Answer:
d = 13 / 5 units
Step-by-step explanation:
Given:
- Point P = (-2 , 3 )
- line : 3x - 4y + 5 = 0
Find:
- Use projection to show the given formula
- use the formula to find the distance from given point and line.
Solution:
- Let line L be defined as:
a*x + b*y + c = 0
- Choose A as fixed point on the line L whose coordinates are ( m , k ).
- Now the coordinates of any point (x , y ) can be given by the line L:
a*( x - m ) + b*( y - k ) + c = 0
- We did that by subtracting the coordinates (x,y) from (m,k).
- The above line can represented by dot product of constants (a , b ) to vector coordinates between points (x,y) to (m,k):
( a , b ) . ( x - m , y - k) = 0
- For above expression to be true i.e the dot product of two vectors is only zero when the vectors are orthogonal. So vector (a , b ) is always perpendicular to every vector in direction of line L
- The vector:
v = ( x - m , y - k)
- It may represent the line connecting the points (x_1 , y_1) and (m, k).
Then the distance d from the point P_1: (x_1, y_1) to the line L is given by the magnitude of the scalar projection of v onto a, because the latter is the length of the side of a right triangle with two of the vertices being P_1 and
( m , k), and the other vertex lying on L So:
d = | component v along x |
d = | v . ( a , b ) | / | (a , b ) |
d = | (x_1 - m , y_1 - k ) . ( a , b)| / sqrt(a^2 + b^2)
d = | ax_1 + by_1 - (a*m + b*k) | / sqrt(a^2 + b^2)
Where point A (m,k) lies on line hence;
a*m + b*k + c = 0
c = - (a*m + b*k)
Hence,
d = | ax_1 + by_1 + c | / sqrt(a^2 + b^2)
- Given point P (-2,3) and line L : 3x - 4y + 5 = 0
where,
(x_1 , y_1) = (-2 , 3 )
(a , b) = ( 3 , -4 )
c = 5
-Evaluate:
d = | 3*(-2) + -4*3 + 5 | / sqrt(3^2 + 4^2)
d = | - 13 | / 5
d = 13 / 5 units
The distance from the point (-2,3) to the line [tex]\(3x - 4y + 5 = 0\)[/tex] is [tex]\(\frac{|3(-2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}}\)[/tex].
To find the distance from a point [tex]\(P_1(x_1, y_1)\)[/tex] to a line [tex]\(ax + by + c = 0\)[/tex], one can use the formula derived from scalar projection:
[tex]\[ \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \][/tex]
Here, [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients of the line, and [tex]\(x_1\)[/tex] and [tex]\(y_1\)[/tex] are the coordinates of the point. The numerator represents the absolute value of the expression [tex]\(ax_1 + by_1 + c\)[/tex], which is the scalar projection of the vector from the origin to the point onto the normal vector of the line. The denominator is the magnitude of the normal vector of the line.
Given the point [tex]\(P_1(-2, 3)\)[/tex] and the line [tex]\(3x - 4y + 5 = 0\)[/tex], we substitute [tex]\(x_1 = -2\)[/tex] and [tex]\(y_1 = 3\)[/tex] into the formula:
[tex]\[ \text{Distance} = \frac{|3(-2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}} \][/tex]
Now, we calculate the numerator:
[tex]\[ 3(-2) - 4(3) + 5 = -6 - 12 + 5 = -13 + 5 = -8 \][/tex]
The absolute value of [tex]\(-8\)[/tex] is [tex]\(8\)[/tex], so the numerator becomes [tex]\(8\)[/tex].
Next, we calculate the denominator, which is the magnitude of the normal vector of the line:
[tex]\[ \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \][/tex]
Finally, we divide the absolute value we found in the numerator by the denominator to get the distance:
[tex]\[ \text{Distance} = \frac{8}{5} \][/tex]
Therefore, the distance from the point (-2,3) to the line [tex]\(3x - 4y + 5 = 0\)[/tex] is [tex]\(\frac{8}{5}\)[/tex].
A government bureau keeps track of the number of adoptions in each region. The accompanying histograms show the distribution of adoptions and the population of each region. a) What do the histograms say about the distributions? b) Why do the histograms look similar? c) What might be a better way to express the number of adoptions?
The histograms indicate that higher population regions tend to have more adoptions. They are similar as adoption rates and population sizes are interlinked. A better representation might be the adoption rate per population quota, which shows comparison between regions clearer.
Explanation:a) The histograms show the "distribution of adoptions" and the "population of each region." We can infer that the distribution of adoptions largely mirrors the population distribution, meaning that regions with larger populations tend to have more adoptions.
b) The histograms look similar because adoption rates and population size are related. If a region has a larger population, it likely has more families, hence more potential for adoption.
c) A better way to express the number of adoptions might be to calculate the adoption rate per population. For example, the number of adoptions per 1,000 or 10,000 population members. This way, it directly relates the number of adoptions to the size of the population, and provides a percentage or ratio rather than absolute numbers. This method can be more helpful in making comparisons between regions.
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The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time.com). The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars.
Private Colleges
53.8 42.2 44.0 34.3 44.0
31.6 45.8 38.8 50.5 42.0
Public Colleges
20.3 22.0 28.2 15.6 24.1 28.5
22.8 25.8 18.5 25.6 14.4 21.8
(a)
Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for private colleges. (Round the standard deviation to two decimal places.)
sample mean $ thousand sample standard deviation $ thousand
Compute the sample mean (in thousand dollars) and sample standard deviation (in thousand dollars) for public colleges. (Round the standard deviation to two decimal places.)
sample mean $ thousand sample standard deviation $ thousand
Answer:
A) Private colleges:
mean = 42.7, SD = 6.72
In thousand dollars:
mean = $42700, SD= $6720
B) Public colleges:
mean = 22.3, SD = 4.53
In thousand dollars:
mean = $22300, SD= $4530
Step-by-step explanation:
A) Private Colleges:
Mean:
Total no. of samples = n =10
Sample values in dollar = x= [53.8, 42.2, 44.0, 34.3, 44.0,31.6, 45.8, 38.8, 50.5, 42.0]
Sum of samples = ∑x= 427
[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{427}{10}\\\\\bar{x}=42.7\\[/tex]
Sample mean in thousand dollars is $ 42700.
Standard Deviation:
Formula for standard deviation of sample data is
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(53.8-42.7)^2+(42.2-42.7)^2+...+(42.0-42.7)^2\\\\\sum(x_i-\bar{x})^2=123.21+0.25+ 1.69+ 70.56+ 1.69+ 123.21+ 9.61+ 15.21+60.84+0.49\\\\\sum(x_i-\bar{x})^2=406.67\\\\\sigma=\sqrt{\frac{406.67}{9}}\\\\\sigma=6.72[/tex]
Standard deviation in thousand dollars is $ 6720.
B) Public Colleges:
Mean:
Total no. of samples = n =12
Sample values in dollar = x= [20.3, 22.0, 28.2, 15.6, 24.1, 28.5,22.8, 25.8, 18.5, 25.6, 14.4, 21.8]
Sum of samples = ∑x= 267.6
[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{267.6}{12}\\\\\bar{x}=22.3\\[/tex]
Sample mean in thousand dollars is $ 22300.
Standard Deviation:
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(20.3-22.3)^2+(22.0-22.3)^2+...+(21.8-22.3)^2\\\\\sum(x_i-\bar{x})^2=4+ 0.09+34.81+ 44.89+ 3.24+ 38.44+0.25+ 12.25+14.44+10.89+62.41+0.25\\\\\sum(x_i-\bar{x})^2=225.96\\\\\sigma=\sqrt{\frac{225.96}{11}}\\\\\sigma=4.53[/tex]
Standard deviation in thousand dollars is $ 4530.
in a dataset with a minimum value of 54.5 and a maximum value of 98.6 with 300 observations, there are 180 points less than 81.2. Find the percentile ofr 81.2
Answer:
60th percentile.
Step-by-step explanation:
When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.
In this problem, we have that:
300 observations, there are 180 points less than 81.2.
180/300 = 0.6
So 60% of the observation are lower than 81.2, which means that 81.2 is the 60th percentile.
The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the other 5 have selected desktops. Suppose that four computers are randomly selected. (a) How many different ways are there to select four of the nine computers to be set up?
(b) What is the probability that exactly three of the selected computers are desktops?
(c) What is the probability that at least three desktops are selected?
Answer:
a) There are 126 different ways are there to select four of the nine computers to be set up.
b) 31.75% probability that exactly three of the selected computers are desktops.
c) 35.71% probability that at least three desktops are selected.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, there are no replacements. Which means that after one of the 9 computers is selected, there will be 8 computers.
Also, the order that the computers are selected is not important. For example, desktop A and desktop B is the same outcome as desktop B and desktop A.
These are the two reasons why the combinations formula is important to solve this problem.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
(a) How many different ways are there to select four of the nine computers to be set up?
Four computers are selected from a set of 9.
So
[tex]T = C_{9,4} = \frac{9!}{4!(9-4)!} = 126[/tex]
There are 126 different ways are there to select four of the nine computers to be set up.
(b) What is the probability that exactly three of the selected computers are desktops?
Desired outcomes:
3 desktops, from a set of 5
One laptop, from a set of 4.
So
[tex]D = C_{5,3}*C_{4,1} = 40[/tex]
Total outcomes:
From a), 126
Probability:
[tex]P = \frac{40}{126} = 0.3175[/tex]
31.75% probability that exactly three of the selected computers are desktops.
(c) What is the probability that at least three desktops are selected?
Three or four
Three:
[tex]P = \frac{40}{126}[/tex]
Four:
Desired outcomes:
4 desktops, from a set of 5
Zero laptop, from a set of 4.
[tex]D = C_{5,4}*C_{4,0} = 5[/tex]
[tex]P = \frac{5}{126}[/tex]
Total(three or four) probability:
[tex]P = \frac{40}{126} + \frac{5}{126} = \frac{45}{126} = 0.3571[/tex]
35.71% probability that at least three desktops are selected.
A fair coin is tossed three times. What is the probability that exactly two heads occur, given that a. the first outcome was a tail b. the first two outcomes were heads c. the first two outcomes were tails
Answer:
a) So 25% probability that exactly two heads occur, given that the first outcome was a tail.
b) 50% probability that exactly two heads occur, given that the first two outcomes were heads.
c) 0% probability that exactly two heads occur, given that the first two outcomes were tails.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
We have the following sample space, that is, the possible outcomes:
In which h is heads and t is tails
h - h - h
h - h - t
h - t - h
h - t - t
t - h - h
t - h - t
t - t - h
t - t - t
What is the probability that exactly two heads occur, given that
a. the first outcome was a tail
Four outcomes in which the first outcome was a tail. They are:
t - h - h
t - h - t
t - t - h
t - t - t
In only 1 one them, exactly two heads occur.
1/4 = 0.25
So 25% probability that exactly two heads occur, given that the first outcome was a tail.
b. the first two outcomes were heads
Two possibilities in which the first two outcomes were heads.
h - h - h
h - h - t
In 1 of them, we have exactly two heads.
1/2 = 0.5
So 50% probability that exactly two heads occur, given that the first two outcomes were heads.
c. the first two outcomes were tails
Two possibilities in which the first two outcomes were tails.
t - t - h
t - t - t
In none of them we have exactly 2 heads.
0% probability that exactly two heads occur, given that the first two outcomes were tails.
The probability of getting exactly two heads in three coin tosses is 5/8.
Explanation:To find the probability of exactly two heads occurring in three coin tosses, we need to consider three different scenarios:
If the first outcome was a tail: In this case, we have two remaining tosses and we need both of them to be heads. The probability of getting a head on any single toss is 1/2, so the probability of getting two heads after a tail is (1/2) * (1/2) = 1/4.
If the first two outcomes were heads: In this case, we have one remaining toss and we need it to be a tail. The probability of getting a tail on the remaining toss is 1/2, so the probability of getting exactly two heads after two heads is (1/2) = 1/2.
If the first two outcomes were tails: In this case, we again have one remaining toss and we need it to be a head. The probability of getting a head on the remaining toss is 1/2, so the probability of getting exactly two heads after two tails is (1/2) = 1/2.
Adding up the probabilities of these three scenarios, the overall probability of getting exactly two heads in three coin tosses is 1/4 + 1/2 + 1/2 = 5/8.
Find the sample space for the experiment.
Two county supervisors are selected from five supervisors, A, B, C, D and E, to study a recycling plan.
Answer:
The sample space is:
[tex]AB,\ AC,\ AD,\ AE,\\BC,\ BD,\ BE\\CD,\ CE\\DE[/tex]
Step-by-step explanation:
The sample space of an experiment is a set of all the possible values that satisfies that experiment.
The five supervisors are A, B, C, D and E.
Two are to selected.
The number of ways to select 2 supervisors from 5 is: [tex]{5\choose 2}=\frac{5!}{2!(5-2)!} =\frac{5!}{2!\times 3!} = 10[/tex]
The sample space will consist of 10 combinations.
The sample space is:
[tex]AB,\ AC,\ AD,\ AE,\\BC,\ BD,\ BE\\CD,\ CE\\DE[/tex]
Among the following, the BEST example of qualitative data would include a. countywide census of speakers of more than one language. b. ethnic composition of a community, by percentage. c. average community income levels, by block. d. field notes recorded during participant observation
Answer:
d. field notes recorded during participant observation
Step-by-step explanation:
Qualitative data are data that can be gathered and described by observation, interviews and evaluation. Such data are not numerically based but are based on features of the phenomenon under study. It is also known as categorical data.
Methods of Qualitative data includes :
Focus groups, keeping records, direct interviews, case studies etc.
Suppose your firm had the following taxable income amounts: 2015 ($5 million) operating loss 2016 $4 million 2017 $4 million 2018 $4 million After you "carry forward" the operating loss, what is the effective taxable income for 2017
Answer:
$7,000,000
Step-by-step explanation:
The taxable income is the amount of money (income earned or unearned) by an individual or an organization that creates a potential tax liability.
The formula is shown below:
Taxable Income Formula = Gross Total Income – Total Exemptions – Total Deductions
Note that $5,000,000 for 2015 is an operating loss which is a deduction.
Gross Total Income = $4000000+$4000000+$4000000 = $12,000,000
Total Exemption = 0
Total Deductions = $5,000,000
Taxable Income Formula = $12,000,000 - $5,000,000 = $7,000,000
The effective taxable income for 2017 after carrying forward the operating loss is -$1 million.
Explanation:The question pertains to a business scenario and involves calculating the effective taxable income after carrying forward an operating loss. To calculate the effective taxable income for 2017, we need to consider the operating losses from previous years. In this case, your firm had a $5 million operating loss in 2015. This loss can be carried forward to offset taxable income in future years.
In 2016, the taxable income is $4 million.In 2017, the operating loss from 2015 can be used to offset the taxable income. Therefore, the effective taxable income for 2017 is $4 million - $5 million = -$1 million. Since the result is negative, there is no taxable income for 2017.Therefore, the effective taxable income for 2017 after carrying forward the operating loss is -$1 million.
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In Saras class, 2/5 of the students ride a bus, and 1/3 ride a car to school. The rest walk to school. Explain how you can find the fraction of students who walk to school. Find the fraction of students who ride a bus or car to school. Draw a diagram and use an equation to find your answer. Find the fraction of students who walk to scool. Draw a diagram and use an equation to find your answer.
Answer:
See explanation
Step-by-step explanation:
In Saras class, [tex]\frac{2}{5}[/tex] of the students ride a bus, and [tex]\frac{1}{3}[/tex] ride a car to school.
The rest
[tex]1-\dfrac{2}{5}-\dfrac{1}{3}=\dfrac{15-6-5}{15}=\dfrac{4}{15}[/tex]
walk to school (simply subtract from one whole the fractions of students).
The fraction of students who ride a bus or car to school is
[tex]\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{6+5}{15}=\dfrac{11}{15}.[/tex]
See attached diagram for pictorial representations (red diagram shows the fraction of students who ride a bus, blue - ride a car, green - walk to school)
The fraction of students who walk to school is [tex]$\boxed{\frac{4}{15}}$[/tex].
To find the fraction of students who walk to school, we first consider the fractions of students who ride a bus or car to school. According to the information given:
- The fraction of students who ride a bus is [tex]$\frac{2}{5}$[/tex].
- The fraction of students who ride a car is [tex]$\frac{1}{3}$[/tex].
To find the total fraction of students who use transportation (bus or car), we add these two fractions together. However, we must first find a common denominator to add them properly. The least common multiple (LCM) of 5 and 3 is 15, so we convert both fractions to have a denominator of 15:
[tex]- $\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$ (for the bus riders)[/tex]
[tex]- $\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$ (for the car riders)[/tex]
Now we can add these fractions:
[tex]- $\frac{6}{15} + \frac{5}{15} = \frac{11}{15}$[/tex]
This represents the fraction of students who either ride a bus or a car to school. To find the fraction of students who walk to school, we subtract this fraction from the whole, which is 1 (since the whole class is represented by 1, or [tex]$\frac{15}{15}$):[/tex]
[tex]- $1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}$[/tex]
Therefore, the fraction of students who walk to school is [tex]$\frac{4}{15}$[/tex].
To visualize this with a diagram, we can imagine a whole circle representing the entire class. We divide this circle into 15 equal parts (since our denominators are 15). We shade 6 parts to represent the bus riders and another 5 parts to represent the car riders. The remaining unshaded parts, which are 4 in number, represent the students who walk to school.
The equation representing the total fraction of students who walk or use transportation is:
[tex]\[ \text{Bus riders} + \text{Car riders} + \text{Walkers} = \text{Whole class} \] \[ \frac{6}{15} + \frac{5}{15} + \text{Walkers} = 1 \][/tex]
Solving for Walkers:
[tex]\[ \text{Walkers} = 1 - \left( \frac{6}{15} + \frac{5}{15} \right) \] \[ \text{Walkers} = 1 - \frac{11}{15} \] \[ \text{Walkers} = \frac{15}{15} - \frac{11}{15} \] \[ \text{Walkers} = \frac{4}{15} \][/tex]
Thus, the fraction of students who walk to school is[tex]$\boxed{\frac{4}{15}}$.[/tex]
The acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors normal to the lines or by the vectors parallel to the lines. Also, note that the vector ai + bj is perpendicular to the line ax + by = c Find the acute angles between the lines: x + √3y = 1 and (1 - √3)x + (1 + √3)y = 8 Thank you!
Answer:
[tex]45^{\circ}[/tex]
Step-by-step explanation:
We are given that two lines equation
[tex]x+\sqrt 3y=1[/tex]...(1)
[tex](1-\sqrt 3)y+(1+\sqrt 3)y=8[/tex]
Compare with the equation of line
ax+by+c=0
[tex]a_1=1,b_1=\sqrt 3[/tex]
[tex]a_2=(1-\sqrt 3),b_2=(1+\sqrt 3)[/tex]
The angle between two lines =Angle between two vectors
The angle between two vector
[tex]a_1i+b_1j[/tex] and
[tex]a_2i+b_2j[/tex]
is given by
[tex]cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}[/tex]
Using the formula
Therefore, the angle between two lines
[tex]cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{\sqrt{(1)+(\sqrt 3)^2}\times \sqrt{(1-\sqrt 3)^2+(1+\sqrt 3)^2}}[/tex]
[tex]cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{\sqrt{1+3}\times\sqrt{1+3-2\sqrt 3+1+3+2\sqrt 3}}[/tex]
[tex]cos\theta=\frac{4}{2\times\sqrt 8}=\frac{2}{2\sqrt 2}[/tex]
[tex]cos\theta=\frac{1}{\sqrt 2}[/tex]
[tex]cos\theta=cos45^{\circ}[/tex]
By using [tex]cos45^{\circ}=\frac{1}{\sqrt 2}[/tex]
[tex]\theta=45^{\circ}[/tex]
Hence, the angle between two lines =45 degree
A company's revenue can be modeled by r -2-23t+64 where r s the revenue (in milions revenue was or will be $8 million. of dollars) for the year that is t years since'2005 Predict when the Predict when the revenue was or will be $8 million. (Use a comma to separate answers as needed Round to the nearest year as needed)
Answer: In year 2008 ( After 3 years since 2005)
Explanation: The given equation is incomplete. The equation for the revenue is assumed to be r = 2-23t + 64
When the revenue reaches $ 8 million, the equation is shown below:
8 = 2 -23t + 64
23t = (2 + 64) / 8
t = 2.52 years
Rounding up the years,
t = 3 years
3 years after 2005 = 2005 + 3 = 2008
In year 2008, the revenue will be $8 million.
A container initially containing10 L of water in which there is 20 g of salt dissolved. A solution containing 4 g/L of salt is pumped into the container at a rate of 2 L/min, and the well-stilled mixture runs out at a rate of 1 L/min. How much salt is in the tank after 40 min
Answer:
[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]
Step-by-step explanation:
For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by [tex] V= 10 +t[/tex]
Since the initial volume is 10 L and the volume increase at a rate of 1L/min.
For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:
[tex] \frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}[/tex]
Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of [tex] \frac{A}{10+t}[/tex]
So then we can reorder the differential equation like this:
[tex] \frac{dA}{dt} +\frac{A}{10+t} =8[/tex]
We find the solution using the integration factor:
[tex] \mu = -\int \frac{1}{10+t} dt = -ln(10+t)[/tex]
And then the solution would be given by:
[tex] A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})[/tex]
And if we simplify this we got:
[tex] A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)[/tex]
And after do the integral we got:
[tex] A= \frac{c}{10+t} +4 (10 +t) [/tex]
And using the initial condition t=0 A= 20 we have this:
[tex] 20 = \frac{c}{10} +40[/tex]
[tex] c= -200[/tex]
So then we have this function for the solution of A:
[tex] A= \frac{-200}{10+t} +4 (10 +t) [/tex]
And now replacinf t= 40 we got:
[tex] A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196 [/tex]
The amount of salt in the tank after 40 minutes is determined by first calculating the total amount of salt added to the tank, then considering the amount of salt-water mixture that exited the tank during this period due to the outgoing flow.
Explanation:This question relates to a typical problem within the field of differential equations. You are asked to find the amount of salt in the tank after 40 minutes. The initial amount of salt in the tank is 20g, and additional salt-water solution is being pumped in at a rate of 2L/min with a salt concentration of 4g/L, hence adding 8g of salt per minute. At the same time, the well-stirred mixture is flowing out at a rate of 1L/min.
Therefore, after 40 minutes:
Additional salt added = 8g/min * 40 min = 320g.
Total salt content now = initial salt content + added salt = 20g + 320g = 340g.
However, within these 40 minutes,40 L of the mixture has flown out. Since the mixture is well-stirred, it retains the same concentration of salt as the mixture in the tank. Hence, if x is the amount of salt left in the tank, it will be on average the same concentration as that which flowed out and can be represented as follows: x/50 = (340-x)/40.
Solving for x we get the amount of salt left in the tank after 40 minutes.
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Following are measurements of soil concentrations (in mg/kg) of chromium (Cr) and nickel (Ni) at 20 sites in the area of Cleveland, Ohio. These data are taken from the article "Variation in North American Regulatory Guidance for Heavy Metal Surface Soil Contamination at Commercial and Industrial Sites" (A. Jennings and J. Ma, J Environment Eng, 2007:587–609).
Cr: 34 1 511 2 574 496 322 424 269 140 244 252 76 108 24 38 18 34 3O 191
Ni: 23 22 55 39 283 34 159 37 61 34 163 140 32 23 54 837 64 354 376 471
(a) Construct a histogram for each set of concentrations.
(b) Construct comparative boxplots for the two sets of concentrations.
(c) Using the boxplots, what differences can be seen between the two sets of concentrations?
Answer:
a) see attached
b) see attached
Step-by-step explanation:
There is existence of outlier in the Ni data and there is none is Cr data.
########################################
# You can try this out in R programming
cr = c(34, 1, 511, 2, 574, 496, 322, 424, 269, 140, 244, 252, 76, 108, 24,
38, 18, 34, 30, 191)
Ni = c(23, 22, 55, 39, 283, 34, 159, 37, 61, 34, 163, 140, 32,
23, 54, 837, 64, 354, 376, 471)
par(mfrow=c(1,2))
hist(cr, col='green')
hist(Ni, col='brown')
par(mfrow=c(1,2))
boxplot(cr, main = 'Boxplot of Cr')
boxplot(Ni, main = 'Boxplot of Ni')
boxplot(cr, Ni)
The question involves constructing histograms and boxplots for two sets of data on soil concentrations of certain metals. After construction, these should be analysed for differences in central tendency, spread, variability, skewness, or presence of outliers.
Explanation:This question seems to be related to Mathematics, specifically Statistics and Data Analysis. Firstly, to construct a histogram, you need to group the data into 'bins' or 'intervals', count how many data points fall into each bin, and then plot these counts as bars in your histogram. It's similar for both Chromium (Cr) and Nickel (Ni) data.
To construct comparative boxplots, it's essential to: calculate the quartiles (Q1, Q2 aka median, and Q3), identify the minimum and maximum points, and define possible outliers. With these statistical measures, you can draw the boxplots for both sets and analyse them. The dataset Cr seems to have several values clustered in the low range while Ni seems to have more spread, which can be verified by the boxplot comparison.
The interpretation of the boxplots should reflect on any differences in central tendency and spread between the two sets of concentrations. The variability, skewness, or outliers could also be significantly different. Any such inferred details from the boxplots may hint towards different contamination levels or heterogeneity in each of the datasets.
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PLEASE HELP!!!! ill give 100 points!!
SHOW ALL WORK PLEASE AND THANK YOU
The area of a triangle is 1/2 x base x height
Area = 1/2 x 24 x 28 = 336 square inches.
Because the corners are rounded the are would actually be just under 336 but no radius was given to get the exact area.
Answer:
yes
Step-by-step explanation:
Determine whether the lines
L1: x=23+6t, y=12+3t, z=19+5t
and
L2: x=-9+7t, y=-7+5t, z=-12+8t
intersect, are skew, or are parallel. If they intersect, determine the point of intersection; if not leave the remaining answer blanks empty.
Do/are the lines:
Point of intersection: (, ,)
Answer:
skew lines
Step-by-step explanation:
Given are two lines in 3 dimension as
[tex]L1: x=23+6t, y=12+3t, z=19+5t\\L2: x=-9+7t, y=-7+5t, z=-12+8t[/tex]
To find out whether parallel or skew or intersect
If parallel direction ratios should be proportional
Direction ratios of I line are 6,3,5 and not proportional to that of II line (7,5,8)
So not parallel
If intersect we must have same point for the two lines
Let us change parameter for II line to s to avoid confusion. If intersecting, then
[tex]23+6t = -9+7s\\12+3t = -7+5s\\19+5t =12+8s\\[/tex]
6t-7s=-32 and 3t-5s = -19
Solving
t=-3 and s =2
Check this with III equation
Left side = 19-15 =4 and right side = 12+16 =28
not equal
So there cannot be any point of intersection.
These two are skew lines
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. What parameter is being tested? Upper H 0H0: sigmaσ equals= 88 Upper H 1H1: sigmaσ not equals≠ 88 What type of test is being conducted in this problem? LeftLeft-tailed test TwoTwo-tailed test RightRight-tailed test
Answer:
parameter tested=σ(population standard deviation)
two tailed
Step-by-step explanation:
We are given that null hypothesis that σ is equal to 88 and alternative hypothesis that σ is not equal to 88. The parameter that is tested here is σ.
σ denotes the population standard deviation. Thus, the population parameter σ is tested here.
The type of test depends on the alternative hypothesis we have taken. The alternative hypothesis states that σ is not equal to 88. It means that σ can either be greater than 88 or less than 88. Thus, the test can either be right tailed or left tailed. This type of test is called as two tailed test.
2. In a recent survey conducted by the International Nanny Association, 4,176 nannies were placed in a job in a given year. Only 24 of the nannies placed were men. Find the probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
Answer:
There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have that:
Desired outcomes:
The number of male nannies selected. 24 of the nannies placed were men. So the number of desired outcomes is 24.
Total outcomes:
The number of nannies selected. 4,176 nannies were placed in a job in a given year. So the number of total outcomes 4176.
Find the probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
[tex]P = \frac{24}{4176} = 0.0057[/tex]
There is a 0.57% probability that a randomly selected nanny who was placed during the last year is a male nanny (a "mannie").
Final answer:
The probability that a randomly selected nanny is a male is 0.57%.
Explanation:
The question asks us to find the probability that a randomly selected nanny who was placed during the last year is a male nanny, also referred to humorously as a "mannie." We are given that a total of 4,176 nannies were placed, and of these, only 24 were men. To find the probability, we use the formula for probability, which is the number of favorable outcomes divided by the total number of possible outcomes.
Here, the number of favorable outcomes is the number of male nannies placed, which is 24. The total number of possible outcomes is the total number of nannies placed, which is 4,176.
Thus, the probability (P) that a selected nanny is male is:
P(male nanny) = Number of male nannies / Total number of nannies
P(male nanny) = 24 / 4,176
P(male nanny) = 0.0057
To express this as a percentage, we multiply by 100:
P(male nanny) = 0.0057 * 100%
P(male nanny) = 0.57%
Therefore, the probability that a randomly selected nanny is a male is 0.57%.
The four sets A, B, C, and D each have 400 elements. The intersection of any two of the sets has 115 elements. The intersection of any three of the sets has 53 elements. The intersection of all four sets has 28 elements. How many elements are there in the union of the four sets
Answer:
the union of the four sets have 1094 elements
Step-by-step explanation:
denoting as N as the number of elements, since
N(A U B) = N(A) + N(B) - N(A ∩ B)
then
N(A U B U C) = N(A U B) + N(C) - N(A U B ∩ C ) = N(A) + N(B) - N(A ∩ B) + N(C) - [ N(A∩C) + N(B ∩ C ) - N(A ∩ B ∩ C )]
then for the union of 4 sets , we have
N (A U B U C U D) = N(A) + N(B) + N(C) +N(D) - N(A ∩ B) - N(A ∩ C) - N(A ∩ D)- N(B ∩ C) - N(B ∩ D) - N(C ∩ D) + N(A ∩ B ∩ C) + N(A ∩ B ∩ D) + N(A ∩ C ∩ D) + N(B ∩ C ∩ D) - N(A ∩ B ∩ C ∩ D)
thus replacing values for the sets, union of 2 sets , union of 3 sets and union of 4 sets
N (A U B U C U D) = ( 4*400 ) - ( 6*115 ) + ( 4*53 ) - 28 = 1094 elements
then the union of the four sets have 1094 elements
At a recent meeting, it was decided to go ahead with the introduction of a new product if "interested consumers would be willing, on average, to pay $20.00 for the product." A study was conducted, with 315 random interested consumers indicating that they would pay an average of $18.14 for the product. The standard deviation was $2.98. a. Identify the reference value for testing the mean for all interested consumers. b. Identify the null and research hypotheses for a two-sided test using both words and mathematical symbols. c. Perform a two-sided test at the 5% significance level and describe the result. d. Perform a two-sided test at the 1% significance level and describe the result. e. State the p-value as either p>0.05, p<0.05, p<0.01, or p<0.001
Answer:
There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $20.00
Sample mean, [tex]\bar{x}[/tex] = $18.14
Sample size, n = 315
Population standard deviation, σ = $2.98
a) Reference value
[tex]\mu = 20[/tex]
b) First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 20.00\text{ dollars}\\H_A: \mu \neq 20.00\text{ dollars}[/tex]
We use Two-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{18.14 - 20.00}{\frac{2.98}{\sqrt{315}} } = -11.078[/tex]
c) Calculating the p-value at 5% significance level
P-Value < 0.00001
Thus,
p<0.001
d) Calculating the p-value at 1% level of significance
P-Value < 0.00001
Thus,
p<0.001
Thus, at both 5% and 1% level of significance, the p value is lower than the significance level, we fail to accept the null hypothesis and reject it.
There is not enough evidence to support the claim that interested consumers would be willing, on average, to pay $20.00 for the product.
What is the minimum number of angles required to determine the Cartesian components of a 3D vector and why?
To determine the Cartesian components of a three-dimensional vector, one typically needs a minimum of two angles. These include the angle in the XY direction and the 'altitude' angle from the XY plane to the vector.
Explanation:To determine the Cartesian components of a three-dimensional vector, a minimum of "two angles" is required. Unlike two-dimensional vectors that need just one angle for direction, 3D vectors are more complex, existing within an x, y, and z Cartesian coordinate system. Therefore, to describe a vector fully in a 3D space, we need direction in the XY plane (measured from the positive x-axis counterclockwise), hence angle one, and the direction from the XY plane to the vector (its 'altitude'), hence angle two.
Let's consider a 3D vector A. The magnitude A and the direction angles can be used to find the components Ax, Ay, Az. Trigonometry helps us here: Ax=Acos(θ), Ay=Asin(φ), and Az=Asin(φ), where θ is the angle with the x-axis in the XY plane, and φ is the angle with the XY plane.
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Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution.
−20x + 30y = −3
8x − 12y = −3
Answer:
No solution.
Step-by-step explanation:
A system has only one solution if it ends at ax = b, in which a is different than 0.
If it ends at 0x = 0, the system has infinitely many solutions.
If it ends at a division by 0, or 0 = constant(different than 0), the system is inconsistent.
Our system is:
−20x + 30y = −3
8x − 12y = −3
I am going to multiply the top equations by 2 and the bottom equation by 5, and add them. So
-40x + 60y = -6
40x - 60y = -15
So
-40x + 40x + 60y - 60y = -6 - 15
0x + 0y = -21
We cannot divide 21 by 0, which means that this system of equations has no solution.
To determine whether using a cell phone while driving in Louisiana increases the risk of an accident, a researcher examines accident reports to obtain data about the number of accidents in which a driver was talking on a cell phone.
a. Is this a randomized experiment or an observational study?
b. What is the population being studied here?
c. Is the number of accidents where the driver was using a cell phone a qualitative or quantitative variable?
d. Is the number of accidents where the driver was using a cell phone an ordinal, nominal, continuous, or discrete variable?
e. What kind of sample is taken if the researcher only examines accident reports in the parish in which he lives?
Answer:
a) Observational study
b) Population of drivers in Louisiana who were involved in an accident as a result of talking on a cell phone while driving.
c) It is quantitative
d) It is discrete
e) The sample of the reports of all drivers who uses phone while driving and were involved in an accidents in the parish where the researcher lives only.
Step-by-step explanation:
a) The study is not done in a controlled environment. It happens randomly. As we must know not all drivers who use phone while driving have accidents. So, it is observational study.
b) We are told in the question the population of interest.
c) Since we are interested in the number of accidents that happens within the population of interest. It is a count data and therefore, it is quantitative.
d) It is count data, thus it is discrete. That is, it cannot take a decimal point. It must be whole number.
e) The kind of sample taken must be from the population of interest.
The study in question is an observational study examining the relationship between cell phone use while driving and accident incidence in Louisiana. It focuses on a quantitative, discrete variable - the count of accidents involving cell phone use - and may employ a convenience sample if only local reports are analyzed.
Explanation:To answer the question of whether using a cell phone while driving increases the risk of an accident in Louisiana, the researcher is conducting an observational study since they are examining existing data and are not manipulating any variables or conducting a randomized experiment.
The population being studied in this research is all drivers in Louisiana, and specifically, those who have been involved in accidents. When examining the number of accidents where the driver was using a cell phone, we are dealing with a quantitative variable since it is a countable number of accidents.
This quantitative variable would be classified as a discrete variable because the number of accidents can only be expressed in whole numbers; a driver cannot be involved in a fraction of an accident.
If the researcher is only examining accident reports in the parish where they live, the sample taken is a convenience sample because it is not randomly selected and might not be representative of the entire population of Louisiana drivers.
Fewer young people are driving. In year A, 67.9% of people under 20 years old who were eligible had a driver's license. Twenty years later in year B that percentage had dropped to 47.7%. Suppose these results are based on a random sample of 1,800 people under 20 years old who were eligible to have a driver's license in year A and again in year B.a. At 95% confidence, what is the margin of error and the interval estimate of the number of nineteen year old drivers in year A?b. At 95% confidence, what tis the margin of error and the interval estimate of the number of nineteen year old drivers in year B?c. Is the margin of error the same in parts (a) and (b)?
Answer:
Case a Case b
margin of error 0.0216 0.0231
Interval estimate (0.7016 , 0.6795) (0.5031 , 0.4569)
margin of error is not same in both cases.
Step-by-step explanation:
a
At 95% confidence interval the interval estimate of number of 20 year old drivers in year A can be computed as
p' ± z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]
= 0.68 ± 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]
= 0.7016 , 0.6795
the margin of error can be written as
z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]
= 1.96 [tex]\sqrt{\frac{0.68(1-0.68)}{1800} }[/tex]
= 0.0216
b
At 95% confidence interval the interval estimate of number of 20 year old drivers in year B can be computed as
p' ± z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]
= 0.48 ± 1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]
= 0.5031 , 0.4569
the margin of error can be written as
z [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]
= 1.96 [tex]\sqrt{\frac{0.48(1-0.48)}{1800} }[/tex]
= 0.0231
c
Sample size is same in case A and B but proportion is different in both cases so margin of error is different in both cases
Fifty pro-football rookies were rated on a scale of 1 to 5, based on performance at a training camp as well as on past performance. A ranking of 1 indicated a poor prospect whereas a ranking of 5 indicated an excellent prospect. The following frequency distribution was constructed. Rating 1 2 3 4 5Frequency 3 7 16 21 3 How many of the rookies received a rating of 4 or better? Number of Rookies?
Answer:
The number of rookies who scored 4 or better is 24.
Step-by-step explanation:
the frequency distribution of for the ranking of pro-football rookies is:
Ranking Frequency
1 3
2 7
3 16
4 21
5 3
Compute the number of rookies who scored 4 or better as follows:
No. of rookies with rankings 4 or more = No. of rookies with rank 4 +
No. of rookies with rank 5
= 21 + 3
= 24
Thus, the number of rookies who scored 4 or better is 24.
With a full tank of gas, if you can drive 485 miles, your car uses 18 miles per gallon. Write an equation to model this situation (use m for miles you can drive and g for gallons in tank).
Answer: m= 18g
Step-by-step explanation:
Let m= miles you can drive
g = g for gallons in tank
We know that there is directly proportional relation between the distance traveled by vehicle and the number of gallons of gas.
Then, rate of miles driven by you per gallon = [tex]\dfrac{m}{g}[/tex]
Since , rate of miles driven by you per gallon = 18 miles per gallon (given)
Then,
[tex]\dfrac{m}{g}=18\\\\ m= 18g[/tex]
Hence, the equation to model this situation : m= 18g
Last week, a coral reef grew 20.2mm taller. How much did it grow in meters?
Answer:
0.0202
Step-by-step explanation:
I got this because there is 1000 millimeters in a meter. if you divide 20.2 by 1000, you get 0.0202.
Have a good day :3
In the field of quality control, the science of statistics is often used to determine if a process is "out of control." Suppose the process is, indeed, out of control and 15% of items produced are defective. (a) If two items arrive off the process line in succession, what is the probability that both are defective? (b) If three items arrive in succession, what is the probability that two are defective?
Answer:
a) P=0.0225
b) P=0.057375
Step-by-step explanation:
From exercise we have that 15% of items produced are defective, we conclude that probabiity:
P=15/100
P=0.15
a) We calculate the probabiity that two items are defective:
P= 0.15 · 0.15
P=0.0225
b) We calculate the probabiity that two of three items are defective:
P={3}_C_{2} · 0.85 · 0.15 · 0.15
P=\frac{3!}{2!(3-2)!} · 0.019125
P=3 · 0.019125
P=0.057375
Using the binomial distribution, it is found that there is a:
a) 0.0225 = 2.25% probability that both are defective.
b) 0.0574 = 5.74% probability that two are defective.
For each item, there are only two possible outcomes, either it is defective, or it is not. The probability of an item being defective is independent of any other item, hence, the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
x is the number of successes. n is the number of trials. p is the probability of a success on a single trial.In this problem, 15% of items produced are defective, hence [tex]p = 0.15[/tex].
Item a:
Two items, hence [tex]n = 2[/tex].
The probability is P(X = 2), then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.15)^{2}.(0.85)^{0} = 0.0225[/tex]
0.0225 = 2.25% probability that both are defective.
Item b:
Three items, hence [tex]n = 3[/tex].
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.15)^{2}.(0.85)^{1} = 0.0574[/tex]
0.0574 = 5.74% probability that two are defective.
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Suppose that a color digital photo has 512 pixels per row and 512 pixels per column, and that each pixel requires two bytes of storage. How many such pictures could you have in your camera if the camera had 6 GB of storage available
Answer:
2,929,687 pictures
Step-by-step explanation:
1 pixel requires 2 bytes of storage
A color digital photo uses 512 pixels (512×2 bytes = 1024 bytes) per row and 512 pixels (1024) bytes per column
Total storage used by 1 picture = 1024 + 1024 = 2048 bytes
Storage capacity my camera = 6GB = 6×10^9 bytes
Number of pictures I can have in my camera = 6×10^9/2048 = 2,929,687 pictures
By calculating the storage required for one 512 x 512 pixel photo and dividing the total storage capacity by this amount, we find that a camera with 6 GB of storage can hold approximately 12,288 such photos.
To calculate how many 512 x 512 pixel color digital photos can be stored in a camera with 6 GB of storage, where each pixel requires two bytes of storage, we follow these steps:
First, find the total number of pixels in one photo:
512 pixels/row × 512 pixels/column = 262,144 pixels/photo.
Next, calculate the storage required for one photo:
262,144 pixels/photo × 2 bytes/pixel = 524,288 bytes/photo.
Since there are 1,024 bytes in one kilobyte (KB) and 1,024 KB in one megabyte (MB), we convert the photo size to megabytes:
524,288 bytes/photo / 1,024 bytes/KB / 1,024 KB/MB
≈ 0.5 MB/photo.
Now, convert 6 GB of storage to MB:
6 GB × 1,024 MB/GB = 6,144 MB.
Finally, divide the total storage by the size of one photo to determine the number of photos:
6,144 MB / 0.5 MB/photo = 12,288 photos.
Therefore, you could store approximately 12,288 such pictures on a camera with 6 GB of storage.
A college sends a survey to members of the class of 2012. Of the 1254 people who graduated that year, 672 are women, of whom 124 went on to graduate school. Of the 582 male graduates, 198 went on to graduate school. What is the probability that a class of 2012 alumnus selected at random is (a) female, (b) male, and (c) female and did not attend graduate school?
Answer:
a) [tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]
b) [tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]
c) [tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]
Step-by-step explanation:
For this case we have a total of 1254 people. 672 are women and 582 are female.
We know that 124 women wnat on to graduate school.
And 198 male want on to graduate school
We can define the following events:
F = The alumnus selected is female
M= The alumnus selected is male
A= Female and attend graduate school
And we can find the probabilities required using the empirical definition of probability like this:
Part a
[tex]P(F) = \frac{672}{1254}=\frac{112}{209}=0.536[/tex]
Part b
[tex]P(M) = \frac{582}{1254}= \frac{97}{209}=0.464[/tex]
Part c
For this case we find the probability for the event A: The student selected is female and did attend graduate school
[tex] P(A) =\frac{124}{672}=\frac{31}{168}=0.185[/tex]
And using the complement rule we find P(A') representing the probability that the female selected did not attend graduate school like this:
[tex] P(A') = 1-P(A) = 1- \frac{124}{672}= \frac{137}{168}=0.815[/tex]
At a used dealership, let X be an independent variable representing the age in years of a motorcycle and Y be the dependent variable representing the selling price of used motorcycle. The data is now given to you.
X = {5, 10; 12, 14, 15}; Y = {500, 400, 300, 200, 100}
(a) Write the regression model.
(b) Estimate the parameters of the model
(c) Write the prediction equation
(d) Calculate SSE.
Answer:
a) Selling price y= a + b (age x)
b)
a= 728.025
b= -38.217
c)
Selling price y = 728.025 - 38.217 age x
d)
SSE=8280.25
Step-by-step explanation:
a)
The regression model can be written as
y=a+bx
Here y=selling price and x is age.
So, the regression model will be
Selling price y= a + b (age x)
b)
We have to find the values of "a" and "b"
[tex]b=\frac{nsumxy-(sumx)(sumy)}{nsumx^{2} -(sumx)^2}[/tex]
sumx=5+10+12+14+15=56
sumy=500+400+300+200+100=1500
sumxy=5*500+10*400+12*300+14*200+15*100=14400
sumx²=5²+10²+12²+14²+15²=690
n=5
[tex]b=\frac{5(14400)-(56)(1500)}{5(690) -(56)^2}[/tex]
b=-12000/314
b=-38.217
ybar=a+bxbar
a=ybar-bxbar
ybar=sumy/n=1500/5=300
xbar=sumx/n=56/5=11.2
a=300-(-38.217)(11.2)
a=300+428.025
a=728.025
c)
Selling price y = a - b(age x)
Selling price y = 728.025 - 38.217 age x
d)
SSE known as sum of square of error can be calculated as
SSE=sum(y-yhat)²
y 500 400 300 200 100
x 5 10 12 14 15
yhat= 728.025 - 38.217 age x 536.940 345.855 269.421 192.987 154.770
y-yhat -36.940 54.145 30.579 7.013 -54.770
(y-yhat)² 1364.56 2931.68 935.08 49.18 2999.75
SSE=sum(y-yhat)²
SSE=1364.56 +2931.68 +935.08 +49.18 +2999.75
SSE =8280.25