Answer:
a) Null hypothesis: [tex]p \geq 0.368[/tex]
Alternative hypothesis: [tex]p<0.368[/tex]
b) A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.
c) A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
Type II error, also known as a "false negative" is the error of not rejecting a null hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.
Part a
On this case we want to test if the proportion of students who enroll in her institution have a lower completion rate (0.398), so the system of hypothesis would be:
Null hypothesis: [tex]p \geq 0.368[/tex]
Alternative hypothesis: [tex]p<0.368[/tex]
Part b
A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.
Part c
A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.
The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion. Making a Type I error would mean wrongly rejecting the null hypothesis, while making a Type II error would mean failing to reject the null hypothesis when it is actually false.
Explanation:(a) The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion of 0.398. The alternative hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is lower than 0.398.
(b) Making a Type I error in this context would mean rejecting the null hypothesis when it is actually true. This would imply that the college has a lower completion rate when in fact it does not.
(c) Making a Type II error in this context would mean failing to reject the null hypothesis when it is actually false. This would imply that the college has the same completion rate as the national proportion when it actually has a lower completion rate.
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What ratio is equivalent to 2 to 17
Answer:
4 to 34
Step-by-step explanation:
Answer:
There are a few correct equivalents to this. Please mark brainliest!!!
Step-by-step explanation:
4:34, 6:51, 8:68, and 10;85
Chrissy walked 1/8 of a mile on Monday and 1/8 of a mile on Tuesday. How many miles did Chrissy walk on Monday and Tuesday combined? Express the answer in lowest terms.
1/16
1/8
1/4
1/2
Answer:
Chrissy walked 1/4 miles on Monday and Tuesday combined.
Step-by-step explanation:
1/8+1/8=2/8
2/8=1/4 you take half of both sides to get the lowest terms using simplification.
________ and currency risks are to key country success factors as land costs and ________ are to key region success factors. Cultural issues; zoning restrictions Labor cost; proximity to customers Land costs; air and rail systems Exchange rates; environmental impact All of the above are accurate relationships.
Country: Cultural issues, currency risks. Region: Land costs, air and rail systems. (Total: 10 words)
In assessing country and region success factors, certain variables hold significant weight. Country success factors encompass aspects such as cultural issues and currency risks, which directly influence a nation's economic stability and investment attractiveness.
Cultural nuances affect business practices and consumer behaviors, while currency risks impact trade competitiveness and profitability.
Conversely, region success factors pivot around considerations like land costs and transportation infrastructure, particularly air and rail systems. Land costs dictate the feasibility of establishing operations or acquiring property within a region, shaping investment decisions.
Meanwhile, efficient air and rail systems facilitate logistical operations and market access, enhancing the competitiveness of businesses located within the region.
These factors collectively contribute to a region's appeal for businesses seeking to establish or expand operations.
Thus, understanding and appropriately navigating these variables are crucial for organizations aiming to optimize their strategic positioning at both the national and regional levels, ensuring long-term success and sustainability in an increasingly complex global landscape.
Assume that you have two dice, one of which is fair, and the other is biased toward landing on six, so that 0.25 of the time it lands on six, and 0.15 of the time it lands on each of 1, 2, 3, 4 and 5. You choose a die at random, and roll it six times, getting the values 4, 3, 6, 6, 5, 5. What is the probability that the die you chose is the fair die? The outcomes of the rolls are mutually independent.
Answer:
0.4038
Step-by-step explanation:
Let A and B be the events
A: “The die is fair”
B: “The die lands on 6 two times out of 6”
We want to determine if the die is fair given that it landed on 6 two times out of 6 tosses, that is P(A | B).
By the Bayes' theorem
[tex]\large P(A|B)=\displaystyle\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}[/tex]
Where [tex]\large A^c[/tex] is the event “the die is not fair”.
Since there are 2 dice,
[tex]\large P(A)=P(A^c)=1/2[/tex]
If the die is fair P(B | A) is the probability of getting exactly two six in a binomial experiment with probability of “success” (land on 6) 1/6 and six repeated trials
[tex]\large P(B|A)=\binom{6}{2}(1/6)^2(5/6)^4=0.2009[/tex]
and [tex]\large P(B|A^c)[/tex] is the probability of getting exactly two six in a binomial experiment with probability of “success” (land on 6) 0.25 and six repeated trials
[tex]\large P(B|A^c)=\binom{6}{2}(0.25)^2(0.75)^4=0.2966[/tex]
hence
[tex]\large P(A|B)=\displaystyle\frac{0.2009*0.5}{0.2009*0.5+0.2966*0.5}=0.4038[/tex]
To find the probability that the chosen die is fair after rolling 4, 3, 6, 6, 5, 5, we use Bayes' theorem and consider the probability of obtaining this sequence with both a fair die and a biased die. We incorporate the prior probability of choosing any die, which is equal, and compute the probability by dividing the product of the probability for the fair die and its prior probability by the sum of both scenarios' probabilities.
We are asked to determine the probability that the die chosen is the fair die after rolling it six times to obtain the values 4, 3, 6, 6, 5, 5. This problem serves as an application of Bayes' theorem and involves calculating the likelihood of obtaining the observed sequence of rolls under two hypotheses: the die is fair, and the die is biased.
For the fair die, the probability of rolling 4, 3, 6, 6, 5, or 5 is
(1/6) for each, as outcomes are equally likely. Therefore, the probability of rolling 4, 3, 6, 6, 5, 5 with the fair die is (1/6)^6.
Conversely, for the biased die, the probabilities are different. It is 0.15 for rolling 4 or 3 and 0.25 for rolling a 6, while it remains 0.15 for a 5. Thus, the probability of rolling 4, 3, 6, 6, 5, 5 with the biased die is (0.15)^2 * (0.25)^2 * (0.15)^2.
By applying Bayes' theorem, we can update our beliefs about whether the fair die was chosen based on the evidence provided by the rolls. The calculation requires the prior probabilities of choosing each die (which is 1/2 since the die is chosen at random) and the likelihood of the observed sequence under each die. The final probability is a fraction with the numerator being the product of the likelihood of the sequence for the fair die and the prior probability of the fair die, and the denominator is the sum of the probabilities for both scenarios (fair and biased dice).
The exact arithmetic is left as an exercise, as the question was not looking for the numerical solution but rather the approach to finding the answer.
"How much do students pay, on average, for textbooks during the first semester in college? From a random sample of 400 students the mean cost was found to be $357.75 ,and the sample standard deviation was $37.89. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean. Compute the confidence interval and describe in words what it means."
Answer: The margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.
Step-by-step explanation:
Since we have given that
Sample size = 400
Mean = $357.75
Standard deviation = $37.89
At 95% confidence level, z = 1.96
We first find the margin of error.
Margin of error is given by
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{37.89}{\sqrt{400}}\\\\=3.71[/tex]
95% confidence interval would be
[tex]\bar{x}\pm \text{margin of error}\\\\=357.75\pm 3.71\\\\=(357.75-3.71,357.75+3.71)\\\\=(354.04,361.46)[/tex]
Hence, the margin of error = 3.71, confidence interval = (354.04, 361.46) and it means that mean cost is lies within the confidence interval.
Final answer:
The margin of error for a 95% confidence interval of the population mean cost of textbooks is approximately $3.71, resulting in a confidence interval of ($354.04, $361.46). This indicates a 95% level of confidence that the true average cost of textbooks lies within this range.
Explanation:
To calculate the margin of error for the 95% confidence interval for the population mean, we use the formula for margin of error (ME): ME = z * (s/√n), where z is the z-score corresponding to the confidence level, s is the sample standard deviation, and n is the sample size. Given that the student sample size is 400, the sample mean cost is $357.75, and the sample standard deviation is $37.89, we first find the z-score for a 95% confidence level, which is approximately 1.96. We then compute the margin of error.
Margin of Error calculation:
ME = 1.96 * ($37.89/√400) = $1.96 * ($37.89/20) = $1.96 * 1.8945 ≈ $3.71
Now that we have the margin of error, we can calculate the 95% confidence interval for the population mean, which is the range where we expect the true population mean to lie. The confidence interval is: ($357.75 - ME, $357.75 + ME), or ($357.75 - $3.71, $357.75 + $3.71), giving us ($354.04, $361.46). This means that we are 95% confident that the true average cost of textbooks during the first semester in college is between $354.04 and $361.46.
We are 95% confident that the population proportion in 2000 that supported preferential hiring of women is about 0.0509 and the population proportion in 2010 is 0.04723. We are 95% confidence that the population proportion in 2000 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2010. We are 95% confidence that the population proportion in 2010 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2000. We are 95% confident that the population proportion in 2010 that supported preferential hiring of women is about 0.0509 and the population proportion in 2000 is 0.04723.
Answer:
What do Americans think about preferential hiring of women? Has there been a change in the past decade? In 2000(group 1) and 2010(Group 2), the General Social Survey asked participants if they would favor or oppose preferential hiring of women. In 2000, out of 849 respondents, 271 said yes. In 2010, out of 696 respondents, 225 said yes. The 95% confidence interval for p1-p2 is (-0.0509, 0.04273). What would be an appropriate conclusion? O We are 95% confident that the population proportion in 2010 that supported preferential hiring of women is about 0.0509 and the population proportion in 2000 is 0.04723. We are 95% confident that the pulation proportion in 2000 that supported preferential hiring of women is about 0.0509 and the population proportion in 2010 is 0.04723. We are 95% confidence that the population proportion in 2010 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2000. O We are 95% confidence that the population proportion in 2000 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2010.
Step-by-step explanation:
A survey was conducted that asked 997 people how many books they had read in the past year. Results indicated that x overbar equals 13.2 books and sequels 18.9 books. Construct a 95 % confidence interval for the mean number of books people read. Interpret the interval.
Answer:
The 95% confidence interval would be given by (12.03;14.37)
Step-by-step explanation:
1) 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=13.2[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=18.9 represent the sample standard deviation
n=997 represent the sample size
2) Calculate the confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=997-1=996[/tex]
Since the confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,996)".And we see that [tex]t_{\alpha/2}=1.96[/tex] and this value is exactly the same for the normal standard distribution and makes sense since the sample size is large enough to approximate the t distribution with the normal standard distribution.
Now we have everything in order to replace into formula (1):
[tex]13.2-1.96\frac{18.9}{\sqrt{997}}=12.03[/tex]
[tex]13.2+1.96\frac{18.9}{\sqrt{997}}=14.37[/tex]
So on this case the 95% confidence interval would be given by (12.03;14.37)
Let A, B, C be events such that P(A) = 0.2 , P(B) = 0.3, P(C) = 0.4
Find the probability that at least one of the events A and B occurs if
(a) A and B are mutually exclusive;
(b) A and B are independent.
Find the probability that all of the events A, B, C occur if
(a) A, B, C are independent;
(b) A, B, C are mutually exclusive.
Answer:
At least one of the events A and B occurs:
(a) 0.5
(b) 0.44
All of the events A, B, C occur:
(a) 0.024
(b) 0
Step-by-step explanation:
Given:
P(A) = 0.2, P(B) = 0.3, P(C) = 0.4
At least one of the events A and B occurs:
(a) If A and B are mutually exclusive events, then their intersection is 0.
Probability of at least one of them occurring means either of the two occurs or the union of the two events. Therefore,
[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=0.2+0.3-0=0.5[/tex]
(b) If A and B are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,
[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=P(A)+P(B)+P(A)P(B)\\P(A\ or\ B)=0.2+0.3-(0.2)(0.3)=0.5-0.06=0.44[/tex]
All of the events A, B, C occur together:
(a) All of the events occurring together means the intersection of all the events.
If A, B, and C are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,
[tex]P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)\\P(A\cap B\cap C)=0.2\times 0.3\times 0.4=0.024[/tex]
(b) If A, B, and C are mutually exclusive events means events A, B, and C can't happen at the same time. Therefore, their intersection is 0.
[tex]P(A\cap B\cap\ C)=0[/tex]
Probability that at least one of the events A and B occurs:
(a) When A and B are mutually exclusive: 0.5 (b) When A and B are independent: 0.44
Probability that all of the events A, B, and C occur:
(a) When A, B, and C are independent: 0.024 (b) When A, B, and C are mutually exclusive: 0
To solve these probability questions, let's understand the given information and the rules of probability related to mutually exclusive and independent events.
Let P(A) = 0.2, P(B) = 0.3, and P(C) = 0.4.
1. Finding the probability that at least one of the events A and B occurs:
(a) When A and B are mutually exclusive:
Mutually exclusive events mean that A and B cannot occur at the same time. Therefore, the probability that at least one of them occurs is given by:
P(A OR B) = P(A) + P(B)
So, P(A OR B) = 0.2 + 0.3 = 0.5
(b) When A and B are independent:
Independent events mean that the occurrence of one event does not affect the occurrence of the other. The probability that at least one of them occurs is given by:
P(A OR B) = P(A) + P(B) - P(A AND B)
For independent events, P(A AND B) is calculated as: P(A AND B) = P(A) × P(B)
So, P(A AND B) = 0.2 × 0.3 = 0.06
Therefore, P(A OR B) = 0.2 + 0.3 - 0.06 = 0.44
2. Finding the probability that all of the events A, B, and C occur:
(a) When A, B, and C are independent:
For independent events, the probability that all of them occur is given by:
P(A AND B AND C) = P(A) × P(B) × P(C)
So, P(A AND B AND C) = 0.2 × 0.3 × 0.4 = 0.024
(b) When A, B, and C are mutually exclusive:
If events are mutually exclusive, it means they cannot all occur together. Therefore, the probability that all of them occur is:
P(A AND B AND C) = 0
The brightness of a picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in ¯x = 317.2 and s = 15.7 measured in microamps. (a) Find a 99% CI on mean current required.
Answer:
The 99% confidence interval for the mean would be (301.064;333.336) mA
Step-by-step explanation:
1) Notation and some definitions
n=10 sample selected
[tex]\bar x=317.2mA[/tex] sample mean for the sample tubes selected
[tex]s=15.7mA[/tex] sample deviation for the sample selected
Confidence = 99% or 0.99
[tex]\alpha=1-0.99=0.01[/tex] significance level
A confidence interval for the mean is used to "places boundaries around an estimated [tex]\bar X[/tex] so that the true population mean [tex]\mu[/tex] would be expected to lie within those boundaries with a confidence specified. If the uncertainty is large, then the interval between the boundaries must be wide; if the uncertainty is small, then the interval can be narrow"
2) Formula to use
For this case the sample size is <30 and the population standard deviation [tex]\sigma[/tex] is not known, so for this case we can use the t distributon to calculate the critical value. The first step would be calculate alpha
[tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], then we can calculate the degrees of freedom given by:
[tex]df=n-1=10-1=9[/tex]
Now we can calculate the critical value [tex]t_{\alpha/2}=3.25[/tex]
And then we can calculate the confidence interval with the following formula
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
3) Calculate the interval
Using the formula (1) and replacing the values that we got we have:
[tex]317.2 - 3.25\frac{15.7}{\sqrt{10}}=301.064[/tex]
[tex]317.2 + 3.25\frac{15.7}{\sqrt{10}}=333.336[/tex]
So then the 99% confidence interval for the mean would be (301.064;333.336)mA
Interpretation: A point estimate for the true mean of brightness level for the tubes in the population is 317.2mA, and we are 99% confident that the true mean is between 301.064 mA and 333.336 mA.
An ANOVA procedure is used for data obtained from five populations. Five samples, each comprised of 20 observations, were taken from the five populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are Select one:
A. 3 and 30
B. 4 and 30
C. 3 and 119
D. 3 and 116
E. None of the above answers is correct
Answer:
E. None of the above answers is correct
Step-by-step explanation:
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have [tex]5[/tex] groups and on each group from [tex]j=1,\dots,20[/tex] we have [tex]20[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
The degrees of freedom for the numerator on this case is given by [tex]df_{num}=df_{within}=k-1=5-1=4[/tex] where k =5 represent the number of groups.
The degrees of freedom for the denominator on this case is given by [tex]df_{den}=df_{between}=N-K=5*20-5=95[/tex].
And the total degrees of freedom would be [tex]df=N-1=5*20 -1 =99[/tex]
On this case the correct answer would be 4 for the numerator and 95 for the denominator.
E. None of the above answers is correct
solve for −5(x+1)=−3(2x−2)
Answer:
x=11
Step-by-step explanation:
According to a survey of American households, the probability that the residents own two cars if annual household income is over $50,000 is 80%. Of the households surveyed, 60% had incomes over $50,000 and 70% had two cars. The probability that the residents of a household do not own two cars and have an income over $50,000 a year is __________?
Answer:
12%
Step-by-step explanation:
The probability that the residents of a household do not own two cars and have an income over $50,000 a year is 100% subtracted by the probability of owning two cars and having an income over $50,000(P(2∩50+) and the probability of not having an income over $50,000 (P(50-))
[tex]P = 1 - P(2\cap 50+) - P(50-)\\P = 1 - (0.6*0.7) - (1-0.6)\\P=1-0.48-0.40\\P=0.12[/tex]
Therefore, the probability that the residents of a household do not own two cars and have an income over $50,000 a year is 12%
Final answer:
To find the probability that a household has an income over $50,000 and does not own two cars, multiply the probability of a household not owning two cars (20%) by the probability of having an income over $50,000 (60%), which equals 12%.
Explanation:
The question is asking us to find the probability that the residents of a household do not own two cars and have an income over $50,000 a year. We know that the probability a household with income over $50,000 owns two cars is 80%, so the probability that such a household does not own two cars is 1 - 0.80 = 0.20 or 20%. Since 60% of households have an income over $50,000, we can calculate the probability that a household has an income over $50,000 and does not own two cars by multiplying these two probabilities: 0.60 * 0.20 = 0.12 or 12%.
Which is an equation in point slope form for the line that passes through the points ( - 1 , 4 ) and ( 3 , - 4 )
Answer:
[tex]y = - 2x + 2[/tex]
Step-by-step explanation:
Explained all in my picture
Answer:
-2
Step-by-step explanation:
y2-y1/x2-x1 = -4-4/3-(-1) = -8/3+1 = -8/4 = -2/1 = -2
Write an expression for
"19 more than a number y."
Answer:
y+19
Step-by-step explanation:
simply because when it says "more than" or "less than" you have to flip it so it wouldnt be 19+y.
For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subintervals. Then take a limit of these sums as n right arrow infinityn → [infinity] to calculate the area under the curve over [a comma b ][a,b]. f (x )equals x squared plus 3f(x)=x2+3 over the interval [0 comma 4 ]
Answer:
See below
Step-by-step explanation:
We start by dividing the interval [0,4] into n sub-intervals of length 4/n
[tex][0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4][/tex]
Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.
Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)
[tex]\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12[/tex]
but
[tex]1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}[/tex]
so the upper sum equals
[tex]\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12[/tex]
When [tex]n\rightarrow \infty[/tex] both [tex]\displaystyle\frac{3}{n}[/tex] and [tex]\displaystyle\frac{1}{n^2}[/tex] tend to zero and the upper sum tends to
[tex]\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}[/tex]
In a Gallup telephone survey conducted on April 9-10, 2013, the person being interviewed was asked if they would vote for a law in their state that would increase the gas tax up to 20 cents a gallon, with the new gas tax money going to improve roads and bridges and build more mass transportation in their state. Possible responses were vote for, vote against, and no opinion. Two hundred ninety five respondents said they would vote for the law, 672 said they would vote against the law, and 51 said they had no opinion.
a. Do the responses for this question provide categorical or quantitative data?
b. What was the sample size for this Gallup poll?
c. What percentage of respondents would vote for a law increasing the gas tax?
d. Do the results indicate general support for or against increasing the gas tax to improve roads and bridges and build more mass transportation?
Answer:
a. categorical
b. 1018
c. 28.98%
d. Most voters interviewed are against the new tax
Step-by-step explanation:
Hello!
There was a poll made to know the voter opinion on a law that would increase the gas tax to 20 cents per gallon, this money is going to use for road and bridges improvement and build more mass transportation in the state.
Data:
295 support
672 against
51 no opinion
a. The study variable is "Opinion of the votes on the new gas tax" Categorized: "support; against; no opinion"
This is a categorical variable, that can take one number on a limited amount of possibles values, assigning each observation to a group of nominal categories based on a qualitative feature.
In this case, each voter is assigned to a category according to their opinion on the new tax.
b. To know the sample size you have to add the subtotals of each category:
n= 295 + 672 + 51 = 1018
c. To calculate the percentage of people supporting the new tax, you have to divide the subtotal by the total and multiply it by 100
[tex](\frac{295}{1018})*100[/tex] = 28.978% ≅ 28.98%
d. To answer this it's better to calculate the percentage of each category:
Support: 28.98%
Against: 66.01%
No opinion: 5.01%
Since the category "against" has the greater percentage, you can say that most voters interviewed are against the new gas tax
I hope it helps!
A cube has edges that are 3 inches in length. How many of these cubes will it take to completely fill a larger cube that has an edge of 1 foot?
64
Step-by-step explanation:
We will divide the volume of the larger cube with that if the smaller cube. However, we’ll first have to convert them to the same SI units;
12 inches = 1 foot
Therefore the volume of the larger cube;
12 * 12 * 12 = 1728 inches cubed
The volume of the smaller cube;
3 * 3 * 3 = 27 inches cubed
Divide the two;
1728/27
= 64
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An inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole lot if AT LEAST 2 of the three items are in perfect condition. If in reality 84% of the whole lot are perfect, what is the probability that the lot will be accepted?
Answer:
We use the Binomial Distribution where p =.84 then probability will be 0.931.
Step-by-step explanation:
P(X≥ 2) = P(X=2) + P (X=3)
= (³₂)(.84)^2(.16) + (³3)(.84)^3(.16)⁰
≅ 0.931
Hence 0.931 is the answer.
The probability that at least two out of three items chosen randomly are in perfect condition, given that the chance for a randomly chosen item being perfect is 84%, is 97% or 0.97 in decimal form.
Explanation:This is a question about the probability that we call binomial probability. Here we have to find the probability that at least two out of three randomly chosen items are in perfect condition while the likelihood of a random item being in perfect condition is 84%.
There are two scenarios in which the lot will be accepted: (1) exactly two of the products are perfect, and (2) all three products are perfect.
1. Two products are perfect (using binomial probability formula):
Prob(Two perfect) = 3C2 * (0.84)^2 * (1-0.84)^1 = 0.38
2. All three are perfect:
Prob(Three perfect) = 3C3 * (0.84)^3 * (1 - 0.84)^0 = 0.59
Thus, adding these probabilities gives the total probability that either two or three products are in good condition, which would result in the lot being accepted:
Total Probability = 0.38 + 0.59 = 0.97 or 97%
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A tree is planted at a point O on horizontal ground. Two points A and B on the ground are 100 feet apart. The angles of elevation of the top of the tree T from the points A and B are 45◦ and 30◦ respectively. The measure of ∠AOB is 60◦ . Find the height of the tree
Final answer:
The height of the tree is determined by the angles of elevation and the distance between points A and B on the ground. Given the 45° and 30° angles of elevation and knowing that the horizontal distance AB is 100 feet, we can infer that the tree height is also 100 feet.
Explanation:
To solve for the height of the tree, we can use the trigonometric properties of right triangles and the given angles of elevation. Since the angles of elevation from points A and B to the top of the tree T are 45° and 30° respectively, and the horizontal distance between A and B is 100 feet, we know that we have two separate right triangles with angle T being the top of the tree and O being the base of the tree. We also have an angle of 60° for ∠AOB which will help us determine the distances OA and OB.
Using the 45° angle from point A, we have a 45°-45°-90° triangle, which means the height of the tree (OT) is equal to OA, the distance from the base of the tree to point A. From point B with a 30° angle, we have a 30°-60°-90° triangle. In a 30°-60°-90° triangle, the height (OT) would be √3 times smaller than the distance OB (the longer side).
Since ∠AOB is 60° we know that triangle AOB is equilateral. Therefore, distances OA and OB are both 100 feet. Now we just need to calculate the height OT in one of the triangles. Taking the 45° triangle, we have OT = OA = 100 feet. Thus, the height of the tree T is 100 feet.
Final answer:
The height of the tree can be determined by using right triangle trigonometry involving the angles of elevation from two points on the ground and the known distance between them. Mathematically, the height of the tree is found to be approximately 28.87 feet.
Explanation:
To determine the height of the tree, we use the provided angles of elevation from points A and B and the distance between these points. The tree, points A and B, and the angles form two right triangles, one with a 45° angle of elevation and the other with a 30° angle of elevation. Since the angle of elevation from point A is 45°, we know that for a 45° right-angled triangle, the tangent is 1, which means that the opposite side (height of the tree) is equal to the adjacent side (distance from A to O).
First, we can calculate the distance from A to O using the angle of 60° between points A and O given as ∠AOB. The distance AO is half of AB because ∠AOB is an equilateral triangle's angle so AO = 50 feet. Next, using the fact that the tan 30° from point B is equal to the height of the tree divided by the distance from B to O, we can calculate the height. Since the distance from A to O is 50 feet, the distance from B to O is also 50 feet. The tangent of 30° is 1/√3, so the tree's height (H) is equal to 50 feet times tan 30°, which is approximately 28.87 feet.
After checking to ensure that the sample data follows the necessary condition for using the t-procedure, we use Minitab to get a 90% confidence interval to estimate mu. We find it to be (118.51 ounces, 124.29 ounces). What does this interval tell us?
Answer:
the 90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces) means that out of all the data 90% of the sample has the weight between the range 118.51 ounces and 124.29 ounces.
Step-by-step explanation:
Data provided in the question:
Confidence level = 90%
90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces)
Now,
The confidence level tells us that out of all the value of the sample the given percentage of confidence level lies within the calculated interval.
Here in the given question,
the 90% confidence interval to estimate mu = (118.51 ounces, 124.29 ounces) means that out of all the data 90% of the sample has the weight between the range 118.51 ounces and 124.29 ounces.
To test the claim (at 1% significance) that the proportion of voters who smoked cannabis frequently was lower among conservatives, the hypotheses were In the hypothesis test about cannabis use by conservatives and liberals, the P-value is extremely small. Which of the following errors is possible in this situation?
A. Type I only
B. Type II only
C. Type I and Type II
D. Neither Type I nor Type II
Answer:
A. TYPE 1 ERROR only
Step-by-step explanation:
In general terms:
‘a hypothesis has been rejected when it should have been accepted’. When this occurs, it is called a type I error, and,
‘a hypothesis has been accepted when it should have been rejected’.
When this occurs, it is called a type II error,
When testing a hypothesis, the largest value of probability which is acceptable for a type I error is called the level of significance of the test. The level of significance is indicated by the symbol α (alpha) and the levels commonly adopted are 0.1,0.05,0.01, 0.005 and 0.002.
A level of significance of 1%,say,0.01 means that 1 times in 100 the hypothesis has been rejected when it should have been accepted.
In significance tests, the following terminology is frequently adopted:
(i) if the level of significance is 0.01 or less, i.e. the confidence level is 9 9% or more, the results are considered to be highly significant, i.e. the results are considered likely to be correct,
(ii) if the level of significance is 0.05 or between 0.05and0.01,i.e.theconfidencelevelis95%or between 95% and 99%, the results are considered to be probably significant, i.e. the results are probably correct,
(iii) if the level of significance is greater than 0.05, i.e. the confidence level is less than 95%, the results are considered to be not significant, that is, there are doubts about the correctness of the results obtained.
February 12, 2009 marked the 200th anniversary of Charles Darwin's birth. To celebrate, Gallup, a national polling organization, surveyed 1,018 randomly selected American adults about their education level and their beliefs about the theory of evolution. In their sample, 325 of their respondents had some college education and 228 were college graduates. Among the 325 respondents with some college education, 133 said that they believed in the theory of evolution. Among the 228 respondents who were college graduates, 121 said that they believed in the theory of evolution.
We want to test, at the 10% level, if there is evidence that the proportion of college graduates that believe in evolution differs significantly from the proportion of individuals with some college education that believe in evolution. Assume that the Pooled proportion (for standard error): = 0.459.
What is the z test statistic for this hypothesis test?
To test if the proportion of college graduates who believe in evolution differs significantly from the proportion of individuals with some college education, we can use the z-test statistic.
Explanation:To test if there is evidence that the proportion of college graduates who believe in evolution differs significantly from the proportion of individuals with some college education who believe in evolution, we can use the z-test statistic. The formula for the z-test statistic is z = (p1 - p2) / sqrt(pq((1/n1) + (1/n2))), where p1 and p2 are the proportions of college graduates and individuals with some college education who believe in evolution, n1 and n2 are the respective sample sizes, and q = 1 - p. Plugging in the given values, we have z = (0.459- 133/325 - 121/228) / sqrt(0.459 * (1-0.459) * ((1/325) + (1/228))). Calculating this expression will give us the z-test statistic.
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You measure 44 textbooks' weights, and find they have a mean weight of 51 ounces. Assume the population standard deviation is 11.8 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places
Answer: (47.51, 54.49)
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= sample size .
[tex]\sigma[/tex] = population standard deviation.
[tex]\overline{x}[/tex] = sample mean
[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)
As per given , we have
[tex]\sigma=11.8\text{ ounces}[/tex]
[tex]\overline{x}=51 \text{ ounces}[/tex]
n= 44
Significance level for 95% confidence = [tex]\alpha=1-0.95=0.05[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]
Now, the 95% confidence interval for the true population mean textbook weight will be :-
[tex]51\pm (1.96)\dfrac{11.8}{\sqrt{44}}\\\\=51\pm(1.96)(1.7789)\\\\=51\pm3.486644\approx51\pm3.49\\\\=(51-3.49,\ 51+3.49)\\\\=(47.51,\ 54.49) [/tex]
Hence, the 95% confidence interval for the true population mean textbook weight. : (47.51, 54.49)
The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flash Drives randomly select a sample of 17 USB flash drives. If the mean weight of the USB flash drives is too heavy or too light, the machinery is shut down for adjustment; otherwise, the production process continues. The last sample showed a mean and standard deviation of 31.9 and 1.8 grams, respectively. Using α = 0.10, the critical t values are _______.
A. reject the null hypothesis and shut down the process.
B. reject the null hypothesis and do not shut down the process.
C. fail to reject the null hypothesis and do not shut down the process) do nothing.
D. fail to reject the null hypothesis and shut down the process.
The critical t values are reject the null hypothesis and shut down the process.
The correct option is (A).
To find the critical t values for a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex], we need to look up the t-distribution table or use statistical software.
Since we have a sample size of 17, the degrees of freedom ( df ) for the t-test will be [tex]\( n - 1 = 17 - 1 = 16 \).[/tex]
For a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom, we find the critical t value.
From the t-distribution table or using statistical software, the critical t value for [tex]\( \alpha = 0.10 \) and \( df = 16 \)[/tex] is approximately 1.337.
So, the critical t values are approximately [tex]\( \pm 1.337 \).[/tex]
Now, we compare the calculated t value for the sample mean to these critical values. If the calculated t value falls beyond these critical values, we reject the null hypothesis.
Since the problem does not provide the calculated t value, we cannot determine whether to reject the null hypothesis or not.
However, based on the information given, we can see that the mean weight of the USB flash drives in the last sample is 31.9 grams, which is heavier than the expected mean weight of 30 grams. If the calculated t value corresponds to a significantly larger value than the critical t value, we may reject the null hypothesis and conclude that the mean weight is significantly different from 30 grams, potentially leading to shutting down the process for adjustment.
Therefore, the correct answer would be:
A. reject the null hypothesis and shut down the process.
Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is: A. reject the null hypothesis and shut down the process.
To find the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and a sample size of n = 17, we can use a t-distribution table or a statistical software.
For a two-tailed test at a significance level of [tex]\( \alpha = 0.10 \)[/tex] and degrees of freedom df = n - 1 = 17 - 1 = 16, we need to find the critical t-values that correspond to the upper and lower tails of the t-distribution.
Using a t-distribution table or a statistical software, the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom are approximately [tex]\( t_{\alpha/2} = \pm 1.746 \).[/tex]
So, the critical t-values are approximately[tex]\( \pm 1.746 \).[/tex]
Now, let's compare the calculated t-value with the critical t-values:
The calculated t-value is given by:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean (31.9 grams)
- [tex]\( \mu \)[/tex] is the population mean (30 grams)
- s is the sample standard deviation (1.8 grams)
- n is the sample size (17)
Substitute the given values:
[tex]\[ t = \frac{31.9 - 30}{\frac{1.8}{\sqrt{17}}} \]\[ t \approx \frac{1.9}{\frac{1.8}{4.123}} \]\[ t \approx \frac{1.9}{0.437} \]\[ t \approx 4.35 \][/tex]
Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is:
A. reject the null hypothesis and shut down the process.
sequences: If tn = 4n−1, find t1, t2, t3 and tn+1 . Express tn+1−tn in its simplest form.
To solve the question, we find the specific terms of the sequence using the given formula and illustrate that the difference between any two consecutive terms of this sequence is consistently 4, displaying the sequence's linear progression.
Explanation:The question requires us to first find the terms t1, t2, t3, and tn+1 of a sequence described by the formula tn = 4n - 1, and then to find the difference tn+1 - tn in its simplest form. To solve this problem, we substitute the appropriate values of n into the given formula and simplify.
t1 = 4(1) - 1 = 3t2 = 4(2) - 1 = 7t3 = 4(3) - 1 = 11To find tn+1, substitute n+1 for n: tn+1 = 4(n+1) - 1 = 4n + 3To find the difference tn+1 - tn in its simplest form, we subtract tn from tn+1:
tn+1 - tn = (4n + 3) - (4n - 1) = 4
The difference between any two consecutive terms in this sequence is 4. This reveals a consistent increment pattern in the sequence, illustrating its linear nature.
A Regional College uses the SAT to admit students to the school. The university notices that a lot of students apply even though they are not eligible. Given SAT scores are normally distributed, have a population mean of 500 and standard deviation of 100, what is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584?
Answer:
0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 100
n = 12
We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(greater than 525 but 584)
Standard error due to sampling =
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{100}{\sqrt{12}}[/tex]
[tex]P(525 < x < 584) = P(\displaystyle\frac{525 - 500}{\frac{100}{\sqrt{12}}} < z < \displaystyle\frac{584-500}{\frac{100}{\sqrt{12}}}) = P(0.866 < z < 2.909)\\\\= P(z \leq 2.909) - P(z < 0.866)\\= 0.998 - 0.807 = 0.191 = 19.1\%[/tex]
[tex]P(525 < x < 584) = 19.1\%[/tex]
0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.
During 2012, global oil consumption grew by 0.7%, to reach 88 million barrels per day. Assume that it continues to increase at this rate, (a) Write the first four terms of the sequence a_n giving daily oil consumption n years after 2011; give a formula for the general term a_n. Round your answers for a_1, a_2, a_3, and a_4 to three decimal places.
a_1 =_____________.
a_2 = _____________.
a-3 = _____________.
a_4 = _____________.
a_n = _____________.
(b) In what year is consumption first expected to exceed 195 million barrels a day? Consumption exceeds 195 million barrels per day during the year
Answer:
(a)
[tex]a_{1} = 88.000 \ million\\a_{2} = 94.160 \ million\\a_{3} = 100.751 \ million\\a_{4} = 107.804 \ million\\a_{n} = 82.243*1.07^n\\[/tex]
(b) 2024
Step-by-step explanation:
Global oil consumption in 2011 is given by:
[tex]C_{2011} =\frac{88}{1.07}=82.243 \ million[/tex]
(a) Assuming a constant growth of 0.7% per year, the formula for the daily oil consumption n years after 2011 is:
[tex]a_{n} = a_{0}*1.07^n\\a_{0} = C_{2011} = 82.243\\a_{n} = 82.243*1.07^n[/tex]
The terms a_1, a_2,a_3 and a_4, corresponding to the global oil consumption in the years of 2012, 2013, 2014 and 2015, respectively, are given by:
[tex]a_{1} = 82.243*1.07^1\\a_{1} = 88.000 \ million\\a_{2} = 82.243*1.07^2\\a_{2} = 94.160 \ million\\a_{3} = 82.243*1.07^3\\a_{3} = 100.751 \ million\\a_{4} = 82.243*1.07^4\\a_{4} = 107.804 \ million\\[/tex]
(b) To find the in year in which consumption reaches 195 million barrels a day, apply logarithmic properties:
[tex]a_{n} = 82.243*1.07^n\\ln(a_{n}) = ln(82.243)+ n*ln(1.07)\\\\n=\frac{ln(195)-ln(82.243)}{ln(1.07)} \\n= 12.760[/tex]
Consumption will reach 195 million barrels, 12.7 years after 2011, round it to the next whole year to find when consumption exceeds 195 million:
[tex]Y = 2011+13 = 2024[/tex]
For the solid S described, do the following:
(a) Sketch the base of S in the xy-plane.
(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.
(c) Compute the volume of S.
1. The base of S is the region lying above the parabola y = x 2 and below the line y = 1 over the interval 0 ≤ x ≤ 1. Cross-sections perpendicular to the x-axis are square
Answer:
(a) and (b) see pictures attached
(c) V = 16/35
Step-by-step explanation:
(a) Sketch the base of S in the xy-plane.
See picture 1 attached
(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.
See picture 2 attached
(c) Compute the volume of S.
The volume is given by the triple integral
[tex]\displaystyle\iiint_{S}zdzdydx[/tex]
The cross-sections perpendicular to the x-axis are squares so
[tex]z=1-x^2[/tex]
The region S is given by the following inequalities
[tex]0\leq x\leq 1\\\\x^2\leq y\leq 1\\\\0\leq z\leq 1-x^2[/tex]
Therefore
[tex]\displaystyle\iiint_{S}zdzdydx=\displaystyle\int_{0}^{1} \displaystyle\int_{x^2}^{1} \displaystyle\int_{0}^{1-x^2} (1-x^2)dzdydx=\\\\\displaystyle\int_{0}^{1}(1-x^2)(1-x^2)(1-x^2)dx=\displaystyle\int_{0}^{1}(1-x^2)^3dx=\displaystyle\frac{16}{35}[/tex]
So the volume V of the solid S is
V=16/35
A one-tailed test is a a. hypothesis test in which rejection region is in one tail of the sampling distribution b. hypothesis test in which rejection region is in both tails of the sampling distribution c. hypothesis test in which rejection region is only in the lower tail of the sampling distribution d. hypothesis test in which rejection region is only in the upper tail of the sampling distribution
Answer:
Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution
Step-by-step explanation:
One Tailed Test:
A one tailed test is a test that have hypothesis of the form[tex]H_0: \bar{x} = \mu\\H_A: \bar{x} < \mu\text{ or } \bar{x} > \mu[/tex]
A one-tailed test is a hypothesis test that help us to test whether the sample mean would be higher or lower than the population mean.Rejection region is the area for which the null hypothesis is rejected.If we perform right tailed hypothesis that is the upper tail hypothesis then the rejection region lies in the right tail after the critical value.If we perform left tailed hypothesis that is the lower tail hypothesis then the rejection region lies in the left tail after the critical value.Thus, for one tailed test,
Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution
Final answer:
A one-tailed test is a hypothesis test where the rejection region is in one tail of the sampling distribution, and is used when the research hypothesis is directional, testing the possibility of a relationship in one direction only. (Option a)
Explanation:
A one-tailed test is a hypothesis test in which the rejection region is in one tail of the sampling distribution. This means that the correct answer to the student's question is a. hypothesis test in which rejection region is in one tail of the sampling distribution. In a one-tailed test, you are testing for the possibility of the relationship in one direction and completely disregarding the possibility of a relationship in the other direction (which would be the criterion of a two-tailed test).
Depending on the research hypothesis, the rejection region could be in the lower tail (option c) or the upper tail (option d) of the probability distribution. For example, if the alternative hypothesis is HA: X > μ, we are performing a right-tailed test and rejects the null hypothesis if the test statistic falls in the upper tail (right side) of the distribution. If the alternative hypothesis is HA: X < μ, it is a left-tailed test and the rejection region would be in the lower tail (left side).
Find the quadratic function f (x )equals ax squared plus bx plus c for which f (1 )equals negative 2, f (negative 3 )equals 26, and f (3 )equals 32.
Answer:
f(x) =4x² + x - 7
Step-by-step explanation:
f(x) = ax² + bx + c
f(1) = a + b + c
f(-3) = 9a - 3b + c
f(3) = 9a + 3b + c
a + b + c = -2 -----(1)
c = -2-a-b -----(2)
9a - 3b + c = 26 ------(3)
9a + 3b + c = 32 -------(4)
(2)->(3)
9a-3b-2-a-b = 26
8a-4b = 28
4(2a-b) = 28
2a-b = 7 ----(5)
(2)->(4)
9a+3b-2-a-b = 32
8a+2b = 34
2(4a+b) = 34
4a+b = 17 ----(6)
(5)+(6)
6a = 24
a = 4
sub a=4 into(6)
4(4)+b = 17
b = 1
sub a=4,b=1 into (2)
c = -2-4-1
c = -7