Answer:
A) The actual error |f(0.5) - P3(0.5)| is 0.0340735. The error using Taylor's error formula R₃(0.5) is 0.2917.
B) Te error in the interval [0.5,1.5] is 0.0340735.
C) The approximate integral using P₃(x) is 1/12
D) The real error between the two integrals is 4.687·10^(-3) while the error using \int_{0.5}^{1.5}|R_{3}(x)dx| is 0.018595
Step-by-step explanation:
A) To determine the error we first have to write the third grade Taylor polynomial P3(x):
[tex]P_n(x)=\displaystyle\sum_{k=0}^N \frac{f^{k}(a)}{k!}(x-a)[/tex] with N=3
[tex]P_3(x)=\displaystyle\sum_{k=0}^3 \frac{f^{k}(1)}{k!}(x-1)=(x-1)^2-\frac{1}{2} (x-1)^3[/tex]
The error for the third grade Taylor polynomial R₃(x) is represented as the next term non-written in the polynomial expression near to the point x:
[tex]R_3(x)=\displaystyle \frac{f^{4}(x)}{4!}(x-a)=\frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4[/tex]
Therefore the real error is:
[tex]Err(0.5)=|f(0.5)-P_3(0.5)|=0.0340735[/tex]
The error of the Taylor polynomial R₃(0.5) is:
[tex]R_3(0.5)=\displaystyle \frac{1}{24} (\frac{2}{0.5^3}+\frac{6}{0.5^4})(0.5-1)^4=0.2917[/tex]
B) The error in the interval [0.5,1.5] is the maximum error in that interval.
This is found in the extremes of the intervals. We analyze what happens in X=1.5:
[tex]Err(1.5)=|f(1.5)-P_3(1.5)|=0.01523[/tex]
[tex]R_3(1.5)=\displaystyle \frac{1}{24} (\frac{2}{1.5^3}+\frac{6}{1.5^4})(1.5-1)^4=\frac{1}{216}=4.63\cdot 10^{-3}[/tex]
Both of these errors are smaller than Err(0.5) and R₃(0.5). Therefore the error in this interval is Err[0.5,1.5] is Err(0.5).
C) The approximation of the integral and the real integral is:
[tex]I_f=\displaystyle\int_{0.5}^{1.5} f(x)\, dx=\int_{0.5}^{1.5} (x-1)ln(x)\, dx=0.08802039[/tex]
[tex]I_{p3}=\displaystyle\int_{0.5}^{1.5} P_3(x)\, dx=\int_{0.5}^{1.5} (x-1)^2-0.5(x-1)^3\, dx=1/12=0.0833333[/tex]
D) The error in the integrals is:
[tex]Errint=|I_f-I_{p3}|=4.687\cdot10^{-3}[/tex]
[tex]Errint(R_3)=\displaystyle\int_{0.5}^{1.5} R_3(x)\, dx=\int_{0.5}^{1.5} \frac{1}{24} (\frac{2}{x^3}+\frac{6}{x^4})(x-1)^4\, dx=0.018595[/tex]
Use the following information to answer the exercise. A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour.What are the dependent and independent variables?
Answer:
Dependent: fee
Independent: time
Step-by-step explanation:
The scenario described in question indicates that fee is the dependent variable and time is independent variable. The reasoning is that fee depends on the time(hour). The fee will increase as the time increases. Whereas change in time cause change in fee. So, fee is the dependent variable and time is independent variable.
Final answer:
In the SCUBA equipment rental situation, the independent variable is the number of hours the equipment is rented, and the dependent variable is the total fee. The equation for the total fee is y = 25 + 12.50x, with $25 representing the y-intercept and $12.50 the slope of the hourly rental cost.
Explanation:
For the SCUBA equipment rental situation at the vacation resort, the independent variable is the number of hours the equipment is rented. Since the renter can decide how many hours to rent the equipment, it is considered independent. On the other hand, the dependent variable is the total fee for renting the equipment because it depends on the number of hours the equipment is rented for.
In this context, the equation expressing the total fee (y) in terms of the number of hours (x) the equipment is rented is:
y = 25 + 12.50x.
Here, $25 is the up-front fee which is also the y-intercept of the equation, while $12.50 per hour is the slope. The y-intercept represents the initial cost of renting the equipment before any hourly charges apply, and the slope represents the rate at which the total cost increases for every additional hour the equipment is rented.
The patriot diner sells 2 cheeseburgers and one soda for $11.00 and 3 hamburgersand 2 sodas for $18.00. What is the cost of a cheeseburger?
Answer: the cost of a cheeseburger is $4
Step-by-step explanation:
Let x represent the cost of one cheeseburgers.
Let y represent the cost of one Soda.
The patriot diner sells 2 cheeseburgers and one soda for $11.00. It means that
2x + y = 11 - - - - - - - - - - - 1
She also sells 3 cheeseburgers and 2 sodas for $18.00. It means that
3x + 2y = 18 - - - - - - - - - - -2
Multiplying equation 1 by 3 and equation 2 by 2, it becomes
6x + 3y = 33
6x + 4y = 36
Subtracting, it becomes.
- y = - 3
y = 3
Substituting y = 3 into equation 1, it becomes
2x + 3 = 11
2x = 11 - 3 = 8
x = 8/2 = 4
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".
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.
Suppose you take the Medical College Admission Test (MCAT) and your score is the 32nd percentile. How do you interpret this result
In Statistics, percentiles are a representation of the relative position of a particular value within a data set. For example, if your exam score is better than k% of the rest of the class. That means your exam score is at the kth percentile.
If your test score is at the 32nd percentile it can be interpreted as follows:
-Your test score is better than only 32 percent of the other scores recorded for the test.
-32 percent of the people who took the admission test have scores which are lower than yours.
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.
For more details, refer to the link:
https://brainly.com/question/19554667
?
1. What is the solution of the system
3x - y = 14,
4x + y = 21
A (-5,3)
B (4, -2)
C (5, 1)
Answer:
The answer to your question is letter C
Step-by-step explanation:
Equations
3x - y = 14 ----------- (I)
4x + y = 21 ----------- (II)
Solve the system of equations by the elimination method.
3x - y = 14
4x + y = 21
7x 0 = 35
Solve for x, divide both sides by 7
7x/ 7 = 35/7
x = 5
Substitute x in equation I
3(5) - y = 14
15 - y = 14
Subtract 15 in both sides
15 - 15 - y = 14 - 15
Simplify
- y = -1
Multiply by -1
y = 1
Solution
(5, 1)
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]
Learn more about Trigonometric identities here:
https://brainly.com/question/33401109
#SPJ12
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]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).
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 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)
Numerical WebAssign will do simple calculations for you, using +, -, *, /, parentheses, and e-notation. For example, if a question asks for half the sum of 246 and 388, you can enter "(246+388)/2". For more complicated calculations you'll need to use a calculator and then enter your result. What is the product of 140 and 80?
Answer:
140·80=11200
Step-by-step explanation:
From exercise we have example for how we find half the sum of 246 and 388, we can enter "(246+388)/2".
Based on this example, we will calculate what is required in the task using a calculator. So we use a calculator to find the product of the following numbers, 140 and 80.
We calculate, and we get
140·80=11200
What type of sampling methods are used in selecting people for exit polls at the polling locations? Select all that apply.
a. Sample random Sampling
b. Cluster Sampling
c. Stratified Sampling
d. Convenience Sampling
e. systematic Sample
Exit polls commonly use simple random sampling, stratified sampling, and systematic sampling to ensure a representative sample of voters. Cluster sampling may be used, but convenience sampling is typically avoided to prevent bias.
Explanation:Exit polls typically employ a variety of sampling methods to ensure a representative sample of voters. These methods can include simple random sampling, where each person has an equal chance of being selected, and stratified sampling, which involves dividing the population into subgroups and sampling from each. Systematic sampling can also be used by selecting every nth person to participate in the poll.
Cluster sampling could be utilized if the population is divided into clusters, and a random selection of clusters is made. Convenience sampling, however, is generally not used in exit polling because it can introduce bias. Instead, methods that provide each individual an equal chance of selection are preferred to obtain a representative sample.
Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t)
Answer:
Since the equation was missing, I solved it with another equation and got an answer of T(0) = <3j / 5 + 4k / 5>.
Please see my explanation. I hope this helps
Step-by-step explanation:
The question asked us to find out unit tangent vector.
Recall unit vector = vector / magnitude of vector
Since the question is missing with an equation. I suppose an equation.
r(t)=Cost i, 3t j, 2Sin(2t) k at t=0
Lets take out differentiation
r'(t) = <(-Sint), 3, 2(Cos(2t)(2))>
r'(t)= <-Sint, 3, 4Cos(2t)>
Now substitute t=0 in the differentiate found above.
r'(0)= <-Sin(0), 3, 4Cos(2*0)>
r'(0)= <0, 3, 4(1)>
r'(0)= <0,3,4>
vector r'(0)=<0i, 3j, 4k>
Now lets find out magnitude of vector
|r'(0)| = [tex]\sqrt{0^{2}+3^{2}+4^{2} }[/tex]
|r'(0)| = [tex]\sqrt{0+9+16}[/tex]
|r'(0)| = [tex]\sqrt{25}[/tex]
|r'(0)| = 5
Unit Tangent Vector
T(0) = <0, 3, 4> / 5
T(0) = <3j / 5 + 4K / 5>
To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector. To find the unit tangent vector, we divide the velocity vector by its magnitude.
Explanation:To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector:
v(t) = -Aw*sin(wt)i + Aw*cos(wt)j
To find the unit tangent vector, we divide the velocity vector by its magnitude:
T(t) = (v(t))/(|v(t)|) = (-Aw*sin(wt)i + Aw*cos(wt)j)/(sqrt((Aw*sin(wt))^2 + (Aw*cos(wt))^2))
So, the unit tangent vector at any point is T(t).
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.
If a linear system has the same number of equations and variables, then it must have a unique solution.
True or false
Answer:
The given statement is false.
Step-by-step explanation:
We are given the following statement:
"If a linear system has the same number of equations and variables, then it must have a unique solution."
The given statement is false because If a linear system has the same number of equations and variables, then it may have unique solution, no solution or infinitely many solution.
There could be three types of solution
no solution, unique solution and infinitely many solutions,depending on the augmented matrix and coefficient matrix of the linear system.
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)
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.
Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.
Answer:
[tex]A)\,\,det(A)=1[/tex]
[tex]B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] [/tex]
Step-by-step explanation:
[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|[/tex]
Expanding with first row
[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1[/tex]
To find inverse we first find cofactor matrix
[tex]C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\[/tex]
[tex]C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\[/tex]
Cofactor matrix is
[tex]C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right][/tex]
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%
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:
Say you want to provide a certain candy for Halloween. You expect around K kids to come to your house, and each kid is to be given three pieces of candy. Each bag of candy you can buy contains N pieces of candy for P dollars. Which algebraic expression will tell you how much should you expect to have to pay (M)
Answer:
Step-by-step explanation:
You expect around K kids to come to your house, and each kid is to be given three pieces of candy. This means that the total number of candies that you would buy is
3 × K = 3K
Each bag of candy you can buy contains N pieces of candy for P dollars.
Therefore,
If N pieces of candy cost $P, then
3K pieces of candy would cost $M
Therefore, the algebraic expression
to tell you how much should you expect to have to pay (M) would be
M = 3KP/N
The algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]
The given parameters are:
Kids = KEach = 3Unit Rate = P for N pieces i.e. P/NThe total amount paid for N pieces of candy is calculated as:
[tex]Total = Kids \times 3 \times Unit\ Rate[/tex]
So, we have:
[tex]Total = K \times 3 \times \frac PN[/tex]
Evaluate the product
[tex]Total = \frac{3KP}N[/tex]
Rewrite the above equation as:
[tex]M= \frac{3KP}N[/tex]
Hence, the algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]
Read more about algebraic expressions at:
https://brainly.com/question/2972832
Determine if the statement is true or false. If a linear system has the same number of equations and variables, then it must have a unique solution. True False Justify your answer.
Answer:
False
Step-by-step explanation:
Consider the equations with the same number of equations and variables as shown below,
Case 1
[tex]x_{1} + x_{2} = 0\\x_{1} + x_{2} = 1[/tex]
This equation has no solution because it is not possible to have two numbers that give a sum of 0 and 1 simultaneously.
Case 2
[tex]x_{1} + x_{2} = 1\\2x_{1} + 2x_{2} = 2[/tex]
This equation has infinitely many possible solutions.
Therefore it is FALSE to say a linear system with the same number of equations and variables, must have a unique solution.
The statement that a linear system with the same number of equations and variables must have a unique solution is false. Other considerations, such as whether the system is consistent or inconsistent and dependent or independent, can impact the amount of solutions a system has.
Explanation:The statement is false. Even if a linear system has the same number of equations and variables, it does not necessarily mean that it will have a unique solution. Rather, whether a system has a unique solution, no solutions, or infinitely many solutions depends on whether the system is consistent or inconsistent and dependent or independent.
For instance, consider two linear equations: x + y = 5 and 2x + 2y = 10. Even though these two equations have the same number of variables and equations, they represent the same line and thus have infinitely many solutions. Similarly, consider the system x + y = 5 and x + y = 6. These two equations have also the same number of equations and variables but they are parallel lines and do not intersect, so this system does not have any solution.
So, the number of variables and equations is not always enough to determine the number of solutions to a linear system.
Learn more about Linear System here:https://brainly.com/question/36898231
#SPJ3
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.
Chemistry is an experimental science in which a consistent system of measurements is used. Scientists throughout the world use the International System of Units (Système International or SI), which is based on the metric system. However, the metric system in not used on an everyday basis in the United States. It is often necessary to convert between the English system, which is used in the United States, and the SI system. One problem with the SI system is the size of the units sometimes turns out to be inconveniently large or small. Therefore it is often necessary to make conversions within the SI system. For this reason, SI units are modified through the use of prefixes when they refer to either larger or smaller quantities.
Part A
In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 22.0mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions:1km=0.6214mi and 1gal=3.78L.
Part B
While in Europe, if you drive 115km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 30.0mi/gal ? Assume that 1euro=1.26dollars.
Part C
A sprinter set a high school record in track and field, running 200.0 m in 20.9s . What is the average speed of the sprinter in kilometers per hour?
Part D
A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece. How many kilograms of dietary fat are in a box containing 4.00lb of candy?
Answer:
(a) 142 km trip requires 15.16 Liters of Gasoline
(b) The money spent is 69.34 euros or $87.37.
(c) The average speed of sprinter is 34.45 km/h.
(d) 4 lb of candy contains 0.559 kg of dietary fat.
Step-by-step explanation:
(a)
Given that;
Mileage = 22 mi/gal
Converting it to km/L
Mileage = (22 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)
Mileage = 9.36 km/L
Now, the gasoline required for 142 km trip:
Gasoline Required = (Length of trip)/(Mileage)
Gasoline Required = (142 km)/(9.36 km/L)
Gasoline Required = 15.16 L
(b)
Given that;
Mileage = 30 mi/gal
Converting it to km/L
Mileage = (30 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)
Mileage = 12.77 km/L
Now, the gasoline required for 142 km trip:
Gasoline Required = (Length of trip)/(Mileage)
Gasoline Required = (115 km/day)(7 days/week)/(12.77 km/L)
Gasoline Required = 63 L/week
Now, we find the cost:
Weekly Cost = (Gasoline Required)(Unit Cost)
Weekly Cost = (63 L/week)(1.1 euros/L)
Weekly Cost = 69.34 euros/week = $87.37
since, 1 euro = $1.26
(c)
Average Speed = (Distance Travelled)/(Time Taken)
Average Speed = 200 m/20.9 s
Average Speed = (9.57 m/s)(3600 s/1 h)(1 km/ 1000 m)
Average Speed = 34.45 km/h
(d)
22.7 g piece contains = 7 g dietary fat
(22.7 g)(1 lb/453.592 g) piece contains = (7 g)(1 kg/1000 g) dietary fat
0.05 lb piece contains = 0.007 kg dietary fat
1 lb piece contains = (0.007/0.05) kg dietary fat
4 lb piece contains = 4(0.007/0.05) kg dietary fat
4 lb piece contains = 0.559 kg dietary fat
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.
An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 6.3 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 130 engines and the mean pressure was 6.5 pounds/square inch. Assume the standard deviation is known to be 0.8. A level of significance of 0.02 will be used. Determine the decision rule. Enter the decision rule.
Answer:
We conclude that the valve performs above the specifications.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 6.3 pounds per square inch
Sample mean, [tex]\bar{x}[/tex] = 6.5 pounds per square inch
Sample size, n = 130
Alpha, α = 0.02
Population standard deviation, σ = 0.8
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 6.3\text{ pounds per square inch}\\H_A: \mu > 6.3\text{ pounds per square inch}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{6.5 - 6.3}{\frac{0.8}{\sqrt{130}} } = 2.85[/tex]
Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]
Decision rule:
If the calculated statistic is greater than the the critical value, we reject the null hypothesis and if the calculated statistic is lower than the the critical value, we accept the null hypothesis
Since,
[tex]z_{stat} > z_{critical}[/tex]
We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.
Thus, we conclude that the valve performs above the specifications.
Emilia and Ferdinand took the same freshman chemistry course: Emilia in the fall, Ferdinand in the spring. Emilia made an 83 on the common final exam that she took, for which the mean was 76 and the standard deviation 8. Ferdinand made a 79 on the common final exam that he took, which was more difficult, since the mean was 65 and the standard deviation 12. The one who has a higher z-score did relatively better. Was it Emilia or Ferdinand?
Answer:
[tex] z = \frac{83-76}{8}=0.875[/tex]
[tex] z = \frac{79-65}{12}=1.167[/tex]
As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Emilia case
Let X the random variable that represent the scores of a test, and we know that
Where [tex]\mu=76[/tex] and [tex]\sigma=8[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Since Emilia made 83 points we can find the z score like this:
[tex] z = \frac{83-76}{8}=0.875[/tex]
Ferdinand case
Let X the random variable that represent the scores of a test, and we know that
Where [tex]\mu=65[/tex] and [tex]\sigma=12[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Since Ferdinand made 79 points we can find the z score like this:
[tex] z = \frac{79-65}{12}=1.167[/tex]
As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.
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.
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?