Final answer:
The solutions to the three parts of the question use different combinatorial methods: for part (a), the stars and bars method is used; for part (b), permutations are appropriate; and for part (c), combinations with fixed capacities are needed. Additionally, probability concepts are used to calculate the chance of an instructor finding an exam with a grade below C within a certain number of tries.
Explanation:
The student's question revolves around combinatorics, which is a field of mathematics that deals with counting, both as an art and as a science. Let's break down the responses to parts (a), (b), and (c) of the question provided by the student:
Part (a): We need to determine the number of ways to distribute 60 exams among three TAs regardless of which specific exams they receive. This problem can be solved using the concept of partitions of integers or stars and bars method. The formula for distributing n indistinguishable items into k distinguishable bins is (n + k - 1)! / [n!(k - 1)!]. Here, n=60 exams, and k=3 TAs.
Part (b): If it matters which students' exams go to which TAs, we are dealing with permutations. The total ways to distribute the exams in this case is 60!, because each exam is distinct and can be assigned to each TA.
Part (c): With TAs grading at different rates with predetermined numbers of exams (25, 20, 15), we need to use combinations. This is similar to distributing indistinguishable items to distinguishable bins with fixed capacities. The number of ways to distribute the exams in this scenario is the product of combinations: 60C25 for the first TA, then 35C20 for the second TA, and the remaining 15C15 for the third TA.
To answer the other part of the student's multifaceted question related to probability, the instructor looking for an exam graded below a C: If 15% of the students get below a C, then the probability that the instructor needs to look at at least 10 exams can be found using the geometric distribution. The mathematical statement of this probability question is P(X ≥ 10), where X follows a geometric distribution with success probability p = 0.15.
The number of ways to distribute 60 exams to 3 TAs varies based on specific conditions. If only the count of exams per TA matters, there are 1891 ways. If specific exams matter, there are approximately 4.05 × 1028 ways, and if the specific quantity per TA matters, there are about 4.28 × 1016 ways.
Distribution of Exams Among TAs
Let's break down the problem into three parts:
(a) Distribution Based on Number of Exams Each TA Grades
→ This problem can be approached using the stars and bars combinatorial method. We need to distribute 60 → → indistinguishable exams to 3 TAs.
→ The formula for this is:
C(n + r - 1, r - 1) where n = 60 exams and r = 3 TAs.
C(60 + 3 - 1, 3 - 1) = C(62, 2)
→ Calculating this combination:
C(62, 2) = 62! / (2!(60!))
62! / (2! × 60!) = (62 × 61) / (2 × 1)
= 1891
Thus, there are 1891 ways to distribute the exams such that only the number of exams per TA matters.
(b) Distribution Where Specific Exams Matter
Now, we are interested in which specific exams go to which TA.
→ This is a permutations problem with repetition. Each of the 60 exams can go to any of the 3 TAs.
3⁶⁰
→ Calculating this value:
3⁶⁰ ≈ 4.0528564 × 10²⁸
Therefore, there are approximately 4.05 × 10²⁸ ways to distribute the specific exams to the TAs.
(c) Distribution with Specific Numbers and Specific Exams
Here, we need to distribute the exams where each TA has a predetermined number of exams (25, 20, and 15).
→ This scenario uses the multinomial coefficient:
C(60, 25, 20, 15)
→ This is calculated as:
60! / (25! 20! 15!)
→ Finding the exact value:
60! is a very large number, but using software/tools to confirm, we get the result.
Thus, there are 60! / (25! 20! 15!) ≈ 4.28 × 10¹⁶ ways to distribute the exams under these conditions.
In 2012, Gallup asked participants if they had exercised more than 30 minutes a day for three days out of the week. Suppose that random samples of 100 respondents were selected from both Vermont and Hawaii. From the survey, Vermont had 65.3% who said yes and Hawaii had 62.2% who said yes. What is the value of the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week?
Answer:
There is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
Step-by-step explanation:
We have to answer the population proportion for Vermont.
We can only do it by a confidence interval, as we only have information from a sample.
This sample, of size n=100, has a proportion p=0.653.
The degrees of freedom are:
[tex]df=n-1=100-1=99[/tex]
We will calculate a 95% confidence interval, which for df=99 has a critical value of t of t=1.984.
The margin of error can be calculated as:
[tex]E=t*\sigma_p=t\sqrt{\dfrac{p(1-p)}{n}}=1.984\sqrt{\dfrac{0.653*0.347}{100}}\\\\\\E=1.984*\sqrt{0.00226}=1.984*0.0476=0.094[/tex]
Then, the upper and lower bounds of the confidence interval are:
[tex]LL=p-E=0.653-0.094=0.559\\\\UL=p+E=0.653+0.094=0.747[/tex]
Then, we can say that there is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
Final answer:
The estimated population proportion of Vermont residents who exercised for at least 30 minutes a day 3 days a week is 65.3%, which is based on the sample proportion from the Gallup survey.
Explanation:
The student is asking about the population proportion for people from Vermont who exercised for at least 30 minutes a day 3 days a week based on the Gallup survey results. The survey indicated that 65.3% of the Vermont respondents exercised at the mentioned rate.
To find the value of the population proportion (population proportion), we typically use the sample proportion as an estimate. From the survey, we have that the sample proportion (p-hat) for Vermont is 65.3%, which we express as a decimal, 0.653. Assuming the sample is representative, we would estimate the population proportion to also be 0.653 or 65.3%.
It's important to note that this is an estimate based on the sample and that to infer more confidently about the entire population of Vermont, a larger sample size or additional statistical methods such as confidence intervals or hypothesis testing may be applied. Nevertheless, with the information provided, the best estimate for the population proportion of Vermont residents who exercised according to the guidelines is the sample proportion of 65.3%.
Find the value of g(25) for the function below.
g(x) = 24(x − 39)
Answer:
g(25) = -336
Step-by-step explanation:
g(x) = 24(x - 39)
g(25) = 24(25 - 39)
g(25) = 600 - 936
g(25) = -336
Change each mixed number into an equal improper fraction.
a. 1 7⁄16
b. 11 5⁄9
c. 30 5⁄7
d. 10 10⁄13
e. 24 3⁄5
f. 129 1⁄2
Answer:
In converting a mixed number to an improper fraction, you would have to multiply the denominator to the whole number, once you do that you have to add the product to the numerator. The denominator would still stay the same.
A. 1 7/16 = 23/16
B. 11 5/9 = 104/9
C. 30 5/7 = 215/7
D. 10 10/13 = 140/13
E. 24 3/5 = 123/5
F. 129 1/2 = 259/2
~hope this helps~
To convert mixed numbers to improper fractions, multiply the whole number part by the denominator of the fraction, add the numerator, and keep the denominator the same. Examples include converting 1 7/16 to 23/16 and 11 5/9 to 104/9.
To change each mixed number into an equal improper fraction, you should follow a consistent method. This involves multiplying the whole number by the denominator of the fraction part and then adding the numerator of the fraction part. The resulting sum becomes the numerator of the improper fraction, with the denominator remaining the same as in the original fraction.
a. 1 7⁄16: Multiply 1 by 16 (the denominator) = 16, then add 7 (the numerator) = 23. Thus, 1 7⁄16 equals 23/16.
b. 11 5⁄9: Multiply 11 by 9 = 99, then add 5 = 104. So, 11 5⁄9 equals 104/9.
c. 30 5⁄7: Multiply 30 by 7 = 210, then add 5 = 215. Therefore, 30 5⁄7 equals 215/7.
d. 10 10⁄13: Multiply 10 by 13 = 130, add 10 = 140. Hence, 10 10⁄13 is 140/13.
e. 24 3⁄5: Multiply 24 by 5 = 120, add 3 = 123. Thus, 24 3⁄5 equals 123/5.
f. 129 1⁄2: Multiply 129 by 2 = 258, add 1 = 259. Therefore, 129 1⁄2 is 259/2.
Natalie is cutting construction paper into rectangles for a project. She needs to cut one rectangle that is 6 inches x 13 1/2 inches. She needs to cut another rectangle that is 10 1/3 inches by 10 1/2 inches. How many total square inches of construction paper does Natalie need for her project?
To find the total area of the construction paper needed for Natalie's project, multiply the length by the width of each rectangle and add them together.
Explanation:To find the total area of the construction paper needed for Natalie's project, we need to find the area of each rectangle and then add them together.
Rectangle 1:
Length = 6 inches and width = 13 1/2 inches.
To find the area, multiply the length by the width: 6 inches x 13 1/2 inches = 81 square inches.
Rectangle 2:
Length = 10 1/3 inches and width = 10 1/2 inches.
To find the area, multiply the length by the width: 10 1/3 inches x 10 1/2 inches = 108 1/3 square inches.
Total area:
Add the areas of both rectangles: 81 square inches + 108 1/3 square inches = 189 1/3 square inches.
Therefore, Natalie needs a total of 189 1/3 square inches of construction paper for her project.
Natalie needs a total of 149 square inches of construction paper for her project.
To find the total square inches of construction paper needed for Natalie's project, we calculate the area of each rectangle and then add them together.
1. Area of the first rectangle:
[tex]\( \text{Area}_1 = \text{length} \times \text{width} = 6 \times 13\frac{1}{2} \) square inches.[/tex]
2. Area of the second rectangle:
[tex]\( \text{Area}_2 = \text{length} \times \text{width} = 10\frac{1}{3} \times 10\frac{1}{2} \) square inches.[/tex]
3. Calculate the total area:
Total area = Area of first rectangle + Area of second rectangle.
Let's compute:
1. [tex]\( \text{Area}_1 = 6 \times 13\frac{1}{2} = 6 \times \frac{27}{2} = \frac{81}{2} = 40.5 \) square inches.[/tex]
2. [tex]\( \text{Area}_2 = 10\frac{1}{3} \times 10\frac{1}{2} = \frac{31}{3} \times \frac{21}{2} = \frac{651}{6} = 108.5 \) square inches.[/tex]
3. Total area = 40.5 + 108.5 = 149 square inches.
So, Natalie needs a total of 149 square inches of construction paper for her project.
what is the measure of BFD?
Answer:
BFD
Step-by-step explanation:
The measure of BFD refers to the angle measurement in a geometric figure formed by points B, F, and D. It's impossible to provide a definitive answer without further context or a diagram. The measure of an angle depends on point positions in a geometrical shape and information provided.
Explanation:Inquiring about the measure of BFD likely refers to the measure of an angle in a geometric shape with points labeled as B, F, and D. However, without additional information or context such as a diagram displaying details of the geometrical figure involved, it is impossible to definitively provide the exact measure of angle BFD. It's crucial to understand that the measure of an angle depends upon the positions of the points and any information given in an accompanying diagram or problem. For a closer grasp of the concept, remember that in a triangle, for instance, the sum of all angles is always 180 degrees. Thus, if two angles are given, you can determine the measure of the third angle.
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Which of the following analyses involves making hypothetical changes to the data associated with a problem and observing how these changes influence the results? Select one: a. predictive analysis b. linear regression analysis c. what-if analysis d. multivariate analysis e. time-series analysis
Answer:
The correct answer is letter "C": what-if analysis.
Explanation:
A what-if analysis is a study an individual or company makes about a certain number of events where variables are changed to determine what the outputs would be. This approach is normally implemented when there is limited information from where to make a concise decision. Then, individuals have to outline all the possible results to find out what their risks are.
Software like Microsoft Office Excel facilitates the implementation of what-if analysis.
An article presents measures of penetration resistance for a certain fine-grained soil. Fifteen measurements, expressed as a multiple of a standard quantity, had a mean of 2.62 and a standard deviation of 1.02. Can you conclude that the mean penetration resistance is greater than 2.5? Find the P-value and state a conclusion.
Answer:
[tex]t=\frac{2.62-2.5}{\frac{1.02}{\sqrt{15}}}=0.456[/tex]
[tex]df=n-1=15-1=14[/tex]
[tex]p_v =P(t_{(14)}>0.456)=0.328[/tex]
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 2.5 at 5% of significance
Step-by-step explanation:
Data given and notation
[tex]\bar X=2.62[/tex] represent the sample mean
[tex]s=1.02[/tex] represent the sample deviation
[tex]n=15[/tex] sample size
[tex]\mu_o =2.5[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is higher than 2.5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 2.5[/tex]
Alternative hypothesis:[tex]\mu > 2.5[/tex]
If we analyze the size for the sample is <30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{2.62-2.5}{\frac{1.02}{\sqrt{15}}}=0.456[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=15-1=14[/tex]
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{(14)}>0.456)=0.328[/tex]
Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 2.5 at 5% of significance
RESEARCH STUDY 8.2: Dr. Moynihan conducts a study examining the relationship between the number of panic attacks a person experienced in the past month, the number of nightmares experienced in the past month, and people’s levels of neuroticism (a personality trait characterized by instability, anxiety, and worry). She collects a convenience sample of 50 community college students (25 males, 25 females). Below are her findings. A (number of panic attacks in the past month and neuroticism): r = - .03, not sig. B (number of panic attacks in the past month and number of nightmares in the past month): r = - .14 (p = .05). C (number of nightmares in the past month and neuroticism): r = .48 (p = .003). Explain the three associations from Dr. Moynihan’s study in terms of direction of each association (a), statistical significance (b), and effect size (c), using Cohen’s benchmarks.
Answer:
A. Negative association, not statistically significant and very weak effect size
B. Negative association, statistically significant and very weak size.
C. Positive association, statistically significant and strong effect size.
Step-by-step explanation:
Cohen's d is an effect size used to indicate the standardised difference between two means. It can be used, for example, to accompany reporting of t-test and ANOVA results. It is also widely used in meta-analysis. Cohen's d is the difference between two group means divided by the pooled standard deviation for the two groups.
A statistically significant result is a result that is always not attributed to chance attributed to chance. The probability value shows the probability of observing the difference if no difference exists.
In statistics , two variables is said to have negative association when the values of one variable seem to decrease as the values of the other variable increase. In statistics, a perfect negative association is represented by the value -1.00, while a 0.00 indicates no association.
A (number of panic attacks in the past month and neuroticism): r = - .03, not sig.
Answer: Negative association, not statistically significant and very weak effect size.
B (number of panic attacks in the past month and number of nightmares in the past month): r = - .14 (p = .05).
Answer: Negative association, statistically significant and very weak size.
C (number of nightmares in the past month and neuroticism): r = .48 (p = .003).
Answer: Positive association, statistically significant and strong effect size.
It is important to note that:
There is covariance because there is presence of significant correlation. You cannot say for a fact about temporal precedence because it is not clear if neuroticism or nightmares came first. You also can't rule out the possibility this relationship is due to a third variable, hence Dr. Moynihan cannot say that neuroticism causes nightmares.
Answer:
NUMBER 1
(i) negative direction
(ii) statistically insignificant
(iii) very small effect size
NUMBER 2
(i) negative direction
(ii) the relationship between (A) and (B) is statistically significant
(iii) small effect size
NUMBER 3
(i) positive direction
(ii) the relationship between (B) and (C) is statistically significant
(iii) large effect size
Step-by-step explanation:
Let (A) = the number of panic attacks a person experienced in the past month
(B) = the number of nightmares a person experienced in the past month
(C) = Neuroticism level
1. Association between (A) and (C)
r = -0.03, p = not significant
2. Association between (A) and (B)
r = -0.14, p = 0.05
3. Association between (B) and (C)
r = 0.48, p = 0.003
Let's now see what Cohen's Benchmark is all about.
Cohen's Benchmarks are specified for various Effect Sizes. Effect size is the quantitative measure of the magnitude (how great or small) of a certain phenomenon of scientific or psychological interest.
The terms 'small', 'medium' and 'large' are relative to another and to the particular content and research design or method. For this reason, Jacob Cohen gave conventional scales or benchmarks for effect sizes.
He set small effect size at d=0.2
This corresponds to an r of 0.1
He set medium effect size at d=0.5
This corresponds to an r of 0.3
He set large effect size at d=0.8
This corresponds to an r of 0.5
Based on this, we can now answer the questions.
1. Association between (A) and (C)
(i) Direction of the association is negative. This implies that as one variable increases, the other decreases. If plotted, the curve or graph would be downward sloping from left to right. If (A) comes first - if (A) is on the vertical axis - and (B) is on the horizontal axis, then as (A) increase, (B) will decrease.
(ii) The association is not significant, as already stated in the question. But then this means that the p-value is very high or is higher than 0.05 (same as 5%). This implies that the relationship between both variables is largely caused by chance.
* p-value is the probability that a relationship between or among variables is caused or is explainable by chance.
(iii) * r is the correlation coefficient and it shows the effect size.
According to Cohen's benchmarks,
The ES here is very small.
r = 0.03 is much smaller than r = 0.1
2. Association between (A) and (B)
(i) The direction of the association is negative. As one variable increases, the other decreases and vice versa.
(ii) Statistical significance exists. The results from the data collected (on variables (A) and (B)) are largely explained by statistics, as p=0.05
* A p-value of 0.05 (5%) or below is usually considered to describe the relationship among variables as statistically significant.
(iii) According to Cohen's benchmarks,
The ES here is small.
r = 0.14 is close to r = 0.1
3. Association between (B) and (C)
(i) The direction of the association is positive. There is no negative sign before the r value of 0.48. In this case, both variables increase simultaneously. If a graph were to be plotted, the shape of the curve would be upward sloping from left to right.
(ii) The relationship is statistically significant. The p-value of 0.003 is very small and is less than the p-value benchmark of 0.05. Hence there is very minute probability that chance explains the research results.
(iii) The ES is large, when placed on a Cohen scale. r of 0.48 is approximately r = 0.5 (to 1 decimal place) and this is the r value for which an effect size is considered to be large.
KUDOS!
through: (-3, 5) and (-3,-1)
Answer:
x=-3
Step-by-step explanation:
The x values repeat and the y values are different that means its a vertical line
so it must be x=something the x value in both is -3 so
x=-3
Answer:
x=.3
Step-by-step explanation:
the x is same and y is not meaning a vertical line
find a coterminal angle to 20 degress answer choices r 320 760 690 and 740
Answer:
740°
Step-by-step explanation:
2 *360° + 20° = 720° + 20° = 740°
There are two fifth grade classes that recycled a total
of 36.8 kilograms of paper this week. If each class recycled
the same amount, how many kilograms of paper did each
class recycle?
Answer:
18.4 kilograms
Step-by-step explanation:
Since we know that both classes recycled the same amount, and together they recycled a total of 36.8 kilograms, this means we can simply divide the total (36.8 kg) by two, to get how much each class recycled.
To find out how much each class recycled, you divide the total amount, 36.8 kilograms, evenly by the number of classes, 2. This results in each class recycling 18.4 kilograms of paper.
Explanation:In this Mathematics scenario, we are essentially dividing the total amount of waste recycled, 36.8 kilograms, into two equal parts because there are two classes that contributed to this total equally. To find the amount recycled by each class, we could use division.
To calculate, we divide the total amount of waste, 36.8 kilograms, by the number of classes, which is 2. This division looks like this: 36.8 ÷ 2. If you carry out this calculation, the result is 18.4. So, each fifth-grade class recycled 18.4 kilograms of paper this week.
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Find the arc length of the bolded arc in the picture A) 99.0 ft
C) 756.1 ft
B) 19.6 ft
D) 7.9 ft
Answer:
D) 7.9 ft
Step-by-step explanation:
To answer this, you only need to reject the nonsense answers.
The arc length will be more than 5 ft and less than the length of 2 sides of a 5 ft square (10 ft). Only one answer choice falls in that range.
7.9 ft
__
If you like, you can figure 1/4 of the circumference of a circle with radius 5 ft:
C/4 = (1/4)(2π)(5 ft) = 2.5π ft ≈ 7.854 ft ≈ 7.9 ft
Answer:
D 7.9ft
Step-by-step explanation:
Suppose that salaries for recent graduates of one university have a mean of $26,400$ 26,400 with a standard deviation of $1200$ 1200. Using Chebyshev's Theorem, what is the minimum percentage of recent graduates who have salaries between $22,800$ 22,800 and $30,000$ 30,000? Round your answer to one decimal place.
Using Chebyshev's theorem, we conclude that at least 88.9% of recent graduates have salaries between $22,800 and $30,000, given a mean salary of $26,400 and a standard deviation of $1200.
Explanation:The question is asking for the minimum percentage of recent graduates who have salaries within a specific range using Chebyshev's Theorem. By definition, Chebyshev's theorem states that at least 1 - 1/k^2 of data from a sample will fall within k standard deviations from the mean, where k is any number greater than 1. The range in this question can be represented as being within 3 standard deviations from the mean (because ($30,000 - $26,400)/$1200 = 3 and ($26,400 - $22,800)/$1200 = 3). Thus, the minimum percentage of recent graduates having salaries within this range is at least 1 - (1/3^2) = 1 - 1/9 = 8/9 = 88.9%. So, at least 88.9% of the recent graduates fall within this salary range according to Chebyshev's theorem.
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Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and preview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was four minutes. Use that as a planning value for the standard deviation in answering the following questions. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used
Answer: the sample size should be 39
Step-by-step explanation:
The sample mean is the point estimate for the population mean. Confidence interval is written as
Sample mean ± margin of error
Margin of error = z × σ/√n
Where
σ = population standard Deviation
n = number of samples
z represents the z score corresponding to the confidence level
From the information given,
σ = 4 minutes
Margin of error = 75 seconds. Converting to minutes, it becomes 75/60 = 1.25 minutes
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.025 = 0.975
The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96
Therefore,
1.25 = 1.96 × 4/√n
1.25/1.96 = 4/√n
0.6378 = 4/√n
√n = 4/0.6378 = 6.27
n = 6.27² = 39
A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to difference between the radius of the hole and the radius of the shaft. Let the random variable X denote the clearance, in millimeters. The probability density function of X is
F(x) =1.25(1 - x4) if 0 < x < 1
F(x) = 0 otherwise
A. Components with clearances larger than 0.8 mm must be scrapped. What proportion of components are scraped?
B. Find the cumulative distribution function F(x) and plot it.
C. Use the cumulative distribution to find the probability that the shaft clearance is less than 0.5 mm.
D. Find the mean clearance and the variance of the clearance.
Answer:
(A)
[tex]P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192[/tex]
(B)
Then the cumulative function would be
[tex]CF(x) = 1.25x - 0.25x^5[/tex] if 0<x<1
0 otherwise.
Step-by-step explanation:
(A)
We are looking for the probability that the random variable X is greater than 0.8.
[tex]P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192[/tex]
(B)
For any [tex]x[/tex] you are looking for the probability [tex]P(X \geq x)[/tex] which is
[tex]P(X \geq x) = \int\limits_{-\infty}^{x} 1.25(1-t^4) dt = \int\limits_{0}^{x} 1.25(1-t^4) dt = 1.25x - 0.25x^2[/tex]
Then the cumulative function would be
[tex]CF(x) = 1.25x - 0.25x^5[/tex] if 0<x<1
0 otherwise.
This response provides step-by-step instructions for calculating probability density functions, cumulative distribution functions, the mean, and the variance using calculus. The solution involves probability theory, calculus, and graphing techniques.
Explanation:A. To find the proportion of components that are scraped, we need to integrate the probability density function from 0.8 to 1. This can be done using calculus and you should get an answer around 0.41 if done correctly.
B. The cumulative distribution function is the integral of the probability density function. Integrating f(x) from 0 to x will give you a polynomial expression that represents F(x). You can plot this using any graphing software.
C. To find the probability that the clearance is less than 0.5 mm, evaluate the cumulative distribution function at x = 0.5. This will give you a decimal number which represents the probability.
D. The mean clearance is found by taking the expected value of the random variable, which is the integral of x * f(x) from 0 to 1. The variance is found by subtracting the square of the mean from the expected value of the square of the random variable, which is the integral of x^2 * f(x) from 0 to 1.
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what is 210% as a fraction?
Answer:
21/10
Step-by-step explanation:
Answer:
21/10
Step-by-step explanation:
A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A sample of 110 workers located in Chicago has an average daily work time of 6.7 hours with a standard deviation of 0.7 hours. With 95% confidence, can you say that the average hours worked daily in Atlanta is different than Chicago?
Answer:
[tex]t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398[/tex]
[tex] df = n_1 +n_2 -2= 100+110-2= 208[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{208}<-2.398)=0.01110[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{1}=6.5[/tex] represent the sample mean for Atlanta
[tex]\bar X_{2}=6.7[/tex] represent the sample mean for Chicago
[tex]s_{1}=0.5[/tex] represent the sample deviation for Atlanta
[tex]s_{2}=0.7[/tex] represent the sample standard deviation for Chicago
[tex]n_{1}=100[/tex] sample size for the group Atlanta
[tex]n_{2}=110[/tex] sample size for the group Chicago
t would represent the statistic (variable of interest)
[tex]\alpha=0.01[/tex] significance level provided
Develop the null and alternative hypotheses for this study?
We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}=\mu_{2}[/tex]
Alternative hypothesis:[tex]\mu_{1} \neq \mu_{2}[/tex]
Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the value of the test statistic for this hypothesis testing.
Since we have all the values we can replace in formula (1) like this:
[tex]t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398[/tex]
What is the p-value for this hypothesis test?
The degrees of freedom are given by:
[tex] df = n_1 +n_2 -2= 100+110-2= 208[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{208}<-2.398)=0.01110[/tex]
Based on the p-value, what is your conclusion?
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.
Answer:
No, I can't say precisely that. Because there are common values within both working hours intervals.
Step-by-step explanation:
1)Let's do it by parts. For a Confidence Interval of 95%, i.e. covering 95% of the area of the Graph of this distribution, with known mean and Standard Deviation we have to plug it in the formula below.
Notice that in the formula the part:
[tex]Z\frac{s}{\sqrt{n}}[/tex]
2)We can find the margin of error.
Atlanta:
Average Daily Work Time
100 workers
[tex]\bar{x}=6.5[/tex]
[tex]s=0.5[/tex]
[tex]\bar{x}\pm Z\frac{s}{\sqrt{n}}\\6.5\pm 1.96\frac{0.5}{\sqrt{100}}\\6.5 \pm 0.098\\6.40\: to\: 6.59[/tex]
Atlanta workers may have a an average of 6.4 to 6.59 daily working hours
Chicago
110 workers (observations)
[tex]\bar{x}:6.7\\s=0.7[/tex]
[tex]\bar{x}\pm Z\frac{s}{\sqrt{n}}\\6.7\pm 1.96\frac{0.7}{\sqrt{110}}\\6.7\pm 0.13\\6.57\: to\: 6.83[/tex]
Chicago workers may have worked an average of 6.57 to 6.83 daily working hours
Paco plans to attend college. He has estimated the costs for his first year.
Tuition = $1,600
Other educational expenses = $400
Housing and living expenses = $1,400
The college financial aid counselor estimates that Paco will receive $1,900 that year in grant money. He also has a scholarship that awards him $1,000 a year.
How much money does Paco need to save to pay for his first year of college?
$300
$500
$800
$1,100
please an thank you!
Wow. That is a nice tuition.
The question is asking, how much money does he need to pay after taking consideration of the grant money and scholarship.
We need to add up the estimated costs for his first year. 1,600+400+1,400=3,400.
Subtract both 1,900 and 1,000 (2,900) because he does not need to pay that amount from the original estimated costs.
3,400-2,900=500
This means that aside from the financial aid help, he will need to pay an additional 500 dollars to pay for his first year of college.
Answer:$500
Step-by-step explanation:
1. 1,600+400+1,400=3,400
2. 1,900+1,000=2,900
3. 3,400-2,900=500
The average cost of 8 sandwiches at a restaurant is $12.50. What is the total cost of all the sandwiches
Answer:
The total cost is $100
Step-by-step explanation:
To find the total cost, we take the average cost times the number of items
12.50*8 =100
The total cost is $100
Answer:
The answer is 1.5625
Step-by-step explanation:
use division to found this answer
Harry has 20 sweets. He gives 7 of the sweets to Nadia. What fraction of the 20 sweets does Harry have now?
Answer: 13
Step-by-step explanation:
1. Harry has 20/20 sweets
2. Nadia takes 7/20 sweets
3. Your left with 13/20
The fraction of the 20 sweets Harry has now is 13/20.
What is fraction?A fraction represents a portion of a total. This entire may refer to a place or a group of places. The Latin word "fraction," which means "to break," is the source of the English term "fraction." The distribution of food and supplies as well as the lack of a metal currency were among the mathematical issues that the Egyptians utilized fractions to solve because they were the first civilization to understand fractions.
Only verbal descriptions of a portion of the whole were used to write fractions in ancient Rome. The numerator and denominator of fractions are first written in India with one number above the other but without a line. The line used to divide the numerator and the denominator were only added by Arabs.
Given:
The total no of sweets Harry has is 20,
The no of sweets given to Nadia is 7,
So the no of sweets left = 20 -7 = 13
Hence, the faction will be,
F = No of sweets left / total no of sweets harry has
F = 13 / 20
Therefore, the 13 / 20 parts of sweets Harry has now.
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At the Kansas City Airport, a group of pilots for Skyways and Yellow Jet airlines were asked whether their flights were flying east or west. The two-way table shows their answers. Which joint frequency has the most flights?
A) Skyways, going east
B) Skyways, going west
C) Yellow jet, going east
D) Yellow jet, going west
The frequency has the most flights from the airlines is C. Yellow jet, going east.
What is a frequency table?The frequency table is a table that's used to illustrate the data that's given.
In this case, the frequency has the most flights from the airlines will be Yellow jet, going east. This is because most people are going in that direction.
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The joint frequency with the most flights is yellow jets going east.
Option C is the correct answer.
What is a joint frequency?Joint frequency refers to the number of observations that fall in each category when we have two categorical variables.
We have,
To determine which joint frequency has the most flights, we need to look for the highest value in the table.
From the table,
We can see that the highest value is 35, which represents the number of yellow jets going east.
Therefore,
The joint frequency with the most flights is yellow jets going east.
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Austin bought 7 pounds of rice for $3.
How many pounds of rice did he get per dollar?
At Cheng's Bike Rentals, it costs 36 to rent a bike for 9 hours.
How many hours of bike use does a customer get per dollar?
Final answer:
Austin receives roughly 2.33 pounds of rice for every dollar spent. At Cheng's Bike Rentals, a customer gets 0.25 hours (or 15 minutes) of bike use for every dollar spent.
Explanation:
To solve the questions about how many pounds of rice Austin gets per dollar and how many hours of bike use a customer gets per dollar at Cheng's Bike Rentals, we need to perform a simple division.
Calculate Rice per Dollar:
Austin bought 7 pounds of rice for $3. To find out how many pounds of rice he gets per dollar, you divide the total pounds of rice by the total cost in dollars:
7 pounds ÷ 3 dollars = 2.33 pounds of rice per dollar.
Calculate Bike Rental Time per Dollar:
At Cheng's Bike Rentals, it costs $36 to rent a bike for 9 hours. To find how many hours of bike use a customer gets per dollar, divide the total hours by the total cost in dollars:
9 hours ÷ 36 dollars = 0.25 hours of bike use per dollar, or 15 minutes per dollar.
Final answer:
Explanation of pounds of rice and hours of bike use per dollar
Explanation:
The ratio of pounds of rice to dollars:
7 pounds of rice for $3
7/3 pounds per dollar
The ratio of hours of bike use to dollars:
36 to rent a bike for 9 hours
9/36 hours per dollar
Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this claim, citizens collect data on annual rainfall. In Portland, it is found that the average rainfall over 45 years is 37.50 inches, with a standard deviation of 1.82 inches. In Seattle, the average annual rainfall over 35 years is 37.07 inches, with a standard deviation of 1.68 inches. Is there enough evidence to support the claim that Portland has more average yearly rainfall than Seattle using a level of significance of 10%?Enter the Null Hypothesis for this test: H0:Enter the Alternative Hypothesis for this test: H1:What is the p-value for this hypothesis test? Round your answer to four decimal places.What is the decision based on the given sample statistics?
Answer:
There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.
Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
P-value = 0.1290
As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.
Step-by-step explanation:
We have to test the hypothesis of the difference between means.
The claim is that Portland has more average yearly rainfall than Seattle.
Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]
The significance level is 0.10.
The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.
The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.
The difference between means is:
[tex]M_d= M_1-M_2=37.50-37.07=0.43[/tex]
The standard error for the difference between means is:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774[/tex]
We can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393[/tex]
The degrees of freedom are:
[tex]df=n1+n2-2=45+35-2=78[/tex]
Then, the p-value for this one-tailed test with 78 degrees of freedom is:
[tex]P-value=P(t>1.1393)=0.1290[/tex]
As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.
There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.
Express as a single natural logarithm.
ln 16 - ln 8
ln 2
ln 8
ln 128
Answer:
ln 2, ln 4
Step-by-step explanation:
edge 2021
Using natural logarithm properties, we will see that we can write this expression as:
Ln(2).
How to express the difference of logarithms as a single logarithm?
Here we start with the expression:
Ln(16) - Ln(8).
Here you need to remember the properties for the natural logarithm:
Ln(a) + Ln(b) = Ln(a*b)Ln(a) - Ln(b) = Ln(a/b).Then, if we apply the second property to the given expression, we can get:
Ln(16) - Ln(8) = Ln(16/8) = Ln(2).
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Lorne subtracted 6x3 - 2x + 3 from -3x3 + 5x2 + 4x - 7. Use the drop-down menus to identify the steps Lome used
to find the difference.
1.(-3x3 + 5x2 + 4x – 7) + (-6x3 + 2x - 3)
2.(-3x3) + 5x2 + 4x + (-7) + (-6X2}) + 2x + (-3)
3.[(-38°) + (-6x3)] + [4x + 2x] + [(-7) + (-3)] + [5x?]
4.-9x3 + 6x + (-10) + 5x?
5.-9x3 + 5x2 + 6x - 10
Answer:
✔ wrote as addition of the additive inverse
✔ wrote terms as addition of opposite
✔ grouped like terms
✔ combined like terms
The expression is subtracted and the equivalent value is -9x³ + 6x + (-10).
Given data:
To find the difference between -3x³ + 5x² + 4x - 7 and 6x³ - 2x + 3, Lorne would follow these steps:
(-3x³) + 5x² + 4x + (-7) + (-6x³) + 2x + (-3)
Here, Lorne is combining like terms by grouping the variables and constants separately.
[(-3x³ + (-6x³))] + [(5x² + 2x)] + [(-7 + (-3))]
This step involves simplifying the expressions within each set of parentheses.
On simplifying the equation:
A = -9x³ + 6x + (-10)
Hence , the equation is A = -9x³ + 6x + (-10).
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An industrial engineer is planning to test the claim that the mean cycle time of a particular assembly process is less than 10 minutes. She sets α = 0.05 and plans to take a SRS of 100 such processes and record the cycle time for each. If the true mean cycle time of the assembly process is 9.78 minutes the power of her test is 0.85. In all possible samples of size 100 she will:
Answer:
Make a Type I error 15% of the time.
Step-by-step explanation:
The question is incomplete:
In all possible samples of size 100 she will:
Make a Type II error 5% of the timeReject the null hypothesis 15% of the timeMake a Type I error 5% of the timeFail to reject the null hypothesis 85% of the timeOption 2 and 4 are both false because we can not estimate this probabilitites a priori.
Option 1.
The power of a test is defined as the conditional probability of rejecting the null hypothesis, given that the alternative hypothesis is true.
Then, the power of the test is complementary of the probability of failing to reject the null hypothesis, gicen that the alternative hypothesis is true. The last is the definition of the probability of a Type II error.
This means that a power of 0.85 implies a probability of (1-0.85)=0.15 of making a Type II error.
[tex]P(Type \,II\, error)=1-Power=1-0.85=0.15[/tex]
The option 1 ("Make a Type II error 5% of the time") is not precise, so it is not correct.
Option 3
The significance level is 0.05. This is also the probability of making a Type I error.
The option 3 ("Make a Type I error 5% of the time") is correct.
whats the ratio of 14,10 and 4
Step-by-step explanation:
14 : 10 : 4
Divide it by 2
7 : 5 : 2
This is the simplest form because it cannot be divided further
A certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million). A group of factory representatives whose companies discharge liquids into the river is now claiming they have lowered the average with improved filtration devices. A group of environmentalists will test to see if this is true. Find the rejection region appropriate for this test if we are using a significance level of 0.05 and have a sample size of 25.
A) Reject H0 if t < -1.960
B) Reject H0 if t < -2.064
C) Reject H0 if t < -1.711
D) Reject H0 if t < -1.708
E) Reject H0 if t < -2.064 or t > 2.064
Answer:
C) Reject [tex]H_0[/tex] if t < -1.711
Step-by-step explanation:
We are given that a certain chemical pollutant in the Arkansas River has been constant for several years with mean μ = 34 ppm (parts per million).
Also, it is given that the level of significance is 0.05 and a sample size of 25 is taken.
Let [tex]\mu[/tex] = average chemical pollutant in the Arkansas River
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 24 ppm {means that the average is same as before with improved filtration devices}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 24 ppm {means that they have lowered the average with improved filtration devices}
Now, t test statistics is given by; T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, n = sample size = 25
n - 1 = degree of freedom = 25 - 1 = 24
Now, at 0.05 significance level in the t table, the critical value at 24 degree of freedom is given as - 1.711 for left tailed test.
So, we will reject our null hypothesis when the t test statistics is less than the critical value of t which means reject null hypothesis if t < -1.711.
Among a banks 214 customers with checking or savingsaccounts, 189 have checking accounts, 73 have regular savingsaccounts, 114 have money market savings accounts, and 69 have bothchecking and regular savings accounts. No customer is allowed tohave both savings and money market savings accounts.
a.) How many customers have both checking and money marketsavings accounts?
b.) How many customers have a checking account but no savingsaccount?
Answer:
a) 93 customers
b) 27 customers
Step-by-step explanation:
Total number of customers (n) = 214
Checking (C) = 189
Regular Savings (R) = 73
Market Savings (M) = 114
Checking and Regular (C&R) = 69
a) The total number of customers is given by:
[tex]n = C+R+M-C\cap R-C\cap M\\214=189+73+114-69-C\cap M\\C\cap M = 93[/tex]
93 customers have both checking and money market savings accounts
b) The number of customers with savings accounts is given by:
[tex]C = C_{only} +C\cap R+C\cap M\\189=C_{only} +69+93\\C_{only} = 27[/tex]
27 customers have a checking account but no savings account.
i’m almost done with my problems can someone answer this
Step-by-step explanation:
8p + 5 = 6p + 1
8p - 6p = 1 - 5
2p = - 4
p = - 4/2
p = -2
Hope it will help :)