Answer:
The coefficient is 3 and the constant is 4 in
relation to the equation mx + c where m is the coefficient of x and c is the constant.
Which of the following statement is true about k-NN algorithm?
k-NN performs much better if all of the data have the same scale
k-NN works well with a small number of input variables (p), but struggles when the number of inputs is very large
k-NN makes no assumptions about the functional form of the problem being solved
A) 1 and 2
B) 1 and 3
C) Only 1
D) All of the above
In this exercise we have to use the knowledge of algorithm to write the correct alternative that best matches, thus we can say that:
Letter D
The computational complicatedness of K-NN increases as the extent or bulk of some dimension of the training basic document file increase and the treasure gets considerably unhurried as the number of examples and free variables increase.
Also, K-NN happen a non-parametric machine intelligence treasure and as such form no assuming possession about the working form of the question at hand.
The invention everything better accompanying information in visible form of the same scale, therefore standard the information in visible form superior to applying the invention happen urged.
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All the statements provided about k-NN (scale importance, efficiency with small input variables, and no assumptions about the functional form of the problem) are true.
Explanation:The question asks which of the given statements is true about the k-NN (k-Nearest Neighbors) algorithm. The correct answer would be 'D) All of the above.' This is because all three statements accurately describe different aspects of how the k-NN algorithm works:
k-NN performs much better if all of the data have the same scale. Applying a suitable scaling method is important as it ensures all features contribute equally. k-NN works well with a small number of input variables (p), but struggles when the number of inputs grows large. This is due to the 'curse of dimensionality,' as the volume of the input space increases exponentially with the number of dimensions, impacting performance and accuracy. k-NN makes no assumptions about the functional form of the problem being solved. This is a key strength of k-NN; it's a non-parametric method that doesn’t make explicit assumptions about the functional form of the data.Learn more about k-NN algorithm here:
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The following sample data represent the gasoline mileages (in miles per gallon) determined for cars in a particular weight class:
25.1 29.0 34.5 35.7 37.9 34.9 24.3 26.6 27.3 32.0 30.0 34.5 35.3 33.5 36.6 34.8 16.2 13.1 24.5 33.6 28.0 33.9 30.7 32.0 37.7 21.1 31.2 35.6 34.4 25.2 35.9 18.3 29.4 29.5 34.8 29.4 26.4 38.8 36.0 28.7 23.4 35.3 33.7 38.1 28.6 34.2 34.8 39.2 39.9 36.8
1. Using 10 mpg as the lower limit of the first class interval, construct a histogram with intervals of width 5 mpg. What is the midpoint of the cell with the largest number of observations (the mode).
2. Is the histogram skewed to the left or right or symmetric?
Answer:
1) Explanation and figure attached below.
2) For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.
Step-by-step explanation:
Part 1
If we order the data from the smallest to the largest we got:
13.1 , 16.2 ,18.3 , 21.1 , 23.4 , 24.3 , 24.5
25.1 , 25.2 , 26.4 , 26.6 ,27.3 ,28 , 28.6 ,
28.7 , 29 , 29.4 , 29.4 , 29.5 , 30 , 30.7
31.2 , 32 , 32 , 33.5 , 33.6 , 33.7 , 33.9
34.2 , 34.4 , 34.5 , 34.5 , 34.8 , 34.8
34.8 , 34.9 , 35.3 , 35.3 , 35.6 , 35.7
35.9 , 36 , 36.6 , 36.8 , 37.7 , 37.9
38.1 , 38.8 , 39.2 , 39.9
For this case if we use 10 mpg as the lower limit and with a width of 5 mpg for the intervals we have the following table:
Interval Frequency Midpoint
[10-15) 1 12.5
[15-20) 2 17.5
[20,25) 4 22.5
[25-30) 12 27.5
[30-35) 17 32.5
[35-40) 14 37.5
Total 50
So the histogram is on the figure attached. The midpoint for the class interval with the largest number of observations is 32.5.
Part 2
For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.
Final answer:
To construct the histogram, determine the class intervals and count the number of observations in each interval. The midpoint of the cell with the largest number of observations is approximately 37.45 mpg.
Explanation:
To construct a histogram with intervals of width 5 mpg, we start by determining the class intervals. Given that the lower limit of the first interval is 10 mpg, the class intervals can be calculated as follows: 10-14.9, 15-19.9, 20-24.9, 25-29.9, 30-34.9, 35-39.9. Next, we count the number of observations falling into each interval:
10-14.9: 3
15-19.9: 6
20-24.9: 5
25-29.9: 7
30-34.9: 9
35-39.9: 10
The cell with the largest number of observations is the one corresponding to the interval 35-39.9 mpg. To calculate the midpoint of this interval, we take the average of the lower and upper limits: (35 + 39.9) / 2 = 37.45. Therefore, the midpoint of the cell with the largest number of observations (the mode) is approximately 37.45 mpg.
A summary measure that is computed to describe a population characteristic from a sample is called a: A) statistic. B) census. C) parameter
Answer:
A. Statistic
Step-by-step explanation:
Usually whole population is difficult to cover so the part of population is considered to assess the aspects of population. The statistic is the quantity which summarizes the sample. The problem illustrate that summary measure is calculated from sample to describe a population. As the summary measure is computed from sample then it is a statistic.
In a box of 25 external hard disks, there are 2 defectives. An inspector examines 5 of these hard disks. Find the probability that there is at least 1 defective hard disk among the 5.
Answer:
0.3667 or 36.67%
Step-by-step explanation:
The probability of getting at least 1 defective hard disk among the 5 (P(X>0)) is equal to 100% minus the probability of getting none defectives (1-P(X=0)).
If 23 out of 25 hard disks are non-defective, the probability is:
[tex]P(X>0) = 1 -P(X=0)\\P(X>0) = 1 -\frac{23}{25}*\frac{22}{24} *\frac{21}{23} *\frac{20}{22} *\frac{19}{21}\\ P(X>0) = 0.3667=36.67\%[/tex]
The probability that there is at least 1 defective hard disk is 0.3667 or 36.67%.
Assume there is a linear relationship between the curb weight of a car (in thousands of kilograms) and its combined mileage (in miles per gallon). The Hummer H2 has a weight of 3 thousand kg and mileage of 9 mpg, while a Ford Taurus sedan weighs 1.9 thousand kg and gets 21 mpg. What is the mileage of a Chevy Cavalier weighing 1.1 thousand kg?
Answer: The mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.
Step-by-step explanation:
if there is a linear relationship between two variables x and y , then we represent this relation in the form of equation as :[tex]y= mx+c[/tex] (*)
, where m = rate of change of y with respect to x
c= Constant or the value of y when x=0.
We assume there is a linear relationship between the curb weight of a car (in thousands of kilograms) and its combined mileage (in miles per gallon).
Let y= combined mileage (in miles per gallon)
x= curb weight of a car (in thousands of kilograms)
When x= 3 thousand kg , y = 9 mpg
⇒ [tex]9= m(3)+c[/tex] [Put values in (*)]
⇒ [tex]9= 3m+c--------(1)[/tex]
When x= 1.9 thousand kg , y = 21 mpg
⇒ [tex]21= m(1.9)+c[/tex] [Put values in (*)]
⇒ [tex]21= 1.9m+c--------(2)[/tex]
Eliminate the equation (1) from (2) , we get
[tex]1.9m-3m=21-9[/tex]
[tex]-1.1m=12\\\Rightarrow\ m=\dfrac{12}{-1.1}=\dfrac{-120}{11}[/tex]
Put value of m in (1) , we gte
[tex]9= 3(\dfrac{-120}{11})+c[/tex]
[tex]9= \dfrac{-360}{11}+c\\\Rightarrow\ c=9+\dfrac{360}{11}=\dfrac{459}{11}[/tex]
Substitute the value of m and c in (*) , we get
[tex]y= \dfrac{-120}{11}x+\dfrac{459}{11}[/tex]
When x= 1.1
[tex]y= \dfrac{-120}{11}(1.1)+\dfrac{459}{11}[/tex]
[tex]y= \dfrac{-120}{11}(\dfrac{11}{10})+\dfrac{459}{11}[/tex]
[tex]y=-12+\dfrac{459}{11}=\dfrac{327}{11}\approx29.72[/tex]
Hence, the mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.
if a pound of almonds costs $8, how many ounces can be bought for $4.60?
The number of ounces that can be bought for $4.60 will be 9.2 ounces.
What is Algebra?The analysis of mathematical representations is algebra, and the handling of those symbols is logic.
Conversion means to convert the same thing into different units.
If a pound of almonds costs $8.
We know that in one pound, there are 16 ounces. Then the cost of each ounce will be
⇒ 8 / 16
⇒ $0.5 per ounce
Then the number of ounces that can be bought for $4.60 will be
⇒ 4.6 / 0.5
⇒ 9.2 ounces
The number of ounces that can be bought for $4.60 will be 9.2 ounces.
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"Use the following information to answer the question. The mean age of lead actresses from the top ten grossing movies of 2010 was 29.6 years with a standard deviation of 6.35 years. Assume the distribution of the actresses' ages is approximately unimodal and symmetric.
In 1993, actress Anna Paquin won an academy award in for the movie ""The Piano."" She was 11-years-old. Finish the statement: ""According to the Empirical Rule, the ages of nearly all lead actresses will be between______________and_____________ years. Anna Paquin was this range when she won the academy award."""
a) 10.6; 48.7; within
b) 23.3: 36.0; within
c) 16.9: 42.3; not within
d) 10.6: 48.7; not within
Answer:
a) 10.6; 48.7; within
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 29.6
Standard deviation = 6.35
"According to the Empirical Rule, the ages of nearly all lead actresses will be between______________and_____________ years.
Nearly all(99.7%) will be within 3 standard deviations from the mean.
So from 29.6 - 3*6.35 = 10.6 years to 29.6 + 3*6.35 = 48.7 years.
Anna was 11-years old, so she was within this range.
So the correct answer is
a) 10.6; 48.7; within
The ages of nearly all lead actresses will be between 16.9 and 42.3 years according to the Empirical Rule. Anna Paquin's age falls within this range when she won the academy award.
Explanation:According to the Empirical Rule, for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations. Since the mean age is 29.6 years and the standard deviation is 6.35 years, we can calculate the range as follows:
One standard deviation: 29.6 ± 6.35 = 23.25 to 35.95
Two standard deviations: 29.6 ± (2 × 6.35) = 16.9 to 42.3
Three standard deviations: 29.6 ± (3 × 6.35) = 10.55 to 48.65
Since Anna Paquin was 11 years old when she won the academy award, her age falls within the range of two standard deviations, which is 16.9 to 42.3 years. Therefore, the correct answer is option c) 16.9: 42.3; not within.
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Write the following in terms of sin θ and cos θ; then simplify if possible. (Leave your answer in terms of sin θ and/or cos θ.) csc θ tan θ.
The expression [tex]Cosec\theta*Tan\theta[/tex] in terms of [tex]Sin\theta[/tex] and [tex]Cos\theta[/tex] in simplified form is [tex]Cosec\theta*Tan\theta = Sin\theta[/tex]
Given expression:
[tex]Cosec\theta*Tan\theta[/tex]
Express [tex]cosec\theta[/tex] and [tex]Tan\theta[/tex] in terms of [tex]Sin\theta[/tex] and [tex]Cos\theta[/tex]
Now,
[tex]Cosec\theta = \dfrac{1}{Sin\theta}[/tex]
[tex]Tan\theta = \dfrac{Sin\theta}{Cos\theta}[/tex]
Substitute the expression of [tex]Cosec\theta[/tex] and [tex]Tan\theta[/tex] in original expression:
[tex]Cosec\theta*Tan\theta = \dfrac{1}{Sin\theta} * \dfrac{Sin\theta}{Cos\theta}[/tex]
[tex]Sin\theta[/tex] is cancel out in the right-hand side of the equation:
[tex]Cos\theta[/tex] can be written as:
[tex]\dfrac{1}{Cos\theta} = \dfrac{1}{\dfrac{1}{Sin\theta} }[/tex]
[tex]= Sin\theta[/tex]
For the given expression:
[tex]Cosec\theta*Tan\theta = Sin\theta[/tex]
The simplified expression is [tex]Cosec\theta*Tan\theta = Sin\theta[/tex]
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The expression csc θ tan θ simplifies to sec θ which further simplifies to [tex]\frac{1}{cos\theta}[/tex].
To express csc θ tan θ in terms of sin θ and cos θ, we start by rewriting each trigonometric function:
csc θ = [tex]\frac{1}{sin\theta}[/tex]tan θ = [tex]\frac{sin\theta}{cos\theta}[/tex]Now, multiplying these together:csc θ tan θ = ([tex]\frac{1}{sin\theta}[/tex]) × ([tex]\frac{sin\theta}{cos\theta}[/tex])The sin θ terms cancel out:csc θ tan θ = [tex]\frac{1}{cos\theta}[/tex]Therefore, the expression simplifies to : [tex]\frac{1}{cos\theta}[/tex]High demand cars that are also in low supply tend to retain their value better than other cars. The data in the table is for a car that won a resale value award
The student's question relates to analyzing trends in car safety award data and understanding supply curve dynamics in the context of business or economics at the college level. It requires statistical analysis and economic reasoning.
Explanation:The student's question involves analyzing data to determine if there has been a change in the distribution of cars that earned top safety picks between the years 2009 and 2013. This question lies within the field of Business or specifically within the area of business statistics, focusing on the analysis of trends over time.
Furthermore, considering the supply of cars and how it changes with price is another key aspect of this question. Using the supplied figures and tables, which is common in business studies, the student is asked to interpret how the quantity supplied of cars changes as the price increases from $20,000 to $22,000—a concept known in economics as the law of supply.
In summary, the question requires the application of statistical analysis to evaluate changes in car safety awards over time and the understanding of basic supply curve dynamics.
Your complete question is: High demand cars that are also in low supply tend to retain their value better than other cars. The data in the table below is for a car that won a resale value award. a. Write a function to represent the change in the percentage of the car's value over time. Suppose that the function is linear for the first 5 years. b. Based on your model, by what percent did the car's value drop the day it was bought and driven off the lot? c. Do you think the linear model would still be useful after 10 years? Why or why not? d. Assume you used months instead of years to write a function. How would your model change?
True or False: Like the edges of a filled-in area, the endpoints of a polygon do not need to conform to snap points.
Answer:
The answer is "False."
Explanation:
An area is considered the amount of space that an object occupies. A filled-area subjects the object to be made up of lines. These lines connect with each other to form an "edge."
The connection of the lines in order to define the object's shape or area are considered "snap points." Remember that "polygons" are made of line segments, where their endpoints meet with each other in order to define its closed shaped. Thus, it needs to conform to "snap points."
This explains the answer.
The given statement "Like the edges of a filled-in area, the endpoints of a polygon do not need to conform to snap points" is false.
The edge of a filled-in area is the boundary of the surface. Area is a two-dimensional space covered by any surface. The boundary of the surface is generally the edges of the area. The edges or the boundary should connect in order to make a closed area or surface.
Similarly, snap points are the end points of the sides of a polygon. These endpoints need to connect with each other in order to make a closed figure. Polygon is a closed figure with n number of sides.
Therefore, the given statement "Like the edges of a filled-in area, the endpoints of a polygon do not need to conform to snap points" is false.
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Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 87, x = 26; 98 percent (0.185, 0.413) (0.202, 0.396) (0.184, 0.414) (0.203, 0.395)
Answer:
(0.185, 0.413)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 87, x = 26, p = \frac{x}{n} = \frac{26}{87} = 0.2989[/tex]
98% confidence interval
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.325[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 - 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.185[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 + 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.413[/tex]
So the correct answer is:
(0.185, 0.413)
The World Health Organization wishes to estimate the mean density of people per square kilometer. They collect data on 56 countries. State the individual, variable, population, sample, parameter, and statistic:_____________.
Individual: A single country.
Variable: The mean density of people per square kilometer of all countries.
Population: Set of all countries.
Sample: Set of the 56 countries on which data is collected.
Parameter: The mean density of people per square kilometer calculated from the population.
Statistic: The mean density of people per square kilometer calculated from the sample.
Explanation:Individual: Individuals are the objects described by a set of data.
Variable: Variables are characteristics of individuals.
Population: Population is all individuals of interest.
Sample: Sample is a subset of the population.
Parameter: Parameter is a characteristic of a population.
Statistics: Statistic is a characteristic of a sample.
The individual is the country, the variable is the density of people per square kilometer, the population is all the countries in the world, the sample is the 56 countries, the parameter is the true mean density of all countries, and the statistic is the estimated mean density from the 56 sampled countries.
Explanation:In the example given, the individual is each of the 56 countries from which the World Health Organization collects data. The variable is the density of people per square kilometer in each of these countries. The population consists of all the countries in the world. The sample includes the 56 countries for which data was collected. The parameter is the true mean density of people per square kilometer for all countries. Finally, the statistic is the estimated mean density of people per square kilometer based on the data from the sample of 56 countries.
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Matrix multiplication and addition
Answer:
matrices can be simple added by ading the adjacent elements of the two matrices
The matrix multiplication is done by multiplying row of first matrix to columns of the other matrix.
each multiplication for each element of the resultant matrix.
Step-by-step explanation:
A sum is invested at 4% continuous interest. This means that its value grows exponentially with k equaling the decimal rate of interest. Find, to the nearest tenth of a year, the time required for the investment to double in value.
Answer:
T = 72/4 = 18years
T = 20years (to the nearest tenth)
Step-by-step explanation:
Using the rule of 72, which is used to estimate the number of years for a given investment to double at a given interest rate.
Doubling time = 72/interest rate
T = 72/r
Rate r (in percentage) = 4%
Time T (in years)
T = 72/4 = 18years
T = 20years (to the nearest tenth)
Suppose that 20% of adults practices a gluten-free diet, 15% of adults practice a dairy-free diet, and 8% of adults practice both of these diets. One adult is selected at random.
a. Given a random individual practices a glute-free diet, what is the probability they practice a dairy-free diet?
b. Are events "practices a gluten-free diet" and "practices a dairy-free diet" independent?
c. Given a random individual practices a dairy-free diet, what is the probability they do not practice a gluten free diet?
Answer:
a. 0.4
b. Not independent events
c. 0.47
Step-by-step explanation:
Let A= Practices Gluten free diet
B= Practices Dairy free diet
A and B= Practices Both diets
P(A)=0.20
P(B)=0.15
P(A and B)=0.08
a.
[tex]P(B/A)=\frac{P(A and B)}{P(A)}[/tex]
[tex]P(B/A)=\frac{0.08}{2}[/tex]
[tex]P(B/A)=0.40[/tex]
b.
The two events are independent if P(B/A) =P(B) or P(A/B)=P(A)
As, P(B/A) ≠P(B)
0.4≠0.15
So, the event gluten free diet and dairy free diet are dependent events.
c.
[tex]P(A'/B)=\frac{P(A' and B)}{P(B)}[/tex]
[tex]P(A'and B)= P(B)-P(A and B)[/tex]
[tex]P(A' and B)=0.15-0.08=0.07[/tex]
[tex]P(A'/B)=\frac{P(A' and B)}{P(B)}[/tex]
[tex]P(A'/B)=\frac{0.07}{0.15}[/tex]
[tex]P(A'/B)=0.47[/tex]
Answer:
a. 0.4
b. Not independent
c. 0.47
Step-by-step explanation:
Determine if the statement is true or false. Any linear system with more variables than equations cannot have a unique solution. True False
Answer:
True
Step-by-step explanation:
A linear system of equations will have a unique solution if and only if the number of variables equal the number of independent equations.
By independent equations we mean the same equation not repeated by multiplying by any constant.
Suppose number of variables are n, we must have the determinant formed by the coefficients non zero to have a unique solutions.
Here the no of equations are less than the number of variables. So we cannot have a unique solution but can have a parametric solution using the number of parameters as n-m where n = number of variables and m = the number of independent equations given.
So the given statement is true.
every 55 students there were 2 professor's
How many professors for 1650 students
Answer:
60 professors
Step-by-step explanation:
for every 55 students there were 2 professors
for 1650 students, there will be (1650 x 2) / 55 =3300/55 = 60 professors
Final answer:
To find the number of professors needed for 1650 students when the ratio is 2 professors for every 55 students, you calculate 2 / 55 = x / 1650, resulting in x = 60. Therefore, 60 professors are needed for 1650 students.
Explanation:
The question asks how many professors would be needed for 1650 students if there is a ratio of 2 professors for every 55 students. To solve this, you can set up a proportion where the number of students is directly proportional to the number of professors. The proportion can be expressed as 2 professors / 55 students = x professors / 1650 students. Solving for x involves cross-multiplying and dividing.
Step 1: Cross-multiply to get 2 × 1650 = 55 × x.
Step 2: Simplify the equation: 3300 = 55x.
Step 3: Divide both sides by 55 to solve for x: x = 3300 / 55.
Step 4: Calculate the result: x = 60. Thus, you would need 60 professors for 1650 students.
In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.1. -.4401 ≤ p1 - p2 ≤ -.13802. -.4401 ≤ p1 - p2 ≤ .13803. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%
Answer:
[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]
[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]
We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]
1. -.4401 ≤ p1 - p2 ≤ -.1380
4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%
Step-by-step explanation:
In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.
1. -.4401 ≤ p1 - p2 ≤ -.1380
2. -.4401 ≤ p1 - p2 ≤ .1380
3. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%
4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%
5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%
Solution to the problem
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]p_1[/tex] represent the real population proportion for San Jose
[tex]\hat p_1 =\frac{30}{73}=0.411[/tex] represent the estimated proportion for San Jos
[tex]n_1=73[/tex] is the sample size required for San Jose
[tex]p_2[/tex] represent the real population proportion for San Francisco
[tex]\hat p_2 =\frac{56}{80}=0.7[/tex] represent the estimated proportion for San Francisco
[tex]n_2=80[/tex] is the sample size required for San Francisco
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]
[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]
We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]
Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.
Based on this the correct options are:
1. -.4401 ≤ p1 - p2 ≤ -.1380
4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%
You draw two cards from a standard deck of 52 cards and do not replace the first one before drawing the second. Find the probability of drawing a 7 for the first card and a king for the second card. Round your answer to the nearest thousandth.
Final answer:
The probability of drawing a 7 followed by a king from a standard 52-card deck without replacing the first card is 4/663, which rounds to approximately 0.006 when rounded to the nearest thousandth.
Explanation:
The student is asking to find the probability of drawing a 7 for the first card and a king for the second card from a standard 52-card deck, without replacing the first card. To solve this, we need to calculate the probability of each event occurring consecutively.
First, the probability of drawing a 7 from a deck of 52 cards is 4/52 or 1/13, because there are 4 sevens in the deck. After drawing a seven, there are now 51 cards left in the deck. Next, the probability of drawing a king from the remaining 51 cards is 4/51, since there are still 4 kings in the deck.
To find the combined probability of both events happening, we multiply these probabilities together:
Probability of drawing a 7 and then a king = (1/13) × (4/51)
Which simplifies to:
Probability = 4/663
Rounded to the nearest thousandth, the probability is approximately 0.006.
Finding the average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11
Answer:
The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.
Step-by-step explanation:
Given a function y, the average rate of change S of [tex]y=g(z)[/tex] in an interval [tex](z_{s}, z_{f})[/tex] will be given by the following equation:
[tex]S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}}[/tex]
In this problem, we have that:
[tex]g(z) = 4z^{2} - 5[/tex]
Between the points( -2,11) and (2,11).
So [tex]z_{f} = 2, z_{s} = -2[/tex]
[tex]g(z_{f}) = g(2) = 4*(2)^{2} - 5 = 11[/tex]
[tex]g(z_{s}) = g(-2) = 4*(-2)^{2} - 5 = 11[/tex]
So
[tex]S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}} = \frac{11 - 11}{2 - (-2)} = \frac{0}{4} = 0[/tex]
The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.
Solve the following differential equations using classical methods. x¨ + 6 ˙x + 8x = 5sin(3t) x¨ + 4 ˙x = t2 ; x(0) = 1; ˙x = 2
Answer:
1. x = (199/130)e^(-2t) - (33/65)e^(-4t) - (18/65)cos3t - (1/65)sin3t
2. x = (191/128) - (63/128)e^(-4t) + (t³/12) - (t²/16) + (t/32)
Step-by-step explanation:
Steps are shown in the attachment.
Law School According to the Law School Admission Council, in the fall of 2007, 66% of law school applicants wereaccepted to some law schooL4 The training program LSATisfaction claims that 163 of the 240 students trained in 2006were admitted to law school. You can safely consider these trainees to be representative of the population of law schoolapplicants. Has LSAfisfaction demonstrated a real improvement over the national average?a) What are the hypotheses?b) Check the conditions and find the P-value.c) Would you recommend this program based on what you see here? Explain.
Answer:
a) [tex]H_{0}: p = 0.66\\H_A: p > 0.66[/tex]
b) P-value = 0.2650
c) No, this programme will not be recommended as there is no real improvement over the national average.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 240
p = 66% = 0.66
Alpha, α = 0.05
Number of students admitted to law school , x = 163
a) First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.66\\H_A: p > 0.66[/tex]
This is a one-tailed(right) test.
Formula:
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{163}{240} = 0.6792[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
[tex]z = \displaystyle\frac{0.6792-0.66}{\sqrt{\frac{0.66(1-0.66)}{240}}} = 0.6279[/tex]
b) Now, we calculate the p-value from the table.
P-value = 0.2650
c) Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.
Thus, there is no real improvement over the national average.
No, this programme will not be recommended as there is no real improvement over the national average.
A researcher wants to determine the impact of soil type on the growth of a certain type of plant. She grows three plants in each of four different types of soil and measures the growth in inches for each plant after one month resulting in the data below.Soil 1 Soil 2 Soil 3 Soil 412.6 12.2 12.2 11.112.6 12,1 0.6 11.714.3 13,9 1 9.6 What null hypothesis is the researcher testing if she runs an ANOVA with this data?a. The mean growth of the plant in each type of soil is the same.b. One type of soil has a higher mean growth for the plant than the others. c. The variability in growth of the plant in each type of soil is the same.d. oil 3 provides a lower mean growth for the plant than the other types of soil.e. The mean growth of the plant is different in each type of soil.
Answer:
a. The mean growth of plant is same for each type of soil.
Step-by-step explanation:
The analysis of variance is statistical procedure used to assess the equality of more than two means by computing two different estimate of variances. Basically when we have more than two means to compare, we can't use t-test or z-test because these procedure will be tedious and time taking. So, the null hypothesis will be the same. Hence the null hypothesis in the given scenario is that "the mean growth of plant is same for each type of soil".
A model for the basal metabolism rate, in kcal/h, of a young man is given by the formula below, where t is the time in hours measured from 5:00 AM. What is the total basal metabolism of this man over a 24 hour period?R(t)=80-0.18cos πt/1224∫ R(t) dt=____?0
Answer:
1920 k calories is the basal metabolism.
Step-by-step explanation:
Formula to get the total basal metabolism has been given by the formula
R(t) = [tex]80-0.18cos\frac{\pi t}{12}[/tex]
Where t = time
Now to calculate the total basal metabolism we will integrate the function with respect to time from 0 to 24 hours.
[tex]\int_{0}^{24}R(t)=\int_{0}^{24}(80-0.18cos\frac{\pi t}{12})dt[/tex]
= [tex][80t-{0.18}\times \frac{sin\frac{\pi t}{12}}{\frac{\pi}{12}}]_{0}^{24}[/tex]
= [tex][(80-0)24-\frac{0.18\times 12}{\pi}(sin2\pi - sin0)][/tex]
= [(80-0)24-(sin2\pi - sin0)]
= 80×24
= 1920 k calories
Therefore, 1920 k calories is the total basal metabolism of the young man.
Ann and Bill play rock-paper-scissors. Each has a strategy of choosing uniformly at random out of rock, paper, and scissors every round (making independent selections each round). Each round can end in a win, loss, or a tie for each player.
a. What is the probability Ann wins the first round. (Remember that a tie is a possibility)
Prob=
b. What is the probability that Ann's first win happens in round #5?
Prob=
C. What is the probability that Ann's first win comes after round #5?
Prob=
Answer:
a) 1/3
b) 0.0658436214
Step-by-step explanation:
Part a
Ann wins first round = (P , R) + (S , P) + (R , S) = 3 possibilities of win
Total outcomes = Wins + Losses + Ties = 9 possibilities
Hence,
P (Ann wins first round) = 3/9 = 1/3
Part b
Ann losses or ties first 4 rounds and wins 5th round
P(Ann loosing or tie in any round) = 6/9 = 2/3
Hence,
P(Ann wins 5th round only) = (2/3)^4 * (1/3) = 0.0658436214
Part c
Final answer:
In rock-paper-scissors, Ann has a 1/3 probability of winning the first round. The probability of Ann's first win in round #5 is (2/3)^4 × (1/3), and the probability of Ann's first win coming after round #5 is 1 - (2/3)^5.
Explanation:
To answer these questions, we will calculate the probabilities based on the uniform random choices made by Ann and Bill in the game of rock-paper-scissors.
Probability Ann wins the first round: Each player has 3 choices, resulting in 9 possible outcomes. There are 3 ways Ann can win (rock beats scissors, paper beats rock, scissors beats paper), so the probability is 3/9 or 1/3.
Probability Ann's first win happens in round #5: We need 4 consecutive non-wins (losses or ties) and a win in the 5th round. The probability of a non-win per round is 2/3, making the probability of four non-wins (2/3)^4. The probability Ann wins the 5th round is 1/3. Therefore, the probability of this scenario is (2/3)^4 × (1/3).
Probability that Ann's first win comes after round #5: This is the probability of Ann not winning in the first 5 rounds. As each round is independent, we raise the non-winning probability to the power of 5, (2/3)^5, and subtract from 1 to get the desired probability.
What is the equation of the line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5)?
Final answer:
To find the equation of a line perpendicular to 2x - 3y = 13 and passing through (-6, 5), first find the slope of the given line, then its negative reciprocal for the perpendicular line, and finally apply the point-slope formula with the new slope and given point.
Explanation:
To find the equation of a line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5), follow these steps:
Rewrite the given equation in slope-intercept form (y = mx + b) to find its slope.
Calculate the slope of the perpendicular line using the negative reciprocal of the given line's slope.
Use the point-slope formula with the slope from step 2 and the given point to find the equation of the perpendicular line.
Rewriting the given equation: 2x - 3y = 13 → -3y = -2x + 13 → y = (2/3)x - 13/3. The slope (m1) is 2/3. The slope of the perpendicular line (m2) is -3/2 (the negative reciprocal of m1). Using the point-slope formula with the perpendicular slope and the point (-6, 5), we get: y - 5 = (-3/2)(x + 6), which simplifies to y = (-3/2)x - 4. This is the equation of the line perpendicular to 2x - 3y = 13 that passes through (-6, 5).
A banner is hung for a party. The
distance from a point on the bottom edge of
the banner to the floor can be determined by
using the function f(x) = 0.25x 2 − x + 9.5,
where x is the distance, in feet, of the point
from the left end of the banner. How high
above the floor is the lowest point on the
bottom edge of the banner? Explain.
Answer:
8.5 feetExplanation:
The function that can determine the distance from a point on the bottom edge of the banner to the floor is:
[tex]f(x)=0.25x^2-x+9.5[/tex]
That function is a quadratic function which means that it is a parabola. Given that the coefficient of the quadratic term (0.25) is positive, the parabola open upwards, and the vertex is the lowest point of the parabola and it represents how high above the floor is the lowest point on the bottom edge of the banner.
So, you need to find the vertex of the parabola.
I will complete squares to find the form A(x -h)² + k, where h and k are the coordinates of the vertex (h, k).
[tex]f(x)=0.25x^2-x+9.5\\ \\ 4(f(x))=4(0.25x^2-x+9.5)\\ \\ 4f(x)=x^2-4x+38\\ \\ 4f(x)-38=x^2-4x\\ \\ 4f(x)-38+4=x^2-4x+4\\ \\ 4f(x)-34=(x-2)^2\\ \\ 4f(x)=(x-2)^2+34\\ \\ f(x)=(1/4)(x-2)^2+8.5[/tex]
Hence, the vertex (h,k) is (2, 8.5), meaning that the lowest point on the bottom edge of the banner is at 2 feet from the left edge of the banner and 8.5 feet above the floor.
The lowest point on the bottom edge of the banner which results in a height of 8.5 feet above the floor.
To find the height above the floor of the lowest point on the bottom edge of the banner using the function f(x) = 0.25x^2 - x + 9.5, we need to determine the vertex of this quadratic function. The vertex form of a quadratic function f(x) = ax^2 + bx + c gives the minimum or maximum value of the function.
Here, the coefficient a is positive (0.25), indicating that the parabola opens upwards and thus has a minimum point. The x-coordinate of the vertex can be found using the formula x = -b / (2a).
Identify the coefficients: a = 0.25 and b = -1.Calculate the x-coordinate of the vertex: x = -(-1) / (2 * 0.25) = 2 feet.Substitute x = 2 back into the function to find the minimum height:
f(2) = 0.25(2)^2 - 2 + 9.5 = 1 - 2 + 9.5 = 8.5 feet.
Therefore, the lowest point on the bottom edge of the banner is 8.5 feet above the floor.
Assume that you are on a cross-country flight at an altitude of 4,500 feet pressure altitude and your course is 275 degrees. The outside air temperature is 10 degrees Celsius, your indicated airspeed is 115 knots, the wind direction is 210 degrees, the wind velocity is 15 knots, and you passed your last checkpoint at 0942. If you have 218 nautical miles remaining, at what time will you arrive at your destination?
First you need to find your True Air Speed (TAS). I used the formula:
[tex]TAS=IAS\sqrt{\frac{\rho_0}{\rho}}\\[/tex]
where [tex]\rho_0[/tex] is the density at mean sea level at 15°C which is 1.225 kg/m^3 and [tex]\rho[/tex] is the density of the air in which the airplane is flying. In this case with the given altitude and air temperature is about: 1.073 kg/m^3.
So TAS is equal to: 123 kt in this case.
To get the time of arrival you need to get your ground speed. For that you need to compensate for the wind speed and direction. With the drift angle you can get your ground speed.
In this case you get ground speed of 115 kt which means to travel 218 nautical miles you need 1 h and 53 min which means you will arrive at: 1135.
Consider the concentration, C, in mg/liter, of a drug in the blood as a function of x, of the amount, in mg, of the drug given and t, the time in hours since the injection. For 0 < = x < = 4 mg and t > = 0 hours, we have C = f(x, t) = te^-t(5-x). Find f(1, 2).Interpret in terms of drug concentration. a. The change in concentration of a 1 mg dose in the blood 2 hours after injection. b. The amount of a 1 mg dose in the blood 2 hours after injection. c. The concentration of a 1 mg dose in the blood 2 hours after injection. d. The concentration of a 2 mg dose in the blood 1 hour after injection. e. The change in concentration of a 2 mg dose in the blood 1 hour after injection.
Answer:
c. the concentration of 1 mg dose in the blood 2 hours after injection.
Step-by-step explanation:
We have, C = f(x, t) = t*e^-t(5-x)
then
C = f(1, 2) = 2*e^-2(5-1)
C = 2*e^-2(4)
C = 2*e^-8 OR C = 2/e^8
The concentration of a drug in blood depends on the dose and the time since administration. The given function allows calculating these concentrations, resulting in a concentration of a 1 mg dose 2 hours after injection of 8e^-2 mg/liter and a 2 mg dose 1 hour after injection of 3e^-1 mg/liter. The change in concentration cannot be determined without the previous concentration information.
Explanation:To find the concentration of the drug, we need to plug the values of x and t into the given function C = f(x, t) = te^-t(5-x). For f(1,2), we replace x with 1 and t with 2. Therefore, C = 2e^-2(5-1) = 2e^-2*4 = 2*4*e^-2 = 8e^-2.
a. The change in concentration of a 1 mg dose in the blood 2 hours after injection. To find the change in concentration, we would need the previous concentration. However, as this information is not provided, we cannot determine the change in concentration for this case.
b. The amount of a 1 mg dose in the blood 2 hours after injection. The amount of the dose remains constant at 1 mg; what changes is its concentration due to distribution and elimination processes.
c. The concentration of a 1 mg dose in the blood 2 hours after injection. With the values provided, we already calculated this to be C = 8e^-2 mg/liter.
d. The concentration of a 2 mg dose in the blood 1 hour after injection. Plain the value of x = 2 and t = 1 into the equation, we get C = 1e^-1(5-2) = 3e^-1 mg/liter.
e. The change in concentration of a 2 mg dose in the blood 1 hour after injection. Similar to part a, without previous concentration information, it is not possible to determine the change in this case.
Learn more about Drug Concentration here:https://brainly.com/question/31595886
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If sinФ = 1/6 . what are the values of cosФ and tanФ? Show steps
sinФ = 1/6
sinФ = opposite/hypotenuse
cosФ = adjacent/hypotenuse
tanФ = opposite/adjacent
SOHCAHTOA
1^(2)+b^(2)=6^(2)
1+b^(2)=36
b^(2)=35
[tex]\sqrt{35}[/tex]
cosФ = [tex]\sqrt{35}[/tex]/6
tanФ = 1/[tex]\sqrt{35}[/tex] = [tex]\sqrt{35}[/tex]/35
Answer:
Cos ø = √35/6
Tan ø = √35/35
Step-by-step explanation:
Sin ø = opposite/hypothenus
Sin ø = 1/6
Pythagoras theorem says
Hyp² = Opp² + Adj²
6² = 1² + A²
A² = 35
A = √35
Hence, cos ø = √35/6
tan ø = 1/√35 = √35/35