Answer:
The outlier for this case would be (19.37 , 337) since this point is far away from the others and this point probably strongly influences the correlation value.
What is the correlation with this point? (Round your answer to two decimal places.)
> cor(x,y)
[1] 0.34
What is the correlation without this point? (Round your answer to two decimal places.)
> cor(x1,y1)
[1] 0.79
Calculate the regression line with the outlier.
y = 35.2 x-523.9
Calculate the regression line without the outlier.
y = 77.9 x -1422.3
Step-by-step explanation:
We have the following data:
Isotope %: 19.90, 20.71, 21.63, 19.84, 20.80, 21.63, 19.46, 20.86, 21.19,20.20, 21.28, 19.37 (representing X)
Silicon : 85,152,226,106,263,233,114,265,186,139,298,337 (representing Y)
Find the single outlier in the data. This point strongly influences the correlation. What is the correlation with this point? (Round your answer to two decimal places.)
We can use the scatter plot in order to see any potential outlier. With the following R code:
> x<-c(19.90, 20.71, 21.63, 19.84, 20.80, 21.63, 19.46, 20.86, 21.19,20.20, 21.28, 19.37)
> y<-c(85,152,226,106,263,233,114,265,186,139,298,337)
> plot(x,y, main="Scatter plot Silicon vs Isotope")
And we can see the plot on the figure attached.
The outlier for this case would be (19.37 , 337) since this point is far away from the others and this point probably strongly influences the correlation value.
What is the correlation with this point? (Round your answer to two decimal places.)
> cor(x,y)
[1] 0.34
What is the correlation without this point? (Round your answer to two decimal places.)
> x1<-x[-12]
> x1
[1] 19.90 20.71 21.63 19.84 20.80 21.63 19.46 20.86 21.19 20.20 21.28
> y1<-y[-12]
> y1
[1] 85 152 226 106 263 233 114 265 186 139 298
> cor(x1,y1)
[1] 0.79
As we can see the correlation changes significantly without the outlier.
c) Is the outlier also strongly influential for the regression line? Calculate the regression line with the outlier. (Round your slope to two decimal places, round your y-intercept to one decimal place.)y = ( ) ? x ( )
We can calculate the regression line with the following R code
> linearmod1<-lm(y~ x)
> linearmod1
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
-523.9 35.2
So our equation would be: y = 35.2 x-523.9
Calculate the regression line without the outlier. (Round your slope to two decimal places, round your y-intercept to one decimal place.)y = ( ) ?( ) x
> linearmod2<-lm(y1~x1)
> linearmod2
Call:
lm(formula = y1 ~ x1)
Coefficients:
(Intercept) x1
-1422.26 77.85
The new equation would be y = 77.9 x -1422.3
So as we can see the outlier also changes significantly the estimation for the slope and the intercept of the linear model
To answer this question one would need to identify the outlier in the data then calculate both the correlation and regression lines with and without this outlier. Changes in these calculations can demonstrate the influence of the outlier.
Explanation:This question involves finding statistical outliers and calculating correlation and regression lines in a dataset. The outlier in a dataset is a data point that is remarkably distinct from the rest of the data. As your question doesn't provide a clear dataset, it's impossible to identify the outlier clearly. However, once identified, to determine if the point strongly influences the correlation you would find the correlation with the outlier and without the outlier and compare these two values.
If the
correlation
drastically changes when the outlier is removed, it can confirm that the point is strong influencing the correlation. The
regression line
could be calculate using the standard formulas for slope and intercept, both with and without the outlier in the data. If the regression line meaningfully shifts after removing the outlier, it indicates that outlier is impactful to the regression line as well. This analytical work typically requires skills in interpreting scatterplots and using statistical programs or calculators.
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A sample of 100 cars driving on a freeway during a morning commute was drawn, and the number of occupants in each car was recorded. The results were as follows: NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Occupants 1 2 3 4 5
Number of Cars 71 16 8 3 2
a. For what proportion of cars was the number of occupants more than one standard deviation greater than the mean? (Round the final answer to two decimal places.)
Answer: 29%
Step-by-step explanation:
I considered the question to be: for what proportion of cars was the number of occupants more than 1 standard deviation and greater than the mean.
The first step is to calculate the weighted mean of the number of occupants in the cars. The next step is to determine the standard deviation using the formula √[ Σ ( xi - μ )² / N ]. Subsequently, identify the number of occupants that are more than one standard deviation greater than the mean.
Explanation:In this question, we're asked to calculate the proportion of cars that had more than one standard deviation above the mean number of occupants. First, we need to calculate the weighted mean (average) of the number of occupants in the cars. Based on the data, the mean can be calculated as:
Mean = (1*71 + 2*16 + 3*8 + 4*3 + 5*2) / 100 = 1.66
Next, we find the standard deviation. The standard deviation tells us how spread out the numbers are from the mean. Calculating standard deviation is a bit involved. The formula is √[ Σ ( xi - μ )² / N ]. Once you have these, look at the number of occupants that are more than one standard deviation greater than this mean.
Note: This is a statistical analysis question that requires knowledge of the concepts of mean and standard deviation.
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Harry notes that the state sales tax went from 2% to 2.5%, which he says is not too bad because it's just a one-half percent increase. But Linda says that it really is bad because it's a 25% increase. Who's right, and why?
We are required to calculate the percentage increase in tax and determine who is right.
The percentage increase in tax is 25% and Linda is very correct
percentage increase = difference in tax /
percentage increase = difference in tax / old tax × 100
old tax = 2%
New tax = 2.5%
Difference = New tax - old tax
= 2.5% - 2%
= 0.5%
percentage increase = difference in tax /
percentage increase = difference in tax / old tax × 100
= 0.5% / 2% × 100
= 0.25 × 100
= 25%
Therefore, the percentage increase in tax is 25% and Linda is very correct
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Suppose you received a score of 95 out of 100 on exam 1 . The mean was 79 and the standard deviation was 8 . If your score on exam 2 is 90 out of 100 , and the mean was 60 with a standard deviation of 15 , then you did:
better on exam 1 .
worse on exam 1 .
the same on both exams.
worse on exam 2
Answer:
You did the same on both exams.
Step-by-step explanation:
To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:
Z = (X - μ)/σ
Where X = score obtained
μ = mean score
σ = standard deviation
For Exam 1:
Z = (95 - 79)/8
= 16/8
Z = 2
For Exam 2:
Z = (90 - 60)/15
= 30/15
Z = 2
The z-scores for both the tests are same hence the third option is correct i.e. you did the same on both exams.
Your performance on exam 1 and exam 2 can be compared using Z-scores, which measure how many standard deviations a score is from the mean. You scored 2 standard deviations above the mean on both exams, so you did the same on both exams.
Explanation:In this question, your performance on exams is being compared relative to the mean of the class scores and their standard deviation. This is a concept in statistics known as Z-scores. The Z-score tells us how many standard deviations an observation (your score) is from the mean. The formula for Z-score is (observation - mean) / standard deviation.
For exam 1 your Z - score is (95-79) / 8 which equals 2. This means you scored 2 standard deviations above the mean on exam 1. For exam 2, your Z-score is (90-60) / 15 which equals 2. Again, this means you scored 2 standard deviations above the mean on exam 2. Because your Z-score for both exams is the same, you did the same on both exams.
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Consider the following vector-valued function:~h(t) =〈2 sin(3t),3 cos(3t),√5 sin(3t)〉0≤t≤2π3This defines a smooth parameterized curve.(a) Find the unit tangent vector~T(t) for 0≤t≤2π3.(b) Find all of the values oftin the interval 0≤t≤2π3where~h(t) and~T(t) areorthogonal.(c) Show that the curve~h(t) lies on a sphere. What is the radius of the sphere?
Answer:
a) h'(t)= (6cos3t,-9sin3t,3[tex]\sqrt[]{5}[/tex]cos3t)
b) t=0.93994736+πn/3
c) Magnitude of h(t) is 3 which is a constant, so h(t) lies on a sphere
Step-by-step explanation:
An elementary school is offering 2 language classes: one in Spanish (S) and one in French (F). Given that P(S) = 50%, P(F) = 40%, P(S ∪ F) = 70%, find the probability that a randomly selected student (a) is taking Spanish given that he or she is taking French; (b) is not taking French given that he or she is not taking Spanish. 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears.
Answer:
(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French = 0.5 .
(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish = 0.6 .
Step-by-step explanation:
We are given that an elementary school is offering 2 language classes ;
Spanish Language is denoted by S and French language is denoted by F.
Also we are given, P(S) = 0.5 {Probability of students taking Spanish language}
P(F) = 0.4 {Probability of students taking French language}
[tex]P(S\bigcup F)[/tex] = 0.7 {Probability of students taking Spanish or French Language}
We know that, [tex]P(A\bigcup B)[/tex] = [tex]P(A) + P(B) -[/tex] [tex]P(A\bigcap B)[/tex]
So, [tex]P(S\bigcap F)[/tex] = [tex]P(S) + P(F) - P(S\bigcup F)[/tex] = 0.5 + 0.4 - 0.7 = 0.2
[tex]P(S\bigcap F)[/tex] means Probability of students taking both Spanish and French Language.
Also, P(S)' = 1 - P(S) = 1 - 0.5 = 0.5
P(F)' = 1 - P(F) = 1 - 0.4 = 0.6
[tex]P(S'\bigcap F')[/tex] = 1 - [tex]P(S\bigcup F)[/tex] = 1 - 0.7 = 0.3
(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French is given by P(S/F);
P(S/F) = [tex]\frac{P(S\bigcap F)}{P(F)}[/tex] = [tex]\frac{0.2}{0.4}[/tex] = 0.5
(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish is given by P(F'/S');
P(F'/S') = [tex]\frac{P(S'\bigcap F')}{P(S')}[/tex] = [tex]\frac{1- P(S\bigcup F)}{1-P(S)}[/tex] = [tex]\frac{0.3}{0.5}[/tex] = 0.6 .
Note: 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears ; This question is incomplete please provide with complete detail.
Solve the system using the substitution or elimination method. How many solutions are there to this system?
Answer:
[tex] -3*(3y+2) + 9y = -6[/tex]
[tex] -9y -6 + 9y = -6[/tex]
[tex]-6=-6[/tex]
So then as we can see we can have infinite solutions.
[tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]
Step-by-step explanation:
Assuming the following system of equations:
[tex] 2x-6y =4[/tex] (1)
[tex] -3x+9y =-6[/tex] (2)
For this case we can use the substitution method in order to find the possible solutions for the system.
If we solve for x from equation (1) we got:
[tex] 2x = 6y +4[/tex]
[tex] x = 3y +2 [/tex] (3)
Now we can replace equation (3) into equation (2) and we got:
[tex] -3*(3y+2) + 9y = -6[/tex]
[tex] -9y -6 + 9y = -6[/tex]
[tex]-6=-6[/tex]
So then as we can see we can have infinite solutions.
And the possible solutions are for a fixed value of x, we can solve y from equation (3) and we got:
[tex] y = \frac{x-2}{3}[/tex]
So the solution would be: [tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]
********Please show work*********
Kenen loves trains, especially those that run on narrow-gauge tracks. (The gauge
of a track measures how far apart the rails are.) He has decided to build a model
train of the Rio Grande, a popular narrow-gauge train.
Use the following information to help him know how big his model should be:
* The real track has a gauge of 3 feet (36 inches).
* His model railroad track has a gauge of 3/4 inches.
*The Rio Grande train he wants to model has driving wheels that measure 44 inches high.
Your Task: With your team, discuss what you know about the model train Kenen will build.
1) What scale factor should he use?
2) What will be the height of the driving wheels of his model?
Since the real track has a gauge of 3 feet but the model railroad track has a gauge of 3/4 inches, the scale factor must be 1/4, because you must follow the rule
"true measurement * scale factor = model measurement"
In fact, we have
[tex]3\cdot\dfrac{1}{4}=\dfrac{3}{4}[/tex]
This implies that the original driving wheels height, 44 inches, must be scaled down to
[tex]44\cdot\dfrac{1}{4}=11[/tex]
inches.
A new drug to treat psoriasis has been developed and is in clinical testing. Assume that those individuals given the drug are examined before receiving the treatment and then again after receiving the treatment to determine if there was a change in their symptom status. If the initial results showed that 2.0% of individuals entered the study in remission, 77.0% of individuals entered the study with mild symptoms, 16.0% of individuals entered the study with moderate symptoms, and 5.0% entered the study with severe symptoms calculate and interpret a chi-squared test to determine if the drug was effective treating psoriasis given the information below from the final examination.
Answer:
Step-by-step explanation:
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: The distribution of severity of psoriasis cases at the end and prior are same.
Alternative hypothesis: The distribution of severity of psoriasis cases at the end and prior are different.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 4 - 1
D.F = 3
(Ei) = n * pi
Category observed Num expected num [(Or,c -Er,c)²/Er,c]
Remission 380 20 6480
Mild
symptoms 520 770 81.16883117
Moderate
symptoms 95 160 24.40625
Severe
symptom 5 50 40.5
Sum 1000 1000 6628.075081
Χ2 = Σ [ (Oi - Ei)2 / Ei ]
Χ2 = 6628.08
Χ2Critical = 7.81
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and Χ2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 6628.08.
We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.58) =less than 0.000001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
We reject H0, because 6628.08 is greater than 7.81. We have statistically significant evidence at alpha equals to 0.05 level to show that distribution of severity of psoriasis cases at the end of the clinical trial for the sample is different from the distribution of the severity of psoriasis cases prior to the administration of the drug suggesting the drug is effective.
The chi-square test is a statistical method that determines if there's a significant difference between observed and expected frequencies in different categories, such as symptom status in this clinical trial. Without post-treatment numbers, we can't run the exact test. However, if the test statistic exceeded the critical value, we could conclude that the drug significantly affected symptom statuses.
Explanation:This question pertains to the use of a chi-squared test, which is a statistical method used to determine if there's a significant difference between observed frequencies and expected frequencies in one or more categories. For this case, the categories are the symptom statuses (remission, mild, moderate, and severe).
To conduct a chi-square test, you first need to know the observed frequencies (the initial percentages given in the question) and the expected frequencies (the percentages after treatment). As the question doesn't provide the numbers after treatment, I can't perform the exact chi-square test.
If the post-treatment numbers were provided, you would compare them to the pre-treatment numbers using the chi-squared formula, which involves summing the squared difference between observed and expected frequencies, divided by expected frequency, for all categories. The result is a chi-square test statistic, which you would then compare to a critical value associated with a chosen significance level (commonly 0.05) to determine if the treatment has a statistically significant effect.
To interpret a chi-square test statistic, if the calculated test statistic is larger than the critical value, it suggests that the drug made a significant difference in the distribution of symptom statuses. If not, we can't conclude the drug was effective.
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Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answer and C denotes a correct answer. b. Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly one correct answer when the guesses are made?
Answer:
a) Probability of picking the first two answers wrong & the third answer correctly in that order, P(WWC) = 0.128
b) All possible outcomes = WWC, WCW, CWW
P(WWC) = 0.128; P(WCW) = 0.128; P(CWW) = 0.128
c) Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = 0.384
Step-by-step explanation:
P(Correct answer) = P(C) = number of correct options/total options available.
That is, P(C) = 1/5 = 0.2
P(Wrong answer) = P(W) = 1 - 0.2 = 0.8 or (number of wrong options)/(total options available) = 4/5 = 0.8
a) Using the multiple rule for independent events,
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
Probability of picking the first two answers wrong & the third answer correctly in that order = 0.128
b) All possible outcomes of picking two wrong answers and one right answer in whichever order for the first 3 questions are (WWC, WCW, CWW)
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
P(WCW) = P(W) × P(C) × P(W) = 0.8 × 0.2 × 0.8 = 0.128
P(CWW) = P(C) × P(W) × P(W) = 0.2 × 0.8 × 0.8 = 0.128
c) Using the addition rule for disjoint events,
Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = P(WWC) + P(WCW) + P(CWW) = 0.128 + 0.128 + 0.128 = 0.384
QED!
The probability of getting exactly one correct answer when guessing on three multiple-choice questions with five possible answers each is [tex]\(\frac{3}{25}\)[/tex].
a. To find the probability that the first two guesses are wrong and the third is correct, denoted as P(WWC), we use the multiplication rule for independent events. Since there are five possible answers for each question and only one correct answer, the probability of guessing wrong on one question is [tex]\(\frac{4}{5}\)[/tex] and the probability of guessing correctly is [tex]\(\frac{1}{5}\)[/tex]. Therefore, the probability of WWC is:
[tex]\[ P(WWC) = P(W) \times P(W) \times P(C) = \left(\frac{4}{5}\right) \times \left(\frac{4}{5}\right) \times \left(\frac{1}{5}\right) = \frac{16}{125} \][/tex]
b. The different possible arrangements of two wrong answers and one correct answer (WWC) are:
1. WWC
2. WWC
3. CWW
4. CW
5. WCW
6. WCW
For each of these arrangements, the probability is the same as calculated in part (a), which is \(\frac{16}{125}\).
c. Since there are six different ways to arrange two wrong answers and one correct answer, and each of these arrangements has the same probability, we multiply the probability of one such arrangement by the number of arrangements to find the total probability of getting exactly one correct
[tex]\[ P(\text{exactly one correct answer}) = 6 \times P(WWC) = 6 \times \frac{16}{125} = \frac{96}{125} \][/tex]
However, we must note that we have counted each arrangement twice because the order of the wrong answers (W) does not matter. Therefore, we need to divide the total by 2 to get the correct probability:
[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{2 \times 125} = \frac{48}{125} \][/tex]
Upon reviewing the calculations, it appears there was an error in the final step. We should not divide by 2 because the arrangements are distinct due to the position of the correct answer (C). The correct total probability is indeed [tex]\(\frac{96}{125}\)[/tex], but since we are looking for the probability of exactly one correct answer, we have to consider that there are three questions and the correct answer could be on any of them. Therefore, we have already accounted for all possible arrangements by multiplying by 6.
The correct probability of getting exactly one correct answer when guessing on three multiple-choice questions is:
[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{125} \][/tex]
However, this is not the final answer. We need to consider that there are three different ways to get exactly one correct answer out of three questions (CWW, WCW, WWC). Since these are mutually exclusive events, we add their probabilities:
[tex]\[ P(\text{exactly one correct answer}) = P(CWW) + P(WCW) + P(WWC) \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = \frac{16}{125} + \frac{16}{125} + \frac{16}{125} \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = 3 \times \frac{16}{125} \][/tex]
[tex]\[ P(\text{exactly one correct answer}) = \frac{48}{125} \][/tex]
Thus, the final correct probability is [tex]\(\frac{48}{125}\)[/tex], which simplifies to [tex]\(\frac{3}{25}\)[/tex].
The histogram displays the number of 2012 births among U.S. women ages 10 to 50 . Each bin represents an interval of two years, and the height of each bin represents the frequency with which the data fall within that interval?
Answer:
Number of births to women below 22 years of age: 830
% of births occurred to women of age between 34 and 36: 6.35%
Step-by-step explanation:
There are two parts to this question:
To calculate the number of births, we look at the histogram below.
We see that each bar has a number on top, suggesting that particular for the age limit. Number of births to women below age of 22, we start adding all numbers below the mark of 22 on x-axis:
⇒ 39+134+29+363 = 830
To calculate the percentage, we divide the number of births in that particular interval by total number of births and then multiply by 100.
To calculate the total number of births, we add all the numbers on the top of the bars:
⇒ 39+ 134+294+363+391+425+460+474+432+343+250+163+98+49+17+4+1 Total births = 3937
[tex]\frac{250}{3937} \times 100\\ 0.0635 \times 100\\= 6.35 \%[/tex]
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]
Data for an economy show that the unemployment rate is 6 percent, the participation rate is 60 percent, and 200 million people 16 years or older are not in the labor force. How many people are in the labor force in this economy
Answer:
300 million people
Step-by-step explanation:
If the participation rate is 60% and 200 million people 16 years or older are not in the labor force, it means that 200 million corresponds to 40% of people 16 years or older. Since 60% of people 16 years or older are in the labor force, the total number of people in the labor force is given by:
[tex]n=\frac{200}{0.4}-200\\ n= 300\ million\ people[/tex]
300 million people are in the labor force in this economy.
Find M. Write your answer in simplest radical
Answer:
(√6/√2)ft
Step-by-step explanation:
cos 45 = m / √6 ft
m = cos 45 x √6 ft
m = (1 / √2) x √6 ft = (√6/√2)ft
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
Let M = {Λ,abb} and L = {bba,ab, a}, what is ML ? ML ={bba, abbbba,abbab,abbba, ab,a} ML ={bba, abbbba,abbab,abba, ab,a} ML ={bbab, abbbba,abbab,abba, ab,a} ML ={ba, abbbba,abbab,abba, ab,a}
Answer:
ML = {bba, ab, a, bbaabb, ababb, aabb}
Step-by-step explanation:
By application of Union of a set.
M = {bba,ab, a}
L = {Λ,abb}
ML = {bba, ab, a, bbaabb, ababb, aabb}
For these types of questions, first click the line tool on the tool palette labelled PFloor, and plot by clicking your mouse for the first end-point -- touching the vertical axis then moving your mouse to the right and clicking again for the second end-point. The new line should intersect both the D1 and S1 lines and have a height greater than 50 as measured on the vertical axis.
To Plotting lines , use the PFloor line tool and click your mouse for the first and second end-points, ensuring that the line intersects both the D1 and S1 lines and has a height greater than 50.
To plot the line described in the question, follow these steps:
Select the line tool on the tool palette labeled PFloor.
Click your mouse for the first end-point on the vertical axis.
Move your mouse to the right and click again for the second end-point.
The new line should intersect both the D1 and S1 lines and have a height greater than 50 on the vertical axis.
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If a confidence interval is given from 45.82 up to 55.90 and the mean is known to be 50.86, what is the margin of error?
Answer:
[tex] ME = \frac{10.08}{2}= 5.04[/tex]
Step-by-step explanation:
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=50.86[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex] \bar X \pm ME[/tex] (1)
Or equivalently:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)
Where the margin of error is given by:
[tex] ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
For this case we have the confidence interval limits given (45.82, 55.90)
We can find the width of the interval like this:
[tex] Width =55.90-45.82= 10.08[/tex]
And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.
[tex] ME = \frac{10.08}{2}= 5.04[/tex]
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).A small island is 3 miles from the nearest point P on the straight shoreline of a large lake. If a woman on the island can row a boat 3 miles per hour and can walk 4 miles per hour, where should the boat be landed in order to arrive at a town 12 miles down the shore from P in the least time? Let x be the distance between point P and where the boat lands on the lakeshore. Hint: time is distance divided by speed.
Answer:
The trip consists of two parts. The rowing part is the hypotenuse of right angled triangle
whose sides are the distance from P to the island, which is 5, and the distance between P and
the landing point of the rowboat on the shore, which is x
so this part of trip is sqrt(25+ x^2)
The 2nd part is the walking part, which is (8-x)
Distance = rate times time (D = rt), so to get the time you have t = D/r. We must divide each
of the trip by the appropriate rate to get the time.
a) T(x) = sqrt(25+x^2)/3 + (8-x)/4
To find minimum time required take derivative of the T(x) function and find it's zeros
T'(x) = x/(3(sqrt(25+x^2)) - 1/4 = 0
x/(3(sqrt(25+x^2)) = 1/4
4x = 3sqrt(25+x^2
16x^2 = 9(25+x^2) = 225 + 9x^2
7x^2 = 225
x^2 = 225/7
x = sqrt(225/7) = 5.669467 miles
T(x) = 3.602386382 hours
Step-by-step explanation:
The point where the boat should be landed can be found by expressing
the distance travelled on the boat and walking as a function of time.
The point where the boat should be landed is the point 3.4 miles from the
point P towards the town.
Reasons:
x represent the distance from point P to the boat landing point.
Therefore, distance of rowing the boat = √((12 - x)² + 3²)
The total time, t, is therefore;
[tex]t = \dfrac{12-x}{4} +\dfrac{\sqrt{x^2 + 3^2} }{3}[/tex]
When the time is minimum, we get;
[tex]\dfrac{dt}{dx} = \dfrac{d}{dx} \left( \dfrac{12-x}{4} +\dfrac{\sqrt{x^2 + 3^2} }{3} \right) = \dfrac{12\cdot \left(-3 + 4 \cdot\dfrac{2 \cdot x }{2\cdot \sqrt{x^2 + 9} } \right)}{144}[/tex]
[tex]\dfrac{12\cdot \left(-3 + 4 \cdot\dfrac{2 \cdot x }{2\cdot \sqrt{x^2 + 9} } \right)}{144} = -\dfrac{1}{4} +\dfrac{x }{3 \cdot \sqrt{x^2 +9} }[/tex]
[tex]\dfrac{x }{3 \cdot \sqrt{x^2 +9} } = \dfrac{1}{4}[/tex]
4·x = 3·√(x² + 9)
16·x² = 9·(x² + 9)
7·x² = 81
[tex]x = \dfrac{9 \cdot \sqrt{7} }{7}[/tex]
x ≈ 3.4 miles.
The point where the boat should be landed is the point approximately 3.4
miles from the point P in the direction of the town.
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A sample of 16 people is taken and their weights are measured. The standard deviation of these 16 measurements is computed to be 5.8. What is the variance of these measurements?
Answer:
The variance of given sample is 33.64 square pounds.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 16
Standard deviation, s = 5.8 pounds
We have to find the variance of the given sample.
Variance is the square of the standard deviation.
[tex]\text{Variance} = (\text{Standard Deviation})^2\\= (5.8\text{ pounds})^2\\=33.64\text{ pound}^2[/tex]
Thus, the variance of given sample is 33.64 square pounds.
Find the probability for the experiment of drawing two marbles (without replacement) at random from a bag containing one green, two yellow, and three red marbles.
1. Both marbles are red.
2. Both marbles are yellow.
3. Neither marble is yellow.
4. The marbles are of different colors.
Final answer:
To find the probability of drawing two marbles without replacement, calculate the probability of each event separately and multiply them together.
Explanation:
To find the probability of drawing two marbles without replacement, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
1. Both marbles are red:
First, we calculate the probability of drawing a red marble on the first draw (3 red marbles out of 6 total marbles). Then, we calculate the probability of drawing a red marble on the second draw, given that the first marble drawn was red (2 red marbles out of 5 total marbles remaining). The probability is (3/6) * (2/5) = 1/5.
2. Both marbles are yellow:
First, we calculate the probability of drawing a yellow marble on the first draw (2 yellow marbles out of 6 total marbles). Then, we calculate the probability of drawing a yellow marble on the second draw, given that the first marble drawn was yellow (1 yellow marble out of 5 total marbles remaining). The probability is (2/6) * (1/5) = 1/15.
A square matrix A is said to be idempotent iff A2 = A. (i) Show that if A is idempotent, then so is I − A. (ii) Show that if A is idempotent, then the matrix 2A − I is also invertible. Hint: Same as before, guess the inverse and check your answer with the definition of inverse.
Answer:
Step-by-step explanation:
Given that A is a square matrix and A is idempotent
[tex]A^2 = A[/tex]
Consider I-A
i) [tex](I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A[/tex]
It follows that I-A is also idempotent
ii) Consider the matrix 2A-I
[tex](2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I[/tex]
So it follows that 2A-I matrix is its own inverse.
An algorithm takes 0.5 seconds to run on an input of size 100. How long will it take to run on an input of size 1000 if the algorithm has a running time that is linear? quadratic? log-linear? cubic?
Answer:
linear: 5s
quadratic: 50s
log-linear: 0.75 s
cubic: 500s
Step-by-step explanation:
Let [tex]t_1,t_2[/tex] be the running time associated with the input of sizes [tex] s_1,s_2[/tex]
If the running time is linear
[tex]t_2 = t_1\frac{s_2}{s_1} = 0.5*\frac{1000}{100} = 0.5*10 = 5s[/tex]
If the running time is quadratic
[tex]t_2 = t_1\left(\frac{s_2}{s_1}\right)^2 = 0.5*\left(\frac{1000}{100}\right)^2 = 0.5*10^2 = 50s[/tex]
If the running time is log-linear
[tex]t_2 = t_1\frac{log(s_2)}{log(s_1)} = 0.5*\frac{log(1000)}{log(100)} = 0.5*1.5 = 0.75s[/tex]
If the running time is cubic:
[tex]t_2 = t_1\left(\frac{s_2}{s_1}\right)^3 = 0.5*\left(\frac{1000}{100}\right)^3 = 0.5*10^3 = 500s[/tex]
Euclidean distance can be used to calculate the dissimilarity between two observations. Let u = (25, $350) correspond to a 25-year-old customer that spent $350 at Store A in the previous fiscal year. Let v = (53, $420) correspond to a 53-year-old customer that spent $4,100 at Store A in the previous fiscal year. Calculate the dissimilarity between these two observations using Euclidean distance.
a. 66.21
b. 88.57
c. 72.28
d. 75.39
Answer:
Option D
75.39
Step-by-step explanation:
When provided with the co-ordinates (x, y) and(a, b) then the distance between them is given by [tex]\sqrt {(x-a)^{2}+(y-b)^{2}}[/tex]
Since u = (25, $350) and v = (53, $420) then the Euclidean distance will be
[tex]\sqrt {(53-25)^{2}+(350-420)^{2}}=75.3923073\approx 75.39[/tex]
The dissimilarity between the two observations using Euclidean distance is approximately 75.39(Option d).
Explanation:To calculate the dissimilarity between two observations using Euclidean distance, we need to find the distance between the corresponding elements in the two observations and then calculate the square root of the sum of their squared differences.
In this case, we have:
u = (25, $350) and v = (53, $420)
The distance between the ages is 53 - 25 = 28.
The distance between the amounts spent is $420 - $350 = $70.
Now we can use the formula for Euclidean distance:
Distance = sqrt((28)^2 + (70)^2) = sqrt(784 + 4900) = sqrt(5684) ≈ 75.39
Therefore, the dissimilarity between the two observations using Euclidean distance is approximately 75.39.
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If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Answer:
C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).
Step-by-step explanation:
Let's introduce the cumulative distribution of (X,Y), X and Y :
F(X,Y)(x,y)=P(X≤x,Y≤y)
FX(x)=P(X≤x) FY(y)=P(Y≤y).Likewise for (s(X),t(Y)), s(X) and t(Y) :
F(s(X),t(Y))(u,v)=P(s(X)≤u
t(Y)≤v)Fs(X)(u)=P(s(X)≤u) Ft(Y)(v)=P(t(Y)≤v).Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :
F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))
The last step is obtained by applying the functions s and t since s preserves order and t reverses it.
This can be further transformed into
F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))
Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.
Now, to transform this into a statement about copulas, note that
C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))
Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,
we get
F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))
The left hand side is the copula C(X,Y), the right hand side still needs some work.
Note that
Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a
and likewise
Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b
Combining all results we obtain for the relationship between the copulas
C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).
According to the 2010 Census, 11.4% of all housing units in the United States were vacant. A county supervisor wonders if her county is different from this proportion. She randomly selects 850 housing units in her county and finds that 129 of the housing units are vacant. Write the null hypothesis and the alternative hypothesis Do a Test of Hypothesis and write the P-value. Write your conclusion: Construct a 95% cl for the true proportion of vacant houses in the supervisor's county. Does the confidence interval support your conclusion. Explain briefly.
Answer:
Null hypothesis:[tex]p=0.114[/tex]
Alternative hypothesis:[tex]p \neq 0.114[/tex]
[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>3.49)=0.00049[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]
And the 95% confidence interval would be given (0.128;0.176).
And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.
Step-by-step explanation:
Data given and notation
n=850 represent the random sample taken
X=129 represent the number of housing units that are vacant.
[tex]\hat p=\frac{129}{850}=0.152[/tex] estimated proportion of housing units that are vacant.
[tex]p_o=0.114[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is equal is 0.114 or not:
Null hypothesis:[tex]p=0.114[/tex]
Alternative hypothesis:[tex]p \neq 0.114[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>3.49)=0.00049[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
Confidence interval
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]
And the 95% confidence interval would be given (0.128;0.176).
And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.
We first set the null and alternative hypothesis and then conduct a z-test for proportions to calculate the z-score and subsequently the p-value. We use the p-value to decide whether to reject the null hypothesis. Finally, we construct a 95% confidence interval for the true proportion of vacant houses and check if this supports our test conclusion.
Explanation:Firstly, define the proportion of vacant housing units in the country as p0 and in the randomly selected county as p. The null hypothesis (H0) states that the county isn't different, so H0: p = p0 = 0.114. The alternative hypothesis (Ha) would be that the county is different, so Ha: p ≠ 0.114.
Let's conduct a Test of Hypothesis using a z-test for proportions. The z-score is calculated as (p - p0) / sqrt((p0 * (1 - p0)) / n), where n represents the sample size. Substituting in your values, the z score will be calculated. This z-score can be used to find the p-value from a standard normal (Z) distribution table.
If the p-value is less than 0.05 (which is α, significance level), we reject the null hypothesis in favor of alternative hypothesis, else we do not reject the null hypothesis. Thus, our conclusion is formulated based on this p-value.
To construct a 95% confidence interval for the true proportion of vacant houses, we use the formula: p ± Z * sqrt((p * (1 - p)) / n). Here, Z will be the Z score corresponding to the desired confidence level, 95% (which is 1.96 for two-tailed test).
If the national proportion (0.114) doesn't lie within this interval, it supports our test conclusion of rejecting the null hypothesis and vice versa.
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A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over the next few weeks, and find that out 10 recommendations, he was correct 3 times. If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successfu
Answer:
There is a 25.52% probability of observating 4 our fewer succesful recommendations.
Step-by-step explanation:
For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]p = 0.553, n = 10[/tex]
If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:
This is
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003[/tex]
[tex]P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039[/tex]
[tex]P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219[/tex]
[tex]P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724[/tex]
[tex]P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567[/tex]
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552[/tex]
There is a 25.52% probability of observating 4 our fewer succesful recommendations.
In a shipment of 58 vials, only 16 do not have hairline cracks. If you randomly select 2 vials from the shipment, what is the probability that none of the 2 vials have hairline cracks?
Answer:
0.0726 or 7.26%
Step-by-step explanation:
When choosing the first vial, there is a 16 in 58 chance that the vial does not have a hairline crack. When choosing the second vial, since on good vial was already picked, there is a 15 in 57 chance that the vial does not have a hairline crack. The probability that none of the 2 vials have hairline cracks is:
[tex]P = \frac{16}{58}*\frac{15}{57}\\P=0.0726[/tex]
There is a 0.0726 or 7.26% chance that none of the 2 vials have a hairline crack.
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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The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1999 to match the purchasing power of $1 in 1909? In other words, how much was that in 1999? (Don't use a $ sign, use 2 decimal places.)
Answer: 13.04
Here are some consumer price indexes from the past 100+ years:
Year CPI
1909 9.1
1919 17.3
1929 17.1
1939 13.9
1949 23.8
1959 29.1
1969 36.7
1979 72.6
1989 118.3
1999 166.6
2009 214.5
2015 238.5
The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1989 to match the purchasing power of $1 in 1909? In other words, how much was that in 1989?