2 , P3 > < P1, P3, P2 > < P2, P1 , P3 >
< P2, P3, P1 > < P3, P2, P1 > < P3, P1, P2 >
P1 is pivotal 3 times, P2 is pivotal 3 times, and P3 is pivotal 0 times.
Player
Times pivotal
Power index
P1
3
3/6 = 50%
P2
3
3/6 = 50%
P3
0
0/6 = 0%
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0