Answer:
Step-by-step explanation:
Given that price, p, and quantity, x, are inearly related
We are given two points on this line as (p,x) = (96.90, 121) and (40.20,310)
Using two point formula we find linear equation as
[tex]\frac{y-121}{310-121} =\frac{0-96.90}{40.20-96.90} \\(x-121)(-56.70)=189(p-96,.90)\\189p+56.70x =25174.80[/tex]
a)[tex]189p+56.70x =25174.80[/tex] is the linear equation
b) A(x)
Substitute to get
189(50.70)+56.70x = 25174.80
x=275 units
c) This is linear function hence no local maxima or minima
d) No maximia or minima
1. Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and Var(X)=2.
Answer:
Var(u|X=1) = 1
Var(Y|X=1) = 1
Var(u|X) = 3
Var(Y|X) = 9
Step-by-step explanation:
First, we need to identify the properties of the variance, so:
1. Var(a) = 0 , if a is constant
2. [tex]Var(aX) = a^{2} V(X)[/tex], Where a is constant and X is a random variable
3. [tex]Var(aX+b)=a^{2} Var(X)+0[/tex], Where a and b are constants and X is a random variable.
4. [tex]Var(X + Y)=Var(X) + Var(Y)[/tex], Where X and Y are random variables and are independents.
Then, if Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and Var(X)=2, the variance of the following cases are calculated as:
Var(u|X=1) = Var( v + X | X = 1) = Var( v + 1 )
= Var(v) + Var(1)
= 1 + 0 = 1
Var(Y|X=1) = Var( 1 + X + u | X = 1 )
= Var (1 + X + v + X|X=1)
= Var(1 + 1 + v + 1)
= Var(3) + Var(v)
= 0 + 1 = 1
Var(u|X) = Var ( v + X | X)
= Var ( v + X)
= Var(v) + Var(X)
= 1 + 2 = 3
Var(Y|X) = Var ( 1 + X + u | X)
= Var ( 1 + X + v + X | X)
= Var ( 1 + 2X + v)
= Var(1) + Var(2X) + Var(v)
[tex]= Var(1)+2^{2}*Var(X)+Var(v)[/tex]
= 0 + 4*2 + 1 = 8 + 1 = 9
A professor rolls a fair, six-sided die. Using the classical method of probability, what is the probability that at least three spots will be showing up on the die
The probability of seeing at least three spots when rolling a fair six-sided die is 2/3, using classical probability concepts. Each possible result, 3, 4, 5, or 6, is mutually exclusive and equally likely, so probabilities add to 2/3.
Explanation:The subject of this question is probability, which is a branch of Mathematics. When a fair six-sided die is rolled, each face (1, 2, 3, 4, 5 and 6) has an equal probability of 1/6. If we want to find the probability that at least three spots will be showing up on the die, we are looking for the probability of getting a 3, 4, 5 or 6.
Since each possibility (getting a 3, 4, 5 or 6) is mutually exclusive and all are equally likely, we can just add these probabilities together. So, the probability of getting at least 3 spots on a roll is (1/6) + (1/6) + (1/6) + (1/6) = 4/6 = 2/3.
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The time rate of change in the temperature T of coffee is proportional to the difference between the fixed temperature M of the air at time t and the temperature of the coffee at time t. Write a differential equation that fits the above description.
Answer:
The required equation would be [tex]\frac{dT}{dt}=k|M-T|[/tex]
Where, k is constant of proportionality.
Step-by-step explanation:
Differential equation : is an equation that contains one or more functions and their derivatives.
Where, derivatives shows the rate of change of a variable with respect to other variable.
Given,
Change in temperature T of coffee with respect to t ( time ) is proportional to difference between the fixed temperature M and temperature of coffee,
i.e. [tex]\frac{dT}{dt}\propto |M - T|[/tex] ( if M > T, Change is M - T if M < T then change is T - M)
[tex]\frac{dT}{dt}=k|M-T|[/tex]
Where,
k = constant of proportionality.
Evaluate the double integral ∬R(2x−y)dA, where R is the region in the first quadrant enclosed by the circle x^2+y^ 2= 4 and the lines x = 0 and y = x, by changing to polar coordinates.
The result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
How to evaluate the double integral?The given parameters are:
[tex]\int\int R(2x-y) dA[/tex]
x^2 + y^2 = 4
Lines x = 0 and y = x
By polar coordinates, we have:
x = rcost and y = rsint
dA = rdrdt
Substitute x = rcost and y = rsint in 2x - y
2x - y = 2rcost - rsint
So, the integral becomes
[tex]\int\int R(2x-y) dA = \int\limits^a_b \int\limits^a_b ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]
The lines x = 0 and y = x imply that the integral varies from 0 to 2 and π/2 to π/4.
So, we have:
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]
Rewrite as:
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2\cos (t) - \sin(t) )\ r^2drdt[/tex]
Split the integral
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} ( 2\cos (t) - \sin(t) ) dt \int\limits^2_0 r^2dr[/tex]
Integrate
[tex]\int\int R(2x-y) dA = [2\sin (t) - \cos(t)]\limits^{\pi/4}_{\pi/2} * [\frac{r^3}{3}]\limits^2_0[/tex]
Expand
[tex]\int\int R(2x-y) dA = [2\sin (\pi/2) + \cos(\pi/2) - 2\sin (\pi/4) - \cos(\pi/4)] * [\frac{2^3 - 0^3}{3}][/tex]
Simplify the above expression
[tex]\int\int R(2x-y) dA = [2*1 + 0 - \sqrt 2 - \frac{\sqrt 2}{2}] * [\frac{8}{3}][/tex]
[tex]\int\int R(2x-y) dA = [\frac{4 - 3\sqrt 2}{2}] * [\frac{8}{3}][/tex]
Evaluate the product
[tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
Hence, the result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
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To evaluate the given double integral in polar coordinates, we first express the given region in terms of polar coordinates. Then, we rewrite the double integral using the polar coordinate expressions and find the limits of integration. Finally, we evaluate the integral using the given limits.
Explanation:To evaluate the double integral ∫∫R(2x−y)dA in polar coordinates, we first need to express the given region R in terms of polar coordinates. The region R is enclosed by the curve x^2+y^2=4, the line x=0, and the line y=x. In polar coordinates, the curve x^2+y^2=4 becomes r^2=4, or r=2. The line x=0 becomes θ=90°, and the line y=x becomes θ=45°. So, the region R is bounded by θ=0° to θ=45° and r=0 to r=2.
Next, we need to express the differential area element dA in polar coordinates. In Cartesian coordinates, dA represents the area element dx dy. In polar coordinates, dA can be expressed as dA=r dr dθ.
Now, we can rewrite the given double integral as ∫∫R(2x−y)dA = ∫∫R(2rcos(θ)−rsin(θ))r dr dθ. Substituting the limits of integration, we have the final form of the double integral as:
∫[0 to 45°] ∫[0 to 2] (2r2cos(θ)−r2sin(θ)) r dr dθ.
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function f has f(10)=20,f′(10)=2 and f′′(x)<0, for x≥10. Which of the following are possible values for f(14)?
f(14) = 20.28 is
f(14) = 20.14 is
f(14) = 20.56 is
Answer:
f(14) = 28
Step-by-step explanation:
The function that satisfy the equations is
f(x) = 3x - x
f(10) = 3*10 - 10 = 30 - 10 = 20
f'(x) = 3 - 1 = 2.
Therefore,
f(14) = 3(14) - 14 = 42 - 14 = 28.
f(14) = 28
Two linear equations are represented by using the tables below. A 2-column table with 4 rows titled Equation A. Column 1 is labeled x with entries negative 2, 0, 3, 4. Column 2 is labeled y with entries negative 8, negative 2, 7, 10. A 2-column table with 4 rows titled Equation B. Column 1 is labeled x with entries negative 3, negative 1, 1, 5. Column 2 is labeled y with entries negative 9, negative 5, negative 1, 7. The data points for equation A are graphed on the coordinate plane below and are connected by using a straight line. On a coordinate plane, a line goes through (0, negative 2) and (2, 4). What is the solution to the system of equations? (–2, –8) (–1, –5) (0, –2) (2, 4)
Answer:
(-1, -5)
Step-by-step explanation:
When you graph the points, you find that the lines intersect at the point (-1, -5). That is the solution to the system of equations.
(x, y) = (-1, -5)
The solution to a system of linear equations is the point where the lines intersect. Given the data points, we can infer the linear relationship of each equation. However, without more information, we cannot calculate the exact point of intersection.
Explanation:The solution to a system of linear equations is the point where the two lines intersect. From the provided data for Equation A and Equation B, we can infer their linear relationship by finding the slope and y-intercept.
For Equation A, the points (0, -2) and (2, 4) gives a slope of (4 - -2) / (2 - 0) = 3 and y-intercept of -2. For Equation B, the points (-3,-9) and (-1, -5) gives a slope of (-5 - -9) / (-1 - -3) = 2 and y-intercept of -3.
As the provided slopes and y-intercepts are different for these two equations, they would intersect at a point. To find this point, we set the two equations equal to each other and solve for x and y. However, with the provided data points, we cannot calculate the exact point of intersection without further information.
So, the solution to the system of equations would be the x and y values at the point of intersection of these two lines.
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Let n be a positive integer. We sample n numbers a1, a2,..., an from the set {1,...,n} uniformly at random, with replacement. We say that picks i and j with i < j are a match if ai = aj.
What is the expected total number of matches?
There is a chance to get one or more number of number of matches.
What is Combination?An arrangement of objects where the order in which the objects are selected does not matter.
Given,
Let n be a positive integer.
sample n numbers a1, a2,..., an from the set {1,...,n} uniformly at random, with replacement.
Where 1<j where ai = aj.
We need to find the expected total number of matches
expected total number of matches=Sum of sample/ Total number of samples
=1+2+3+4+...n/n
=n(n+1)/n
=n+1
Hence, there is a chance to get one or more number of number of matches.
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To determine the expected total number of matches when picking n numbers from the set {1,...,n} with replacement, we find that each pair of picks has a 1/n chance of a match. Multiplying this by the total number of distinct pairs gives us an expectation of (n-1)/2 matches.
To calculate the expected total number of matches when n numbers are sampled from the set {1,...,n} uniformly at random with replacement, we consider the probability of a match for each pair of picks (i,j) with i < j. Since each number has a 1/n chance of being picked, the probability that two picks are a match, P(match), is similarly 1/n. Assuming all pairs are independent, the expected number of matches for a specific pair is thus 1/n.
Because there are a total of n(n-1)/2 such pairs in our selection (since we can choose 2 out of n elements in a combinatorial fashion), the expected total number of matches E(total matches) is the sum of the expectations for all pairs. Therefore, E(total matches) = n(n-1)/2 * (1/n).
Upon simplifying the expression, we obtain E(total matches) = (n-1)/2. Thus, this is the expected total number of matches for the given scenario.
"From a standard deck of cards, find the number of different 5-card hands that are made up of 3 spades and 2 diamonds"
Answer:
[tex]^{13}C_3 \times ^{12}C_2 = 22308[/tex]
Step-by-step explanation:
We have 5 spaces. In our hand.
_ _ _ _ _
a standard deck of 52 cards contains 13 cards for each suit.
so we have 13 spades and 13 diamonds in total. There arrangements don't matter (it doesn't matter if the first card is 9 of spades or the second is 9 of spades, all of these will be counted as one arrangement)
[tex]^{13}C_3[/tex]: to choose the 3 spade card from a total of 13 spades.
[tex]^{12}C_2[/tex]: to choose the 2 diamond cards from a total of 13 diamonds.
and that is it!
we're gonna multiply the two values.
[tex]^{13}C_3 \times ^{12}C_2[/tex]
A physical fitness researcher devises a test of strength and finds that the scores are Normally distributed with a mean of 100 lbs and a standard deviation of 10 lbs. What is the minimum score needed to be stronger than all but 5% of the population
Answer:
116.45 is the minimum score needed to be stronger than all but 5% of the population.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 10
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.05
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 100}{10})=0.05[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 100}{10})=0.05[/tex]
[tex]=P( z \leq \displaystyle\frac{x - 100}{10})=0.95 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z<1.645) = 0.95[/tex]
[tex]\displaystyle\frac{x - 100}{10} = 1.645\\x =116.45[/tex]
Hence, 116.45 is the minimum score needed to be stronger than all but 5% of the population.
Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below are the weights in pounds of 1111 players randomly selected from the roster of a championship sports team. Are the results likely to be representative of all players in that sport's league? 278 303 186 292 276 205 208 236 278 198 208
a.Find the mean.The mean is ? pound(s).(Type an integer or a decimal rounded to one decimal place asneeded.)b. Find the median.The median is ? pound(s).(Type an integer or a decimal rounded to one decimal place asneeded.)c. Find the mode.
Answer: a. 242.5 pounds
b. 236 pounds
c . 208 and 278 pounds
The results are unlikely to be representative of all players in that sport's league because players randomly selected from championship sports team not the whole league.
Step-by-step explanation:
Given : Listed below are the weights in pounds of 11 players randomly selected from the roster of a championship sports team.
278 303 186 292 276 205 208 236 278 198 208
Mean = [tex]\dfrac{\text{Sum of weights of all players}}{\text{Number of players}}[/tex]
[tex]=\dfrac{278+303+186+292+276+205+208+236+278+198+208}{11}\\\\=\dfrac{2668}{11}\approx242.5\text{ pounds}[/tex]
For median , first arrange weights in order
186, 198 , 205 , 208, 208 , 236 , 276 , 278 ,278, 292 , 303
Since , number of data values is 11 (odd)
So Median = Middlemost value = 236 pounds
Mode = Most repeated value= 208 and 278
The results are unlikely to be representative of all players in that sport's league because players randomly selected from championship sports team not the whole league.
The mean, median and mode values of the distribution are :
242.5, 236 and (208 and 278) respectively.Given the data :
278, 303, 186, 292, 276, 205, 208, 236, 278, 198, 208Arrange the data in ascending order :
186, 198, 205, 208, 208, 236, 276, 278, 278, 292, 303The mean = ΣX/ n
Where, n = sample size = 11 Mean = (186 + 198 + 205 + 208 + 208 + 236 + 276 + 278 + 278 + 292 + 303) / 11 Mean = 2668 / 11 Mean = 242.545The median ;
Since the data has been arranged : 1/2(n + 1)th 1/2(11 + 1)th = 1/2(12)th = 6th value The 6th value in the ordered data is 236 Hence, the median = 236The mode :
The mode of the distribution is the most frequently occurring value :The most frequently occurring values are : 208 and 278 with frequency values of 2.Therefore ;
Mean = 242.5 pounds
Median = 236 pounds
Mode = 208 and 278
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Find the solution of the given initial value problem. ty' + 6y = t2 − t + 1, y(1) = 1 6 , t > 0
Final answer:
To solve the given initial value problem, we'll use the method of integrating factors. After rearranging the equation to standard form, we'll find the integrating factor and proceed to solve for y.
Explanation:
To find the solution of the given initial value problem, we'll use the method of integrating factors. First, we'll rearrange the equation to the standard form: y' + (6/t)y = t - 1/t + 1. Now, let's identify the integrating factor, which is e^(∫6/t dt) = e^(6ln|t|) = t^6. Multiply both sides of the equation with t^6 to get t^7y' + 6t^5y = t^7 - t^6 + t^6. Now, we can integrate both sides and solve for y.
Consider the first five steps of the derivation of The Quadratic function
Answer:
Step-by-step explanation:
pick like termsx² + b²/4a² = -c / a + b²/4a²
b²/4a² = (b/2a)²x² + (b/2a)² = -c/a + (b/2a)²
(x + b/2a)² = -c/a + (b/2a)² = -c / a + b²/4a² = (-4ac+ b²)/4a²
(x + b/2a)² = (-4ac+ b²)/4a²
square root both sides√{(x + b/2a)²} = √{(-4ac+ b²)/4a²}
x + b/2a = √(-4ac+ b²) / √(4a²) = √(-4ac+ b²) / 2a = √( b²-4ac) / 2a
x + b/2a = √( b²-4ac) / 2a
subtract b/2a from both sidesx + b/2a -b/2a = {√( b²-4ac) / 2a } -b/2a
x = -b/2a + {√( b²-4ac) / 2a }
the l.c.m is the samex = {-b±√( b²-4ac)}/2a
A quadratic equation is an equation of the sort; ax^2 + bx + c =0. It can be solved by the formula method.
What is a quadratic equation?A quadratic equation is an equation of the sort; ax^2 + bx + c =0. One of the ways of solving a quadratic equation is the formula method which is being derived here.
From the step shown in the image in the question;
Collecting like terms;
x² + b²/4a² = -c / a + b²/4a²
x² + (b/2a)² = -c/a + (b/2a)²
We can now write;
(x + b/2a)² = -c/a + (b/2a)²
Hence;
(x + b/2a)² = (-4ac+ b²)/4a²
Taking the square root of both sides and solving for x
x =-b±√( b²-4ac)/2a
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construct a boxplot for the following data and comment on the shape of the distribution representing the number of
Answer:
Check the boxplot below,plus the comments
Step-by-step explanation:
1) Completing the question:
Construct a boxplot for the following data and comment on the shape of the distribution representing the number of games pitched by major league baseball’s earned run average (ERA) leaders for the past few years.
30 34 29 30 34 29 31 33 34 27 30 27 34 32
2) Arranging the distribution orderly to find the 2nd Quartile (Median):
27 27 29 29 30 30 30 31 32 33 34 34 34 34
Since n=14, dividing the sum of the 14th and 15th element by two:
[tex]Md=Q_{2}=\frac{30+31}{2} =30.5[/tex]
3) Calculating the Quartiles, the Upper and the Lower one comes:
[tex]Q_{1}=29.25\\Q_{3}=33.75[/tex]
4) Boxplotting (Check it below)
5) Notice that since the values are very close, then the box is not that tall. The difference between the Interquartile Range is not so wide what makes it shorter. Check the values on the table below.
To construct a boxplot, organize the data, calculate the median, Q1 and Q3, and then draw the boxplot. The shape of the distribution can be assessed by observing the boxplot - if it's symmetric, positively skewed, or negatively skewed.
Explanation:To construct a boxplot for a given set of data, follow these steps:
Arrange the data in ascending order. This makes it easier to identify the quartile numbers.Calculate the median of the data set. This is the middle number when the data is listed in numerical order. If the data set has an even number of observations, the median is the average of the two middle numbers.Calculate Q1 and Q3. Q1 is the median of the lower half of the data, not including the median if the data set has an odd number of observations. Q3 is the median of the upper half.Construct a box over Q1 and Q3, making sure the lines (whiskers) extend to the smallest and largest observations that are not outliers.With respect to the shape of the distribution, by closely looking at the boxplot you can determine if the distribution is symmetric (box is centered about the median), positively skewed (box is shifted towards the lower end of the scale) or negatively skewed (box is shifted towards the upper end).
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A client is prescribed to receive 1/2 cup of a dietary supplement 3 times a day. How many Tablespoons of the supplement will the client be taking each day?
Answer: There are 1 and half table spoons of the supplement that the client will be taking each day.
Step-by-step explanation:
Since we have given that
Quantity of cup of a dietary supplement = [tex]\dfrac{1}{2}[/tex]
Number of times a day = 3
So, the number of tablespoons of the supplement that the client will be taking each day would be
[tex]3\times \dfrac{1}{2}\\\\=\dfrac{3}{2}\\\\=1\dfrac{1}{2}[/tex]
Hence, there are 1 and half table spoons of the supplement that the client will be taking each day.
Casey needed to move 23 huge boxes from his truck to the loading dock. His forklift could only hold three boxes at once. How many times did Casey have to visit the loading dock?
Casey visited that loading dock 8 times.
Determination of the number of times the forklift visited the loading dockIn order to determine the number of times the forklift visited the loading dock, divide the total number of boxes by the total number of boxes the forklift can carry at once.
Number of visits = 23 / 3 = 7 2/3 = 8 times
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Final answer:
Casey had to visit the loading dock 8 times in total to move all 23 boxes, given that his forklift can carry 3 boxes per trip.
Explanation:
To calculate how many times Casey had to visit the loading dock to move 23 huge boxes with his forklift, which can only hold three boxes at once, we use division.
We divide the total number of boxes by the number of boxes the forklift can carry per trip. Since 23 divided by 3 gives us 7 with a remainder of 2, it means Casey made 7 full trips carrying 3 boxes each and one additional trip to carry the remaining 2 boxes.
Therefore, Casey had to visit the loading dock 8 times in total.
What’s the answer I will do 50 points first person to answer
Answer: 1. Route ABC is a right triangle.
2. Route CDE ia not a right triangle.
3. Distance HJ= 23.32miles
4. Distance GE = 17
5. Missing length= 35
6. The sides 16, 60, 62 do NOT belong to a right triangle.
Step-by-step explanation:
Going by Pythagoras theorem, for triangle to be proven to be a right triangle, the condition below must be satisfied.
Hypotenuse² = Opposite² + Adjacent²
For question 1,
Hyp =13, opp = 5, Adj is 12
Going by Pythagoras rule.
Since 13²= 5² + 12²
Then triangle ABC is a right triangle.
For question 2,
Using the same Pythagoras theorem to prove,
In triangle CDE,
Hyp= 22, opp= 18, Adj = 14
Since 22² is not = 18² + 14²
then CDE is not a right triangle.
For question 3,
For triangle HIJ, since it is confirmed to be a right triangle, then we use the Pythagoras theorem to calculate the missing side.
Longest side if the triangle= IJ = hypotenuse = 25
HI = 9.
IJ² = HI² + HJ²
HJ²= IJ² - HI²
HJ² = 25² - 9²
HJ² = 625 - 81
HJ= √544
HJ = 23.32miles
For question 4,
FGE is also shown to be a right triangle and the missing side GE is the longest side which is also the hypotenuse.
FG= 8, FE =15
Using the Pythagoras theorem,
Hyp² = FG² + FE²
GE² = 8² + 15²
GE² = 289
GE = √289
GE = 17.
For question 5,
The hypotenuse is given as 37, one side is given as 12, let's call the missing side x
Going by Pythagoras theorem,
37² = 12²+ x²
x²= 37² - 12²
x²= 1225
x=√1225
x=35.
The missing side is 35inches.
For number 6,
The numbers given are 16, 60, 62
To know if three sides belong to a right angle, we simply put them to test using Pythagoras theorem.
It is worthy of note that the longest side is the hypotenuse.
This brings us to the equation to check below that since:
62² Is not = 60² + 16²
Then the side lengths 16, 60, 62 do not belong to a right angle.
Which equation could be used to find the number of days, d, in h hours?
Answer: In my opinion, I would say b is the correct answer
Step-by-step explanation:
the number of ways 8 cars can be lined up at a toll booth would be computed from
a. 8 to the 8th power
b. (8)*(8)
c. 8!
d. 8!/7!1!
Answer: c. 8!
Step-by-step explanation:
We know , that if we line up n things , then the total number of ways to arrange n things in a line is given by :-
[tex]n![/tex] ( in words :- n factorial)
Therefore , the number of ways 8 cars can be lined up at a toll booth would be 8! .
Hence, the correct answer is c. 8! .
Alternatively , we also use multiplicative principle,
If we line up 8 cars , first we fix one car , then the number of choices for the next place will be 7 , after that we fix second car ,then the number of choices for the next place will be 6 , and so on..
So , the total number of ways to line up 8 cars = 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 8!
Hence, the correct answer is c. 8! .
Solve the given differential equation by finding, as in Example 4 from Section 2.4, an appropriate integrating factor.4xy dx (4y 6x2) dy
Answer:
y=1/6 · ln |x|+c .
Step-by-step explanation:
From Exercise we have the differential equation
4xy dx= (4y6x²) dy.
We calculate the given differential equation, we get
4xy dx= (4y6x²) dy
xy dx=6yx² dy
6 dy=1/x dx
∫ 6 dy=∫ 1/x dx
6y=ln |x|+c
y=1/6 · ln |x|+c
Therefore, we get that the solution of the given differential equation is
y=1/6 · ln |x|+c .
The differential equation presented can be rewritten and solved using an integrating factor. The integrating factor in this case is 1, yielding the solution x^2 y = C.
Explanation:To solve the given differential equation, we first need to rewrite it in a recognizable form namely, in the form Mdx + Ndy = 0. The given differential equation can be rewritten as 4xy dx + (4y - 6x2 ) dy = 0.
Now, we need to find an integrating factor which is e^∫(M_y - N_x) / N dx. In this case, M = 4xy, N = 4y - 6x2, M_y = 4x and N_x = -12x. Substituting these values into the equation ∫(M_y - N_x) / N dx, we find that the integrating factor is e^0=1.
With the integrating factor being 1, the given differential equation can be rewritten as d(x2y) = 0. Integrating both sides of this equation, we get x2y = C.
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An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A) = 0.2 and P(B) = 0.5.
(a) If the Asian project is not successful, what is the probability that the European project is also not successful?
______
Explain your reasoning.
a. Since the events are independent, then A' and B' are independent, too.
b. Since the events are independent, then A' and B' are mutually exclusive.
c. Since the events are not independent, then A' and B' are mutually exclusive.
d. Since the events are independent, then A' and B' are not independent.
(b) What is the probability that at least one of the two projects will be successful?
(c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?
Answer:
a)
0.5
option A
b)
0.6
c)
0.1
Step-by-step explanation:
The event A and B are independent so
P(A∩B)=P(A)*P(B)
P(A∩B)=P(0.2)*P(0.5)=0.10
a)
We have to find P(B'|A')
P(B'|A')=P(B'∩A')/P(A')
P(A)=0.2
P(A')=Asian project is not successful=1-P(A)=1-0.2=0.8
P(B)=0.5
P(B')=Europe project is not successful=1-P(B)=1-0.5=0.5
P(B'∩A')=Europe and Asia both project are not successful=P(A')*P(B')=0.8*0.5=0.4
P(B'|A')=P(B'∩A')/P(A')=0.4/0.8=0.5
This can be done by another independence property for conditional probability
P(B|A)=P(B)
P(B'|A')=P(B')
P(B'|A')=0.5
b)
Probability of at least one of two projects will be successful means that the probability of success of Asia project or probability of success of Europe project or probability of success of Europe and Asian project which is P(AUB).
P(AUB)=P(A)+P(B)-P(A∩B)
P(AUB)=0.2+0.5-0.1
P(AUB)=0.6
c)
Probability of only Asian project is successful given that at least one of the two projects is successful means that probability of success of project Asia while the project Europe is not successful denoted as P((A∩B')/(A∪B))=?
P((A∩B')/(A∪B))=P((A∩B')∩(A∪B))/P(A∪B)
P((A∩B')∩(A∪B))=P(A∩B')*P(A∪B)
P(A∩B')=P(A)*P(B')=0.2*0.5=0.10
P((A∩B')∩(A∪B))=0.1*0.6=0.06
P((A∩B')/(A∪B))=0.06/0.6=0.1
Final answer:
The probability of European project being unsuccessful is 40%. The probability that at least one project will be successful is 60%. Given that at least one project is successful, the probability that only the Asian project is successful is approximately 16.7%.
Explanation:
The student is learning about independent and mutually exclusive events in probability.
The question involves calculating the probability of certain outcomes given two independent events, A and B, with known probabilities P(A) = 0.2 and P(B) = 0.5.
(a) Probability of Both Projects Being Unsuccessful
Since events A and B are independent, events A' (Asian project not successful) and B' (European project not successful) are also independent. Therefore, the correct reasoning is:
a. Since the events are independent, then A' and B' are independent, too.
The probability of both A' and B' occurring is P(A')P(B') = (1 - P(A))(1 - P(B)) = (1 - 0.2)(1 - 0.5) = 0.8 imes 0.5 = 0.4.
Thus, the probability that the European project is also not successful given the Asian project is not successful is 0.4 or 40%.
(b) Probability that At Least One Project Will Be Successful
To find the probability that at least one project will be successful, we need to calculate 1 - P(A' AND B').
This gives us 1 - P(A')P(B') = 1 - 0.4
= 0.6 or 60%.
(c) Probability that Only the Asian Project Is Successful Given At Least One Is Successful
First, we calculate the probability of only the Asian project being successful, which equals P(A)P(B') = 0.2 x 0.5 = 0.1 or 10%.
Next, we calculate the probability of at least one project being successful, which we found to be 60%.
Therefore, the conditional probability is P(A and not B) / P(at least one successful), which equals 0.1 / 0.6 ≈ 0.167 or 16.7%
Sketch the region enclosed by x+y2=12x+y2=12 and x+y=0x+y=0. Decide whether to integrate with respect to xx or yy, and then find the area of the region.
Answer:
[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]
Step-by-step explanation:
For this case we have these two functions:
[tex] x+y^2 = 12[/tex] (1)
[tex] x+y=0[/tex] (2)
And as we can see we have the figure attached.
For this case we select the x axis in order to calculate the area.
If we solve y from equation (1) and (2) we got:
[tex] y = \pm \sqrt{12-x}[/tex]
[tex] y = -x[/tex]
Now we can solve for the intersection points:
[tex] \sqrt{12-x} = -\sqrt{12-x}[/tex]
[tex] 12-x = -12+x[/tex]
[tex] 2x=24 , x=12[/tex]
[tex] \sqrt{12-x} =-x[/tex]
[tex] 12-x = x^2[/tex]
[tex] x^2 +x -12=0 [/tex]
[tex] (x+4)*(x-3) =0[/tex]
And the solutions are [tex] x =-4, x=3[/tex]
So then we have in total 3 intersection point [tex] x=12, x=-4, x=3[/tex]
And we can find the area between the two curves separating the total area like this:
[tex] \int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx[/tex]
[tex] \int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx[/tex]
We can separate the integrals like this:
[tex] \int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx[/tex]
For this integral [tex] \int_{-4}^3 |\sqrt{12-x} dx [/tex] we can use the u substitution with [tex]u = 12-x[/tex] and after apply and solve the integral we got:
[tex] \int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}[/tex]
The other integral:
[tex] \int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}[/tex]
And for the other integral:
[tex]2\int_{3}^{12}\sqrt{12-x} dx[/tex]
We can use the same substitution [tex] u = 12-x[/tex] and after replace and solve the integral we got:
[tex]2\int_{3}^{12}\sqrt{12-x} dx =36[/tex]
So then the final area would be given adding the 3 results as following:
[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]
The task is to sketch the region enclosed by the parabola [tex]x + y^2 = 12[/tex]and the line [tex]x + y = 0,[/tex] and then find the area of that region. The area can be found by integrating with respect to y, using vertical slices through the region and the points of intersection as limits.
Explanation:The student has asked to sketch the region enclosed by two equations and then find the area of that region. It looks like there may be some confusion with the equations provided, as they seem to contain typos. Assuming the correct equations are [tex]x + y2 = 12[/tex] and [tex]x + y = 0[/tex], we can proceed by first graphing these two curves.
After sketching, you would see that the curve [tex]x + y2 = 12[/tex] forms a parabola that opens to the right, and the line [tex]x + y = 0[/tex]is a straight line that has a negative slope and passes through the origin. The region enclosed by these two would resemble a 'cap' shape, with the point of intersection providing the limits for integration.
To find the area of this region, you could choose to integrate with respect to y since the equations can be easily solved for x, and the structure of the parabola suggests that vertical slices would be easier to work with. The integration limits would be the y-values where the curves intersect. After setting up the integral, you would calculate the definite integral to find the area.
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Translate the following English statements into a logical expression with the same meaning.
a. All friendly people at HTS are knowledgeable.
b. Nobody at HTS is friendly, helpful, and knowledgeable.
c. Someone at HTS is helpful.
d. There is no one at HTS who is both friendly and helpful.
e. No friendly person at HTS is helpful.
Answer:
C makes most sence
Step-by-step explanation:
(1 point) Find an equation of the largest sphere with center (5,3,5)(5,3,5) and is contained in the first octant. Be sure that your formula is monic. Equation:
Answer:
x^2+y^2+z^2-10x-6y-10z +50 =0
Step-by-step explanation:
Given that a sphere is contained in the first octant
Centre of the sphere is given as (5,3,5)
Since this is contained only in the first octant radius should be at most sufficient to touch any one of the three coordinate planes
When it touches we can get the maximum sphere
We find that y coordinate is the minimum of 3 thus radius can be atmost 3 so that then only it can touch y =0 plane i.e. zx plane without crossing to go to the other octants.
Hence radius =3
Equation of the sphere would be
[tex](x-5)^2 +(y-3)^2+(z-5)^2 = 3^2\\x^2+y^2+z^2-10x-6y-10z +50 =0[/tex]
Residents of the town of Maple Grove who are connected to the municipal water supply are billed a fixed amount monthly plus a charge for each cubic foot of water used. A household using 1700 cubic feet was billed $126, while one using 2500 cubic feet was billed $150. Write an equation for the total cost of a resident's water as a function of cubic feet of water used.
Answer:
Cost function C(x) where x is the cubic feet of water used.
[tex]C(x) = 75 + 0.03x[/tex]
Step-by-step explanation:
We are given the following in the question:
The municipal water supply are billed a fixed amount monthly plus a charge for each cubic foot of water used.
Let x be the fixed amount and y be the cost for each cubic foot of water used.
A household using 1700 cubic feet was billed $126.
Thus, we can write:
[tex]x + 1700y = 126[/tex]
A household using 2500 cubic feet was billed $150.
[tex]x + 2500y = 150[/tex]
Solving the two equation by eliminating x by subtracting the equations, we get,
[tex]x+2500y-(x + 1700y) = 150-126\\800y = 24\\\\y =\dfrac{24}{800} = 0.03\\\\x+1700(0.03) = 126\\\Rightarrow x = 126-(1700)(0.03) = 75[/tex]
Thus, the fixed cost is $75 and the cost per cubic foot of water is $0.03
Cost function can be written as:
[tex]C(x) = 75 + 0.03x[/tex]
where x is the cubic feet of water used.
Final answer:
To derive the total cost equation for the water bill, a system of linear equations was solved to find the fixed charge and the variable rate per cubic foot. The cost function is C(x) = 75 + 0.03x, where C represents the total cost and x represents cubic feet of water used.
Explanation:
To find the equation that represents the total cost of a resident's water bill as a function of the cubic feet of water used, we first need to establish the fixed charge and the variable charge per cubic foot. We do this by setting up a system of linear equations based on the information provided about the two households.
Let F be the fixed monthly charge and V be the variable charge per cubic foot. We can set up two equations based on the given information:
For the household using 1700 cubic feet with a bill of $126: F + 1700V = 126For the household using 2500 cubic feet with a bill of $150: F + 2500V = 150Subtracting the first equation from the second gives us:
800V = 24
Dividing both sides by 800, we find:
V = 0.03
Now, we plug the value of V back into the first equation to find F:
F + 1700(0.03) = 126
F + 51 = 126
F = 126 - 51
F = 75
Therefore, the equation for the total cost C, in dollars, as a function of x, the cubic feet of water used, is:
C(x) = 75 + 0.03x
Consider the time series:xₜ= β₁+ β₂t + wₜwhere β1 and β2 are known constants and wₜ is a white noise process with variance σ ²_w.(a) Determine whether xₜ is stationary.(b) Show that the process yₜ= xₜ − xₜ₋₁ is stationary.(c) Show that the mean of the moving averagevt = [1 /2q + 1] (j=-q→q) Σ xₜ₋ⱼis β₁ + β₂t, and give a simplified expression for the autocovariance function.
Answer:
Please refer to the attachment below.
Step-by-step explanation:
Please refer to the attachment below for explanation.
The summer monsoon rains bring 80 % of India's rainfall and are essential for the country's agriculture. Records going back more than a century show that the amount of monsoon rainfall varies from year to year according to a distribution that is approximately Normal with mean 852 millimeters (mm) and standard deviation 82 mm. Use the 68 ‑ 95 ‑ 99.7 rule to answer the questions. (a) Between what values do the monsoon rains fall in the middle 95 % of all years?
Answer:
95% of monsoon rainfall lies between 688 mm and 1016 mm.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 852 mm
Standard Deviation, σ = 82 mm
We are given that the distribution of monsoon rainfall is a bell shaped distribution that is a normal distribution.
68 ‑ 95 ‑ 99.7 rule
Also known as Empirical rule.It states that all data lies within the three standard deviation of the mean for a normal distribution.About 68% of data lies within one standard deviation of meanAbout 95% of data lies within two standard deviation of mean.About 99.7% of data lies within three standard deviation of mean.We have to find the monsoon rains fall in the middle 95 % of all years.
By the rule 95% of data lies within two standard deviation of mean.Thus,
[tex]\mu + 2\sigma = 852 + 2(82) = 1016\\\mu - 2\sigma = 852 - 2(82) = 688[/tex]
Thus, 95% of monsoon rainfall lies between 688 mm and 1016 mm.
Using the Empirical Rule, we can determine that 95% of the time, the monsoon rains in India will fall between the amounts of 688 mm and 1016 mm. This is determined by subtracting and adding two standard deviations from the mean amount of rainfall.
Explanation:The question pertains to the principle of the Normal Distribution curve in statistics, specifically using the 68 - 95 - 99.7 rule. Also known as the Empirical Rule, this principle suggests that for a Normal Distribution: approximately 68% of data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations. Given that the mean monsoon rainfall is 852 mm and standard deviation is 82 mm, we can calculate the range for the middle 95% of all years.
To find this range, we add and subtract two standard deviations from the mean. Therefore for two standard deviations (164 mm, since 82 mm x 2 = 164 mm), our range is: 852 mm - 164 mm = 688 mm and 852 mm + 164 mm = 1016 mm. Therefore, in 95% of all years, the monsoon rainfall in India falls between 688 mm and 1016 mm.
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Which of the following best represents the highest potential for nonresponse bias in a sampling strategy? Describe why this option should be considered nonresponse a. Surveying a population on Sunday mornings for a new needs assessment b. Submitting a post online advertising the need for participants in a new study c. Asking people leaving a local election to take part in an exit poll d. Posting a leaflet in the elevator of a university asking for students to take part in a paid study
Answer:
c. Asking people leaving a local election to take part in an exit poll
Step-by-step explanation:
Asking people leaving a local election to take part in an exit poll best represents the highest potential for nonresponse bias in a sampling strategy because of the importance of the local election compared to the exit polls.
It is worthy of note that nonresponse bias occurs when some respondents included in the sample do not respond to the survey. The major difference here is that the error comes from an absence of respondents not the collection of erroneous data. ...
Oftentimes, this form of bias is created by refusals to participate for one reason or another or the inability to reach some respondents.
Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify your answer.
Answer:
p ∈ IR - {6}
Step-by-step explanation:
The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2
is all R2 ⇔
[tex]u\neq 0_{R2}[/tex]
[tex]v\neq 0_{R2}[/tex]
And also u and v must be linearly independent.
In order to achieve the final condition, we can make a matrix that belongs to [tex]R^{2x2}[/tex] using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.
Let's make the matrix :
[tex]A=\left[\begin{array}{cc}3&1&p&2\end{array}\right][/tex]
We used the first vector ''u'' as the first column of the matrix A
We used the second vector ''v'' as the second column of the matrix A
The determinant of the matrix ''A'' is
[tex]Det(A)=6-p[/tex]
We need this determinant to be different to zero
[tex]6-p\neq 0[/tex]
[tex]p\neq 6[/tex]
The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that [tex]p\neq 6[/tex]
We can write : p ∈ IR - {6}
Notice that is [tex]p=6[/tex] ⇒
[tex]u=(3,6)[/tex]
[tex]v=(1,2)[/tex]
If we write [tex]3v=3(1,2)=(3,6)=u[/tex] , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.
Jesse takes a 3-day kayak trip and travels 72 km south from Everglades City to a camp area in Everglades National Park. The trip to the camp area with a 2-km/hr current takes 9 hr less time than the return trip against the current. Find the speed that Jesse travels in still water.
Answer: The speed that Jesse travels in still water is 6 km/hr.
Step-by-step explanation:
Let the speed that Jesse travels in still water be 'x'.
Distance = 72 km
The trip to the camp area with a 2-km/hr current takes 9 hr less time than the return trip against the current.
Speed of current = 2 km/hr
According to question, we get that
[tex]\dfrac{72}{x-2}-\dfrac{72}{x+2}=9\\\\\dfrac{x+2-(x-2)}{x^2-4}=\dfrac{9}{72}\\\\\dfrac{4}{x^2-4}=\dfrac{1}{8}\\\\32=x^2-4\\\\32+4=x^2\\\\x^2=36\\\\x=\sqrt{36}\\\\x=6[/tex]
Hence, the speed that Jesse travels in still water is 6 km/hr.
Jesse's speed in still water is determined by setting up a system of equations using the distance equals rate times time formula for both downstream and upstream travel. By accounting for the time difference and the current speed, we solve for the variable representing Jesse's speed in still water.
Explanation:Jesse takes a kayak trip traveling 72 km with and against a current, and we need to find Jesse's speed in still water. Let's denote the speed in still water as v (km/hr) and the current speed as 2 km/hr. The trip downstream increases Jesse's speed to (v + 2) km/hr, and upstream decreases it to (v - 2) km/hr.
Using the distance equals rate times time formula (d = rt), we can write the following equations for the time taken downstream (td) and upstream (tu):
72 = (v + 2)td72 = (v - 2)tuGiven that it takes 9 hours less to travel downstream, we have tu = td + 9.
By solving these linear equations, we find the system:
td = 72 / (v + 2)td + 9 = 72 / (v - 2)Combining these gives us:
72 / (v + 2) + 9 = 72 / (v - 2)
By solving this equation, we find the value of v, Jesse's speed in still water.
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If a linear system has four equations and seven variables, then it must have infinitely many solutions. This statement is false, could you explain why and give an example?
Answer:
False. See the explanation below.
Step-by-step explanation:
We need to proof if the following statement "If a linear system has four equations and seven variables, then it must have infinitely many solutions." is false.
And the best way to proof that is false is with a counterexample.
Let's assume that we have seven random variables given by [tex]a_1, a_2, a_3, a_4, a_5, a_6, a_7[/tex] and we have the following four equations given by the following system:
[tex] a_1 +a_2 +a_3 +a_4 +a_5 +a_6 +a_7 =1[/tex] (1)
[tex] a_1 +a_2= 0[/tex] (2)
[tex] a_3 +a_4 +a_5 =1[/tex] (3)
[tex] a_6 +a_7 =1[/tex] (4)
As we can see we have system and is inconsistent since equation (1) is not satisfied by equation (2) ,(3) and (4) if we add those equations we got:
[tex] a_1 +a_2 +a_3 +a_4 +a_5 +a_6 +a_7 = 0+1+1= 2 \neq 1[/tex]
So then we can have a system of 7 variables and 4 equations inconsistent and with not infinitely solutions for this reason the statement is false.
A linear system can have more variables than equations, this is referred to as an underdetermined system. It's often believed that such systems have infinitely many solutions, but it's not necessarily the case. Such a system could also have no solutions if the system is inconsistent, signifying that not all underdetermined systems yield infinite solutions.
Explanation:Contrary to the common belief, the statement that a linear system with more variables than equations must have infinitely many solutions is not always true. A system having more variables than equations is termed as underdetermined. Indeed, such systems often have infinite solutions, but not necessarily.
A system could also be inconsistent, meaning there are no solutions. As an example, consider the system of equations where:
x + y + z = 1 x - y + z = 2 2x + y + z = 3 2x - y + z = 4Even though there are more variables (3) than equations (4), there are still no solutions as these equations contradict each other.
This emphasizes the point that, for a linear system to have infinitely many solutions, it must have at least one free variable and must be consistent in the first place.
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