Given:
An archer hits a target 50% of the time.
To find:
The experimental probability that the archer hits the target exactly four of the next five times.
Solution:
It is given that an archer hits a target 50% of the time. It means the probability of hitting the target is
[tex]p=\dfrac{50}{100}[/tex]
[tex]p=0.5[/tex]
The probability of not hitting the target is
[tex]q=1-p[/tex]
[tex]q=1-0.5[/tex]
[tex]q=0.5[/tex]
Binomial distribution formula:
[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]
We need to find the probability that the archer hits the target exactly four of the next five times. So, [tex]n=5,r=4,p=0.5,q=0.5[/tex].
[tex]P(x=4)=^5C_4(0.5)^4(0.5)^{5-4}[/tex]
[tex]P(x=4)=\dfrac{5!}{4!(5-4)!}(0.5)^4(0.5)^{1}[/tex]
[tex]P(x=4)=5(0.5)^{5}[/tex]
[tex]P(x=4)=0.15625[/tex]
Therefore, the experimental probability that the archer hits the target exactly four of the next five times is 0.15625.
What is the time complexity of binary search?
a) O(N^2)
b) O(N)
c) O(NLogN)
d) None
The time complexity of the binary search is c) O(NLogN)
An effective searching technique that uses sorted arrays is a binary search. The search space is continually divided in half until the target element is located or the search space is empty, using a divide-and-conquer strategy. It makes the necessary adjustments to the search bounds at each stage by comparing the target element to the centre element of the active search area.
The divide-and-conquer strategy is logarithmic, hence binary search has an O(log n) time complexity. This implies that number of operations needed to discover the target element rises at a logarithmic rate as input size increases. When compared to linear search algorithms with O(n) complexity, where each element must be evaluated sequentially, binary search is significantly more efficient for big arrays due to its logarithmic time complexity.
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Let C[0, 1] have the inner product (f,g) = f(x)g(x)dx. For u = x and v= x + 1 find the following: a) ||f|| b) |lg|| c) (f,g) d) Find the angle between u and v.
The function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3). The angle between u and v is arccos(1/√5).
a) The norm f of a function f in the vector space C[0, 1] with the given inner product is defined as the square root of the inner product of the function with itself. In this case, f = sqrt((f, f)) = sqrt(integral of f(x) * f(x) dx over the interval [0, 1]). For the function u = x, the norm is u = sqrt(integral of x^2 dx from 0 to 1), which evaluates to sqrt(1/3). For the function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3).
b) The inner product space C[0, 1] induces a norm on the set of functions, and we can define the distance between two functions as the norm of their difference. In this case, the norm of the difference between the functions u and v is |u - v| = sqrt((u - v, u - v)) = sqrt(integral of (x - (x+1))^2 dx from 0 to 1), which simplifies to sqrt(1/3). Therefore, |u - v| = sqrt(1/3).
c) The inner product (f, g) between the functions f and g is defined as the integral of their pointwise product over the interval [0, 1]. For the functions u = x and v = x + 1, (u, v) = integral of (x * (x + 1)) dx from 0 to 1, which evaluates to 7/6.
d) The angle between two functions u and v in an inner product space can be computed using the definition of the inner product and the norms of the functions. The angle theta between u and v is given by the equation cos(theta) = (u, v) / (u * v). In this case, cos(theta) = (7/6) / (sqrt(1/3) * sqrt(5/3)), which simplifies to 1/√5. Therefore, the angle between u and v is arccos(1/√5).
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Write the equation of the line based on the given information explain the answer in slope intercept form
Answer:
y = 2/3 x - 3
Step-by-step explanation:
substitute the given slope and the coordinate into the slope-intercpet form:
y = mx + b to solve for b,
-1 = 2/3(3) + b
=> b = -3
y = 2/3 x - 3
a hiker goes 6 miles east and then turns south. if the hiker finishes 7.2 miles from the starting point. how far south did the hiker go?
Answer:
1.2
Step-by-step explanation:
You have to subtract 7.2 - 6 = 1.2
East and south direction motion are perpendicular to each other. The hiker went approx 4 miles in south.
How are directions related?East or west are perpendicular (forming 90° ) to north or south directions.
North is 180° to south, and east is 180 degrees to west.
What is Pythagoras Theorem?If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
where |AB| = length of line segment AB.
Considering the diagram attached below, we deduce that the triangle ABC is right angled triangles, and thus, its sides' lengths will follow Pythagoras theorem.
The movement the hiker did in the south direction is the length of the side BC, denoted by |BC| = x miles (assume)
Then, by applying Pythagoras theorem, we get:
[tex]|AC|^2 = |AB|^2 + |BC|^2\\(7.2)^2 = 6^2 + x^2\\51.84 = 36 + x^2\\\\\text{Subtracting 36 from both the sides and taking root, we get}\\\\x = \sqrt{15.86} \approx 3.98 \approx 4 \: \rm miles[/tex]
(took positive root since distance is a non-negative quantity)
Thus, the hiker went approx 4 miles in south.
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Drag each sign or value to the correct location on the equation. Each sign or value can be used more than once, but not all signs and values will
be used.
The focus of a parabola is (-4,-5), and its directrix is y=-1. Fill in the missing terms and signs in the parabola's equation in standard form.
Answer: (x+4)² = -8 (y+3)
Step-by-step explanation: I got this correct on Edmentum.
The equation of a line in the form y = a number creates this kind of line.
Answer:
a diagonal one??? like its linear uhm okay here i have a picture one sec
Step-by-step explanation:
Determine if the following system is Lyapunov stable x=tx
The system x = tx is Lyapunov stable.
To determine if the system x = tx is Lyapunov stable, we need to check if it satisfies the Lyapunov stability criterion. The criterion states that for every ε > 0, there exists a δ > 0 such that if ||x(0) - xe|| < δ, then ||x(t) - xe|| < ε for all t > 0.
Let's analyze the given system x = tx:
Taking the derivative of x with respect to t, we get:
dx/dt = x
This is a first-order linear ordinary differential equation.
The general solution to this equation is:
x(t) = C×[tex]e^{t}[/tex]
where C is the constant of integration.
Now, let's consider the initial condition x(0) = x0, where x0 is some constant.
Using this initial condition, we can solve for the constant C:
x(0) = C × e⁰
x0 = C
So, the specific solution to the initial value problem is:
x(t) = x₀ × [tex]e^{t}[/tex]
Now, we can check if the system is Lyapunov stable.
For any ε > 0, let's consider the case where δ = ε. Then, if ||x(0) - xe|| < δ, we have:
||x(0) - xe|| = ||x₀ - x₀ × [tex]e^{t}[/tex]|| = ||x₀(1 - [tex]e^{t}[/tex])||
Since e^t > 1 for t > 0, we can rewrite the expression as
||x₀(1 - [tex]e^{t}[/tex])|| = x₀ × ||[tex]e^{t}[/tex] - 1||
Now, we want to show that ||x(t) - xe|| < ε for all t > 0. We have
||x(t) - xe|| = ||x₀ × [tex]e^{t}[/tex] - x₀ × [tex]e^{t}[/tex]|| = 0
Since ||x(t) - xe|| = 0 for all t > 0, it satisfies the condition ||x(t) - xe|| < ε.
Therefore, the given system x = tx is Lyapunov stable.
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The question is incomplete the complete question is :
Determine whether the origin of the following system is Lyapunov stable: x= tx. Use Lyapunov's stability criterion .
the impressionist period in music was about creating
Answer:
The Impressionist period was distinctive because composers were focused on creating an impression through building atmospheres, pictures, and sound worlds with music. Composers Debussy and Ravel were at the forefront of this ground-breaking musical style.
Find the general solution of the third order equation, in real form without using Laplace Transform ỹ + 343y = 0
Given equation is y''' + 343y = 0 which is a third-order linear differential equation. To find the general solution of the equation, we can use the characteristic equation of the differential equation as follows;
Let y = erx be the trial solution of the differential equation, where e is the exponential function and r is an unknown constant to be determined. Substituting the trial solution into the differential equation, we have; y''' + 343y = 0 y' = rerx, y'' = rerx, y''' = rerx. Substituting into the differential equation, we have;r³erx + 343erx = 0. Factorizing out erx, we have erx(r³ + 343) = 0For erx ≠ 0;r³ + 343 = 0r³ = -343r = (-343)¹/³ = -7 (r = -7, since we are working with real forms of the equation). Therefore, the general solution of the third order linear differential equation is;
y = c₁e^-7x + c₂e^-7x cos(12.124x) + c₃e^-7x sin(12.124x) where c₁, c₂, and c₃ are arbitrary constants.
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a manufacturer is designing a new container for their chocolate-covered almonds. Their original container was a cylinder with the height of 18 cm and a diameter of 14 cm. the new container can be modeled by a rectangular prism with a square base and will contain the same amount of chocolate-covered almonds. if the new container's height is 16 cm, determine and state, to the nearest tenth of a centimeter, the side length of the new container if both containers contain the same amount of almonds
Answer:
answer is 24
Step-by-step explanation:
The side length of the rectangular prism with a square base is 13.2 cm if the original container was a cylinder with the height of 18 cm and a diameter of 14 cm.
What is a cylinder?In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.
We know the volume of the cylinder is given by:
[tex]\rm V = \pi r^2 h[/tex]
The radius of the cylinder r = 14/2 = 7 cm
The volume of the cylindrical container V = π(7)²(18) = 882π cubic cm
Let's suppose the side length is L:
Then the volume of the rectangular prism is the same as the cylindrical container:
882π = L×L×16 (h = 16 cm)
882π = 16L²
L = 13.15 ≈ 13.2 cm
Thus, the side length of the rectangular prism with a square base is 13.2 cm if the original container was a cylinder with the height of 18 cm and a diameter of 14 cm.
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Please help! Will give Brainliest! SHOW ALL WORK
Factor:
2x^(2)+ay-ax^(2)-2y
Answer:
(2 - a) (x² - y)
Step-by-step explanation:
2x² + ay - ax² - 2y
=> 2x² - 2y + ay - ax²
=> 2 (x² - y) + -a (-y + x²)
=> 2 (x² - y) + -a (x² - y)
=> (2 - a) (x² - y)
The area of a rectangle
is 100 square feet. If the
width is 25 feet. What is the
Length?
solve for x round to your nearest tenth
Answer:
x ≈ 5.98Steps:
tan(50.1°) = x/5
1.19.. = x/5
x = 5 × 1.19 ..
x = 5.97.. ≈ 5.98
An urban economist wishes to estimate the proportion of Americans who own their house. What size sample should be obtained if he wishes the estimate to be within 0.02 with 90% confidence if
(a) he uses a 2010 estimate of 0.669 obtained from the U.S Census Bureau?
(b) he does not use any prior estimate.?
(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.
n = 1.497.87 or 1,498
(b) The researcher should use a sample size of 1,692 if the proportion is unknown.
n = 1,691.06 or 1,692
Sample Size:When a researcher is designing a sampling study to measure a population parameter, like the proportion, the sample size must be determined. The key to calculating sample size is the margin of error the researcher desires. The smaller the margin of error, the larger the sample size needed.
(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.
To find sample size for the proportion use the equation:
[tex]n=\frac{Z^2\pi (1-\pi )}{e^2}[/tex]
Where:
n is the sample sizeZ is the Z score which from a Z table, or Excel, is found to be 1.6449 for 90% confidence level.[tex]\pi[/tex] = the population proportion = 0.669e = is the margin of error. = 0.02Substituting values:
[tex]n=\frac{1.6449^2*0.50*0.331}{0.02^2}[/tex]
n = 1.497.87 or 1,498
(b) The researcher should use a sample size of 1,692 if the proportion is unknown.
Use the same equation as in part (a). However, to maximize [tex]\pi (1-\pi )[/tex] use 0.5 for [tex]\pi[/tex].
Substituting the values:
[tex]n=\frac{1.6449^2*0.50*0.500}{0.02^2}[/tex]
n = 1,691.06 or 1,692
Always round up because it is impossible to have a fraction of an observation.
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Question is in picture
Mr. Singh needed to understand the trend of his students’ latest test scores. Find the mean and median for each set of data, and then determine which gives a better picture of the scores.
Scores: 89, 99, 80, 5, 90, 0, 100, 95
Answer:
See explanations below
Step-by-step explanation:
Mean is the average of the data
Mean = sum of data/sample size
Sum of data = 89+99+80+5+90+0+100+95
Sum of data = 558
Sample size = 8
Mean = 558/8
Mean = 69.75
Median is the data value in the middle after rearrangement
Rearrange
0 , 5, 80) 89, 90, (95, 99 100
Median = 89+90/2
Median = 89.5
The parameter that gives a better picture of the scores is the Mean since all the given datas are taken into consideration
The mean test score is
✔ 69.75
.
The median test score is
✔ 89.5
.
Which measure of center best represents the test data?
✔ median
I hope this helps!
True or False (50 question)
_____ 1. The sum of the probabilities can give one and sometimes exceed one
_____ 2. The alpha value indicates the % of error allowed in the investigation
_____ 3. The critical value is obtained from the formulas applied to each Test
_____ 4. H0 is rejected when the manual value is greater than the critical value
_____ 5. All probability is between zero and 1
_____ 6. The ANOVA Test uses the entire bell
_____ 7. There are 2 types of hypotheses
_____ 8. The null hypothesis may posit that there is no significant difference between the statistic and the parameter or between 2 parameters
_____ 9. When α =.05 then you allow 95% error in the study
____ 10. The critical value is where the rejection zone for H0 begins
. ____ 11. In the binomial distribution r can be greater than n.
____ 12. If H0 ≤ 30 then H1 cannot be determined
____ 13. In the normal distribution the probability is in agreement to the total deviations you find
____ 14. In an incompatible event the probabilities must all give to one
____ 15. In Classical Probability the sample space is always known.
____ 16. In Binomial when n =30, the Pr can give greater than one.
____ 17. All probability is in Pr ≤ 1 and zero
____ 18. In the Binomial Distribution the Pr (0) is part of the probabilities.
____ 19. In Poison Distribution, N is known and its average N is used.
____ 20. Incompatible events are mutually exclusive.
____ 21. The [Pr (Oc) + Pr (Not Oc)] can give greater than one, sometimes.
____ 22. Incompatible events can be seen in classical probability
______23. The variable "eye color" is a qualitative variable, nominal
______24. The "scatter diagram" illustrates the relationship between 2 variables.
______25. In the range of grouped data you subtract the extreme values.
______26. The independent variable is manipulated by the researcher.
______27. In Poison, with lambda = 20, then Pr (X >1) = 1 – [Pr (0, 1)]
______28. In the Binomial distribution, with n=10, the Pr (r > 1) = 1 –[Pr (0, 1)]
______29. The Pr (Oc) + Pr (Not Oc) = 1
______30. In Normal distribution the sum of the Probabilities equals one.
_______31. In the Binomial Distribution r can be greater than n.
_______32. The bar chart is used to illustrate the relationship of nominal qualitative variables
_______33. In Binomial Distribution, with n =25, the sum of all Pr = 1
_______3. 4. In the Poison Distribution, with lambda = 30, the Pr (X = 31) is > 1
_______35. The null and alternate hypotheses are mutually exclusive.
_______36. Let a and b be independent events, Pr(No Oc) = 1 –[Pr(a) Pr (b)].
_______37. Let A and B be Dependent Events, Pr(A)Pr (B/A) = Pr(B)Pr (A/B)
_______38. The Correlation Coefficient can be negative
_______39. The Scatter Plot illustrates the dispersion of the data
_______40. All Pr (Oc) give to one.
_______41. When Z is negative also the probability found
_______42. In Normal Dist when you have Pr(Z ≤ 3) = 1- [Pr (Z=3)]
_____43. The arithmetic average is sought by adding all the data/(n-1)
_____44. The location of the median is found by looking for (n + 1) / 2
_____45. Median of ungrouped data does not use outliers
_____46. In the Normal Dist Z = total of the events
_____47. When Z = 1, the probability = .5 - .3413
_____48. Variance in grouped data is divided by n + 1
_____49. If r > 1 then the variables are directly proportional
_____50. Every probability is a proper fraction
1. False: The sum of the probabilities can never exceed one.
2. False: The alpha value indicates the maximum level of error allowed in the investigation.
3. True: The critical value is obtained from the formulas applied to each test.
4. False: H0 is rejected when the test statistic value is greater than the critical value.
5. True: All probability is between zero and one.
6. False: The ANOVA test uses only a portion of the bell curve.
7. True: There are two types of hypotheses.
8. True: The null hypothesis may posit that there is no significant difference between the statistic and the parameter or between two parameters.
9. True: When α = 0.05, it means that the level of significance is 5%, not that you allow 95% error in the study.
10. True: The critical value is the point at which the rejection zone for H0 begins.
11. False: In binomial distribution, r cannot be greater than n.
12. False: The value of H1 depends on the alternative hypothesis.
13. False: In a normal distribution, the probabilities can never exceed one.
14. True: In incompatible events, the sum of probabilities must be one.
15. True: In classical probability, the sample space is always known.
16. False: When n = 30 in a binomial distribution, Pr can never be greater than one.
17. True: All probability values are between 0 and 1.
18. True: In binomial distribution, Pr(0) is one of the probabilities.
19. True: In Poisson distribution, N is known, and its average N is used.
20. True: Incompatible events are mutually exclusive.
21. False: The sum of Pr(Oc) and Pr(Not Oc) can never exceed one.
22. True: Incompatible events can be seen in classical probability.
23. True: The variable "eye color" is a qualitative variable, nominal.
24. True: The "scatter diagram" illustrates the relationship between two variables.
25. True: In the range of grouped data, you subtract the smallest value from the largest value.
26. True: The independent variable is manipulated by the researcher.
27. False: In Poisson distribution, Pr(X > 1) = 1 - Pr(0) - Pr(1).
28. False: In binomial distribution, Pr(r > 1) = 1 - Pr(0) - Pr(1).
29. True: The sum of Pr(Oc) and Pr(Not Oc) is always one.
30. True: In normal distribution, the sum of probabilities equals one.
31. False: In binomial distribution, r cannot be greater than n.
32. True: The bar chart is used to illustrate the relationship of nominal qualitative variables.
33. True: In binomial distribution, the sum of all Pr is 1.
34. False: In the Poisson distribution, Pr(X = 31) is always less than or equal to 1.
35. False: The null and alternative hypotheses are not mutually exclusive.
36. False: The given formula is only true for independent events.
37. True: The given formula is only true for dependent events.
38. True: The correlation coefficient can be negative.
39. True: The scatter plot illustrates the dispersion of the data.
40. True: The sum of all Pr(Oc) is always one.
41. True: When Z is negative, the probability is also negative.
42. False: In normal distribution, Pr(Z ≤ 3) = Pr(Z = 3) + Pr(Z < 3).
43. False: The arithmetic average is found by adding all the data and dividing by n.
44. True: The location of the median is found by looking for (n + 1) / 2.
45. True: Median of ungrouped data does not use outliers.
46. False: In normal distribution, Z is equal to the number of standard deviations away from the mean.
47. False: When Z = 1, the probability is 0.3413, not 0.5 - 0.3413.
48. True: Variance in grouped data is divided by n - 1.
49. False: If r > 1, then the variables are not directly proportional.
50. True: Every probability is a proper fraction.
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how might larcom’s initial impression of the mill have been different if she had started as a machine worker?
If Harriet Hanson Robinson (Larcom) had started her career in the mill as a machine worker instead of as a doffer, her initial impression of the mill may have been significantly different.
Here are a few ways her perspective might have been altered:
Physical Experience: As a machine worker, Larcom would have been directly involved in operating the machinery. She would have experienced the physical demands and potentially dangerous conditions associated with working with heavy machinery. This hands-on experience might have given her a greater appreciation for the risks and challenges faced by the workers. Social Interaction: Machine workers often worked in close proximity to each other, tending to the machines and coordinating their tasks. By working alongside her fellow machine workers, Larcom would have had more direct contact and interaction with her peers. This could have provided her with a deeper understanding of the camaraderie, unity, and challenges faced by the workers as a community.
Skill Development: Operating the machines required a certain level of technical skill and knowledge. If Larcom had started as a machine worker, she would have gained expertise in machine operation and maintenance. This technical knowledge could have given her a different perspective on the machinery, its intricacies, and its impact on the workers. Perspective on Management: As a machine worker, Larcom might have had more direct interactions with mill managers and overseers. This could have given her insights into the management practices, decision-making processes, and power dynamics within the mill. Such firsthand experiences might have influenced her initial impression of the mill's management and their treatment of workers.
Overall, starting as a machine worker would have provided Larcom with a different vantage point and firsthand experience of the mill's operations. This could have shaped her understanding of the work environment, the challenges faced by the workers, and the dynamics between management and labor.
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1 3/4 cups of flour for one batch but he wants to make 1 1/2 batches. How many more cups of flour will he need
Answer:
[tex]2\frac{5}{8} cups = 1\frac{1}{2}\ batches[/tex]
Step-by-step explanation:
Given
[tex]1\frac{3}{4} cups = 1\ batch[/tex]
Required
Number of cups for [tex]1\frac{1}{2}[/tex] batches
[tex]1\frac{3}{4} cups = 1\ batch[/tex]
Multiply both sides by [tex]1\frac{1}{2}[/tex]
[tex]1\frac{1}{2} * 1\frac{3}{4} cups = 1\ batch * 1\frac{1}{2}[/tex]
[tex]1\frac{1}{2} * 1\frac{3}{4} cups = 1\frac{1}{2}\ batches[/tex]
Express fractions as improper fractions
[tex]\frac{3}{2} * \frac{7}{4} cups = 1\frac{1}{2}\ batches[/tex]
[tex]\frac{21}{8} cups = 1\frac{1}{2}\ batches[/tex]
Express fraction as improper fraction
[tex]2\frac{5}{8} cups = 1\frac{1}{2}\ batches[/tex]
Find the perimeter of rectangle with length 6 inches and width 3 inches.
Answer:
18
Step-by-step explanation:
Formula
P = 2 * (L + W)
Solution
L = 6
W = 3
P = 2 (6 + 3)
P = 2 (9)
P = 18
Use cylindrical coordinates. ∭_e x^2 dv .where E … solid thati x2 dV, where E is the solid that lies within the cylinder x^2 + y^2=4, above the plane z = 0, and below the cone z^2 4x^2 + 4y^2
The value of the triple integral ∭_E x² dV over the given solid E is 64π/5.
To evaluate the triple integral ∭_E x^2 dV, where E is the solid that lies within the cylinder x² + y² = 4, above the plane z = 0, and below the cone z² = 4x² + 4y², we can express the integral in terms of cylindrical coordinates.
In cylindrical coordinates, we have:
x = r cos(theta)
y = r sin(theta)
z = z
The limits for the cylindrical coordinates are as follows:
0 ≤ r ≤ 2 (limits for the radius)
0 ≤ theta ≤ 2π (limits for the angle)
0 ≤ z ≤ sqrt(4r²) = 2r (limits for the height, as per the cone equation)
Now let's express the integral in cylindrical coordinates:
∭_E x² dV = ∭_E (r cos(theta))² r dz dr d(theta)
Let's evaluate the integral step by step:
∫[0 to 2π] ∫[0 to 2] ∫[0 to 2r] (r³ cos²(theta)) dz dr d(theta)
We can simplify the innermost integral with respect to z:
∫[0 to 2π] ∫[0 to 2] [r³ cos²(theta)z] |[0 to 2r] dr d(theta)
= ∫[0 to 2π] ∫[0 to 2] (r³ cos²(theta)(2r)) dr d(theta)
= 2 ∫[0 to 2π] ∫[0 to 2] (2r⁴ cos²(theta)) dr d(theta)
Now, let's integrate with respect to r:
2 ∫[0 to 2π] [(1/5) r⁵ cos²(theta)] |[0 to 2] d(theta)
= 2 ∫[0 to 2π] [(1/5)(32 cos²(theta))] d(theta)
= (64/5) ∫[0 to 2π] cos²(theta) d(theta)
To evaluate the remaining integral, we can use the identity cos²(theta) = (1 + cos(2theta))/2:
(64/5) ∫[0 to 2π] [(1 + cos(2theta))/2] d(theta)
= (64/10) ∫[0 to 2π] (1 + cos(2theta)) d(theta)
= (64/10) [(theta + (1/2)sin(2theta))] |[0 to 2π]
= (64/10) [(2π + (1/2)sin(4π)) - (0 + (1/2)sin(0))]
= (64/10) (2π + 0 - 0)
= (64/10) (2π)
= 128π/10
= 64π/5
Therefore, the value of the triple integral ∭_E x² dV over the given solid E is 64π/5.
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What’s the answer for number 1 ????
Answer:
there are 8 packages left
Step-by-step explanation:
120÷12=10. So there are 10 packages of pencils to begin with. She loses 24 pencils which means that she lost two packages worth of pencils (12+12=24). So, 2 packages-10 packages=8 packages.
Use the appropriate substitution to find the explicit solution
to the Bernoulli DE dy/dt-y=e^(-2t)*y^2
[tex]y = (e^{(-2t)} + C . e^{(-t)})^{(-1)}[/tex]. is the explicit solution to the given Bernoulli differential equation. we used an appropriate substitution to transform it into a linear differential equation.
The given Bernoulli differential equation is in the form [tex]dy/dt - y = e^{(-2t)} . y^2[/tex]. Bernoulli equations can be transformed into linear equations by using a substitution. In this case, we can let [tex]v = y^{(-1)[/tex]be our substitution.
Differentiating v with respect to t gives dv/dt = -y^(-2) * dy/dt. Substituting this into the original equation, we have:
[tex]-dv/dt - v = e^{(-2t)}.[/tex]
Now, we have transformed the Bernoulli equation into a linear first-order differential equation. We can solve this equation using standard methods. Multiplying through by -1, we get [tex]dv/dt + v = -e^{(-2t)}[/tex].
The integrating factor for this linear equation is given by [tex]e^{ \[ \int_{}^{}1\,dt \] = e^t[/tex]. Multiplying both sides of the equation by e^t, we have:
[tex]e^t. (dv/dt + v) = -e^{(t-2t)[/tex].
Simplifying further, we get [tex]d/dt (e^t.v) = -e^{(-t)[/tex].
Integrating both sides with respect to t, we obtain:
[tex]e^t . v = \[ \int_{}^{} -e^{(-t)} \,dt \][/tex].
Evaluating the integral, we get:
[tex]e^t . v = e^{(-t) }+ C[/tex],
where C is the constant of integration.
Finally, solving for v, we have:
[tex]v = e^{(-2t)} + C .e^{(-t)[/tex].
Since [tex]v = y^{(-1)[/tex], we can take the reciprocal of both sides to find the explicit solution for y:
[tex]y = (e^{(-2t) }+ C .e^{(-t)})^{(-1)}[/tex].
This is the explicit solution to the given Bernoulli differential equation.
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Use these equations to find ∂z/∂x and ∂z/∂y for the following.
x8 + y8 + z5 = 6xyz
For the given equation x⁸ + y⁸ + z⁵ = 6xyz, the value of partial derivatives are ∂z/∂x = (6yz - 8x⁷) / (5z⁴ - 6xz) and ∂z/∂y = (6xz) / (8y⁷ - 6xy)
To find the partial derivatives ∂z/∂x and ∂z/∂y for the equation x⁸ + y⁸ + z⁵ = 6xyz, we need to differentiate the equation with respect to x and y while treating y and z as constants.
Taking the partial derivative with respect to x (∂z/∂x), we differentiate each term separately:
8x⁷ + 0 + 5z⁴ (∂z/∂x) = 6yz + 6xz(∂z/∂x)
Simplifying the equation, we get:
8x⁷ + 5z⁴ (∂z/∂x) = 6yz + 6xz(∂z/∂x)
Now, let's take the partial derivative with respect to y (∂z/∂y):
0 + 8y⁷ + 0 (∂z/∂y) = 6xz + 6xy(∂z/∂y)
Simplifying further:
8y⁷ (∂z/∂y) = 6xz + 6xy(∂z/∂y)
To find the values of ∂z/∂x and ∂z/∂y, we need to isolate the partial derivatives:
∂z/∂x = (6yz - 8x⁷) / (5z⁴ - 6xz)
∂z/∂y = (6xz) / (8y⁷ - 6xy)
These equations give the partial derivatives of z with respect to x and y for the given equation x⁸ + y⁸ + z⁵ = 6xyz.
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Find the critical t-value that corresponds to 95% confidence. Assume 15 degrees of freedom.
The critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom is approximately 2.131.
To find the critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom, we can use a t-table or a statistical software.
Using a t-table or a statistical software, we look up the critical t-value for a two-tailed test with a confidence level of 95% and 15 degrees of freedom.
For a two-tailed test, we divide the desired confidence level by 2 to account for both tails of the t-distribution. In this case, we divide 95% by 2, which gives us 0.475.
Looking up the critical t-value for a confidence level of 0.475 and 15 degrees of freedom, we find that the critical t-value is approximately 2.131.
Therefore, the critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom is approximately 2.131.
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Classify the following expression by degree and number of terms. 6 - 3x - 2
Answer:
in terms of degree it is linear equation
in terms of number of terms it is binomial
Among various populations of plants or animals, diseases spread exponentially. Use the function y = 6000(1 - e^-0.154)
to model the spread of the Avian Bird Flu disease among a flock of 6000 chickens on a chicken farm, with t equal to
the number of days since the first case of the disease. How many birds will be infected with the Flu after 6 days?
3560 chickens
C. 1280 chickens
356 chickens
d. 516 chickens
a.
Answer: A. 3560 chickens
Step-by-step explanation:
Plugged in the 6 days into the equation provided
Birds will be infected with the Flu after 6 days will be 3560 chickens
What is exponential?An exponential function is a mathematical function of the following form: f ( x ) = a^x. where x is a variable, and a is a constant called the base of the function.
Given function: y = 6000(1 - [tex]e^{-0.15t}[/tex])
Total birds are =6000
Now, take t= 6 days.
So, y = 6000(1 - [tex]e^{-0.15t}[/tex])
y= 6000(1- [tex]e^{-0.15 * 6}[/tex])
= 6000 (1-0.40656)
= 6000*0.59343
= 3560.58204
= 3560 birds
Hence, 3560 birds get infected in 6 days.
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i need to know what -2(3x)+56
Answer:
-6x + 56
Step-by-step explanation:
-2(3x) + 56
-6x + 56
Given the differential equation: dy/dx +xy = 3x- e+ 2 with the initial condition y(O) = 1, find the values of y corresponding to the values of Xo+0.1 and Xo+0.2 correct to four decim
the approximate values of y at x = X₀ + 0.1 and x = X₀ + 0.2 are:
y ≈ 0.9282 at x = 0.1
y ≈ 0.8281 at x = 0.2
To solve the given differential equation, we will use an appropriate method, such as the Euler's method, to approximate the values of y at specific points.
The Euler's method uses the equation:
y₍ₙ₊₁₎ = yₙ + h * f(xₙ, yₙ),
where:
yₙ is the value of y at xₙ
h is the step size,
f(x, y) is the derivative of y with respect to x (i.e., dy/dx), and
y₍ₙ₊₁₎ is the approximation of y at the next point x₍ₙ₊₁₎ = xₙ + h.
Let's apply Euler's method to find the values of y at x = X₀ + 0.1 and x = X₀ + 0.2, where X₀ is the initial condition x = 0 and y(X₀) = 1.
Given the differential equation:
dy/dx + xy = 3x² - [tex]e^y[/tex] + 2
Rewriting the equation in the form:
dy/dx = -xy + 3x² - [tex]e^y[/tex] + 2
We have the initial condition:
y(0) = 1
Using Euler's method with a step size of h = 0.1:
1. At x = X₀ + 0.1:
y₁ = y₀ + h * [ -x₀ * y₀ + 3 * x₀² - [tex]e^y_0[/tex] + 2 ]
= 1 + 0.1 * [ -(0) * (1) + 3 * (0)² - e¹ + 2 ]
= 1 + 0.1 * (0 - e + 2)
= 1 + 0.1 * (-e + 2)
= 1 + 0.1 * (-2.7183 + 2)
= 1 + 0.1 * (-0.7183)
= 1 - 0.07183
≈ 0.9282
2. At x = X₀ + 0.2:
y₂ = y₁ + h * [ -x₁ * y₁ + 3 * x₁² - [tex]e^y_1[/tex] + 2 ]
= 0.9282 + 0.1 * [ -(0.1) * (0.9282) + 3 * (0.1)₂ - [tex]e^{0.9282}[/tex] + 2 ]
= 0.9282 + 0.1 * (-0.009282 + 0.003 - [tex]e^{0.9282}[/tex] + 2)
≈ 0.8281
Therefore, the approximate values of y at x = X₀ + 0.1 and x = X₀ + 0.2 are:
y ≈ 0.9282 (correct to four decimal places) at x = 0.1
y ≈ 0.8281 (correct to four decimal places) at x = 0.2
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Plz help ASAP and thank you!
Answer:
Plot attached
Step-by-step explanation:
We want to locate √36 and ∛36 on the given number line
Now, √36 = 6.
And ∛36 = 3.3
I have attached a plot number line both values as we are told in the question