It should be noted that the profit allocated to Veronica Visentin based on the contribution will be $35,675
How to calculate the valueZoe, Karen, and Veronica each withdraw $30,100 cash as a "year-end bonus."
Zoe's year-end withdrawal: $30,100
Karen's year-end withdrawal: $30,100
Veronica's year-end withdrawal: $30,100
Profit Allocation:
Total profit for 2020: $109,000
Profit-sharing ratio: Zoe (3), Karen (2), Veronica (3)
Zoe's share: ($109,000 / 8) * 3 = $40,875
Karen's share: ($109,000 / 8) * 2 = $27,250
Veronica's share: ($109,000 / 8) * 3 = $40,875
2020 Ending Capital Balances:
Zoe Moreau: Initial contribution + Share of profit - Year-end withdrawal
= $51,100 + $40,875 - $30,100 = $61,875
Karen Krneta: Initial contribution + Share of profit - Year-end withdrawal
= $10,100 + $27,250 - $30,100 = $7,250
Veronica Visentin: Initial contribution + Share of profit - Year-end withdrawal
= $24,900 + $40,875 - $30,100
= $35,675
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x/2 + 4 < 18
What is the value of x?
And what does the point on the number line look like?
Someone help me
Worth 29 points
Answer:
x<28
Step-by-step explanation:
Isolate x
First, subtract 4 on both sides
x/2+<14
Then, multiply both sides by 2 to get x alone
x<28
On a number line, there would be an open circle (not filled in dot) on 28, and the entire left side of the number line would be filled in
Answer:
x<28
Step-by-step explanation:
x/2+4<18
multiply the 2 on both sides to get rid of it
x+8<36
isolate the x
x<28
on the number line, it's an open circle with the arrow pointing to the left.
Let k be a constant and consider the function f(x,y,z) = kx? - kry + y2 -2yz - 22. (Thus, for example, if k = 4, then f(xy.z) = 4x2 - 4xy + 2y2-2yz -22) For what values (if any) of the constant k does / have a (nondegenerate) local maximum at (0.0.0)? For what values of k does / have a (nondegenerate) local minimum at (0.0.0)? Be sure to explain your reasoning.
The values of k for which the function f(x, y, z) = kx² - kry + y² - 2yz - 22 has a nondegenerate local maximum at (0, 0, 0) are when k > 0.
To find the critical points of the function, we need to calculate the partial derivatives with respect to each variable:
∂f/∂x = 2kx ∂f/∂y = -kr + 2y - 2z ∂f/∂z = -2y
2kx = 0 => x = 0 (Equation 1) -kr + 2y - 2z = 0 => r = y - z (Equation 2) -2y = 0 => y = 0 (Equation 3)
From Equation 3, we can see that y = 0. Substituting this into Equation 2, we get:
r = 0 - z r = -z (Equation 4)
The Hessian matrix is given by:
H = | ∂²f/∂x² ∂²f/∂x∂y ∂²f/∂x∂z | | ∂²f/∂y∂x ∂²f/∂y² ∂²f/∂y∂z | | ∂²f/∂z∂x ∂²f/∂z∂y ∂²f/∂z² |
Calculating the second-order partial derivatives:
∂²f/∂x² = 2k ∂²f/∂y² = 2 ∂²f/∂z² = 0 ∂²f/∂x∂y = 0 ∂²f/∂y∂z = -2 ∂²f/∂z∂x = 0
Thus, the Hessian matrix becomes:
H = | 2k 0 0 | | 0 2 -2 | | 0 -2 0 |
D = ∂²f/∂x² ∂²f/∂y² ∂²f/∂z² + 2∂²f/∂x∂y ∂²f/∂y∂z ∂²f/∂z∂x - (∂²f/∂x² ∂²f/∂y∂z ∂²f/∂z∂x + ∂²f/∂y² ∂²f/∂z∂x ∂²f/∂x∂y ∂²f/∂z²)
Substituting the partial derivatives we calculated earlier:
D = (2k)(2)(0) + 2(0)(-2)(0) - (2k)(-2)(0) - (2)(0)(0) D = 0
If the determinant D is zero, the second derivative test is inconclusive. In such cases, we need to consider the eigenvalues of the Hessian matrix.
To find the eigenvalues, we solve the characteristic equation:
det(H - λI) = 0
where λ is the eigenvalue and I is the identity matrix. Substituting the values from the Hessian matrix:
| 2k-λ 0 0 | | 0 2-λ -2 | | 0 -2 -λ |
The characteristic equation becomes:
(2k - λ)((2 - λ)(-λ) - (-2)(0)) - (0)((2 - λ)(-2) - (0)(0)) = 0 (2k - λ)(λ² - 2λ) = 0
From this equation, we can see that one eigenvalue is (2k - λ) = 0, which implies λ = 2k.
For our case, we have one eigenvalue (λ = 2k). Thus, the sign of λ depends on the value of k.
When k < 0, the point (0, 0, 0) is a nondegenerate local minimum. When k = 0, the second derivative test is inconclusive, and further analysis would be required to determine the nature of the critical point.
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Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30.
a. Find the probability that the number of successes is greater than 310.
P(X ˃ 310) = _____ (round to four decimal places as needed and show work)
b. Find the probability that the number of successes is fewer than 250.
P(X ˂ 250) = _____ (round to four decimal places as needed and show work)
P(X < 250) = P(X ≤ 249) = 0 (approximately) Hence, P(X ˃ 310) = 0 and P(X ˂ 250) = 0.
Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30. The probability that the number of successes is greater than 310 and the probability that the number of successes is fewer than 250 are to be found.
Solution: a)We know that P(X > 310) can be found using normal approximation.
We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.
Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630. Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.
Using the normal approximation formula, z = (X - μ) / σwhere X = 310, μ = np and σ = √(npq), we getz = (310 - 270) / √(900*0.30*0.70)z = 4.25
Using the z-table, the probability of z being greater than 4.25 is almost zero.
Therefore, P(X > 310) = P(X ≥ 311) = 0 (approximately)
b)We know that P(X < 250) can be found using normal approximation. We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.
Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630.
Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.
Using the normal approximation formula,z = (X - μ) / σwhere X = 250, μ = np and σ = √(npq), we getz = (250 - 270) / √(900*0.30*0.70)z = -4.25Using the z-table, the probability of z being less than -4.25 is almost zero.
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Given data: n = 900, P = 0.30.
a. The probability that the number of successes is greater than 310 is 0.0000.
b. The probability that the number of successes is fewer than 250 is 0.0174.
a. The formula for finding probability of binomial distribution is:
P(X > x) = 1 - P(X ≤ x)
P(X > 310) = 1 - P(X ≤ 310)
Mean μ = np
= 900 × 0.30
= 270
Variance σ² = npq
= 900 × 0.30 × 0.70
= 189
Standard deviation
σ = √σ²
= √189
z = (x - μ) / σ
z = (310 - 270) / √189
z = 4.32
Using normal approximation,
P(X > 310) = P(Z > 4.32)
= 0.00001673
Using calculator, P(X > 310) = 0.0000(rounded to four decimal places)
b. P(X < 250)
Mean μ = np
= 900 × 0.30
= 270
Variance σ² = npq
= 900 × 0.30 × 0.70
= 189
Standard deviation
σ = √σ²
= √189
z = (x - μ) / σ
z = (250 - 270) / √189
z = -2.12
Using normal approximation, P(X < 250) = P(Z < -2.12) = 0.0174.
Using calculator, P(X < 250) = 0.0174(rounded to four decimal places).
Therefore, the probability that the number of successes is greater than 310 is 0.0000 and the probability that the number of successes is fewer than 250 is 0.0174.
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Park trails and their elevation:
Sand trail has a -2 feet elevation
Cactus Trail has 15 feet elevation
Southern Trail has a -12 feet elevation
Rocky Trail has 42 feet elevation
Chi hiked the Rocky Trail What is the opposite of the elevation of the Rocky Trail?
Answer:
fjekwnkewgnelwnlgnendndj
Step-by-step explanation:
I NEED HELP WITH MATH PLS
screenshot is posted below
Answer: The correct answer is A or B
`
Step-by-step explanation:
Let R be a commutative ring with 1. An element x ER is nilpotent if x=0 for some n E N. (a) Prove that the set N(R) := {x ER: x is nilpotent} is an ideal of R. (b) Prove that N(R/N(R)) = 0.
(a) To prove that the set N(R) = {x ∈ R: x is nilpotent} is an ideal of the commutative ring R with 1.
We need to show that it satisfies the two conditions of being an ideal: closure under addition and closure under multiplication by elements of R.
To demonstrate closure under addition, let x and y be nilpotent elements in N(R). This means that there exist positive integers m and n such that xm = 0 and yn = 0.
We want to show that x + y is also nilpotent. By expanding (x + y)^k using the binomial theorem, we can see that each term involves a product of powers of x and y. Since both x and y are nilpotent, their product is also nilpotent.
Therefore, the sum (x + y) raised to a sufficiently high power will result in zero, showing that x + y is indeed nilpotent. Hence, N(R) is closed under addition.
To prove closure under multiplication by elements of R, let x be a nilpotent element in N(R) and r be any element in R. We aim to show that rx is nilpotent. Since x is nilpotent, there exists a positive integer m such that xm = 0.
When we raise rx to a sufficiently high power, (rx)^k, it can be expanded as r^k * x^k. Since x^k is zero due to x being nilpotent, the product r^k * x^k is also zero. Therefore, rx is nilpotent, and N(R) is closed under multiplication by elements of R.
Hence, N(R) satisfies both conditions of being an ideal, and thus, it is an ideal of the commutative ring R.
(b) To prove that N(R/N(R)) = 0, we want to show that every element in R/N(R) is not nilpotent.
Let [x] be an element in R/N(R), where [x] represents the equivalence class of x modulo N(R). Our goal is to demonstrate that [x] is not nilpotent, meaning it is not equal to the zero element in R/N(R).
Suppose, for contradiction, that [x] = 0 in R/N(R). This would imply that x belongs to N(R), the set of nilpotent elements in R. However, if x is an element of N(R), it means that x is nilpotent, and by definition, there exists some positive integer n such that xn = 0. This contradicts our assumption that [x] = 0, since it would imply that x is not nilpotent.
Therefore, our assumption that [x] = 0 leads to a contradiction, and we conclude that every element in R/N(R) is not nilpotent.
Consequently, N(R/N(R)) = 0, indicating that the set of nilpotent elements in the quotient ring R/N(R) is empty.
In summary, we have shown that N(R/N(R)) = 0 and established that N(R) is an ideal of the commutative ring R.
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In real-life applications, statistics helps us analyze data to extract information about a population. In this module discussion, you will take on the role of Susan, a high school principal. She is planning on having a large movie night for the high school. She has received a lot of feedback on which movie to show and sees differences in movie preferences by gender and also by grade level. She knows if the wrong movie is shown, it could reduce event turnout by 50%. She would like to maximize the number of students who attend and would like to select a PG-rated movie based on the overall student population's movie preferences. Each student is assigned a classroom with other students in their grade. She has a spreadsheet that lists the names of each student, their classroom, and their grade. Susan knows a simple random sample would provide a good representation of the population of students at their high school, but wonders if a different method would be better. a. Describe to Susan how to take a sample of the student population that would not represent the population well. b. Describe to Susan how to take a sample of the student population that would represent the population well. c. Finally, describe the relationship of a sample to a population and classify your two samples as random, cluster, stratified, or convenience.
a. To take a sample of the student population that would not represent the population well, Susan could use a biased sampling method.
For example, she could choose students only from specific classrooms or grade levels that she believes have a certain movie preference, or she could select students based on her personal biases or preferences. This would introduce sampling bias and potentially skew the results, leading to a sample that does not accurately reflect the overall student population.b. To take a sample of the student population that would represent the population well, Susan should use a random sampling method. Random sampling ensures that every student in the population has an equal chance of being selected for the sample.c. A sample is a subset of the population that is selected for analysis to make inferences about the entire population. The relationship between a sample and a population is that the sample is used to draw conclusions or make predictions about the population as a whole.To know more about Random samples:- https://brainly.com/question/30759604
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John is cutting 3 wooden sticks to build part of a kite frame. The part he is building must be a right triangle.
Select all the possible lengths, in inches, of the sticks John could cut to make a right triangle.
A. 6, 8, 10
B. 2, 5, 10
C. 2, 3, 5
D. 12, 16, 20
E. 3, 4, 13
Answer:
I found 2 answers for this one
Step-by-step explanation:
B- 2,5,10
D-12,16,20
The possible length in inches of the stick that will make a right angle triangle are as follows
6, 8, 1012, 16, 20What is a right angle triangle?A right angle triangle has one of its angles as 90 degrees.
The right angle triangle must obeys the Pythagorean theorem.
c² = a² + b²
where
c = hypotenusea and b are the other legs.Therefore, the sides that makes a right angle are as follows:
6, 8 , 10 : 6² + 8² = 10²
12, 16, 20 : 12² + 16² = 20²
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a cylinder has a volume of 500cm³ and a diameter of 18cm. which of the following is the closest to the height of the cylinder
Step-by-step explanation:
Volume of Cylinder =
[tex]500 {cm}^{3} = \pi {r}^{2} h[/tex]
given d = 18
r = 1/2 x d = 9cm,
[tex]\pi( {9}^{2} )h = 500 \\ 81\pi \: h = 500 \\ h = \frac{500}{81\pi} cm[/tex]
I will leave the answer in terms of Pi as I am not sure how you want to leave your answer as.
PLEASE PLEASE PLEASE HELP 7 points
Answer:
a) 2x+(x+36)=90
Step-by-step explanation:
b) A1+A2=90°. (A=angle)
2x+(x+36)=90
2x+x+36=90
3x+36=90
3x=90-36
x=54/3
x=18
then A1=2x=2*18=36°
A2=x+36=18+36=54°
what are some good editing apps i use alight motion and capcut
:))))))
Step-by-step explanation:
videochamp, picsart
Picsart , Inshot , Gandr , Photo lab and Viva video.
The radius of a circle is 8 inches. What is the area?
r=8 in
Give the exact answer in simplest form.
Answer:
Radius = 8 inches
Area = [tex]\pi {r}^{2} [/tex]
[tex]area = \pi( {8}^{2})[/tex]
[tex] = 64\pi \: square \: inches[/tex]
A = [tex]201.06 ^{2} \: inches[/tex]
Step-by-step explanation:
Hope it is helpful....
Apply the properties of exponents to determine which of these numerical expressions
are equivalent to 5^12. Select all that apply.
Very confused and forgot the rules to figuring this out.
Answer:
Second One-
[tex] {5}^{14}. {5}^{ - 2} [/tex]
Fifth One-
[tex] {5}^{6} \: . \: {5}^{6} [/tex]
Sixth One-
[tex] \sqrt{ {5}^{24} } [/tex]
Seventh One-
[tex] {5}^{11} \: . \: 5 [/tex]
Introduction
Scientists have established a timeline of events after the Big Bang, based on astronomical observations and our understanding of the physical laws of the universe, such as gravity and the speed of light. In this lab activity, you will gather evidence to support the Big Bang theory.
Problem:
How can models demonstrate theories of our expanding universe?
Hypothesis:
Review the virtual lab demonstration in the lesson and stop the video when prompted to formulate a hypothesis. Hypothesize (or predict) what will happen to the distances between the labeled circles when you blow up the balloon ¼ full, ½ full, and ¾ full. Remember to include independent and dependent variables in your hypothesis.
The carbon dioxide represents how galaxies will spread out.
Materials:
Watch the virtual lab demonstration video within the lesson. No additional materials are needed.
Variables:
For this investigation:
List the independent variable(s):
List the dependent variable(s):
List the controlled variable(s):
Procedures:
1. Watch the virtual lab demonstration video within the lesson and record your observations in Table 1.
2. Using your expanding universe data from Table 1, construct a line graph using the volume of the below on the X axis and the distance between points on the Y axis. Be sure to include units and add titles to the graphs. Refer to the graph example and graphing tutorial in the lesson if needed.
3. Complete the Questions and Conclusion section of the lab report.
Data and Observations:
Table 1: Expanding Universe Observations
Galaxies Distance: Uninflated balloon (centimeters)
Distance: ¼ full (centimeters) Distance: ½ full (centimeters) Distance: ¾ full (centimeters)
A to B
A to C
A to D
B to C
B to D
C to D
Construct a line graph using the expanding the universe data from table 1. The volume will be plotted on the x-axis. The distance between the points will be plotted on the y-axis. Be sure to include units and add titles to the graph. Refer to the graph example and graphing tutorial in the lesson if needed.
Place your graph here.
Questions and Conclusion
1. How does the density and distribution of your “stars” change as the balloon expands?
2. How does your expanding balloon model represent an expanding universe?
3. What are some shortcomings of using this model as a replica of universe expansion?
4. How does the model you created help to show that the Steady State theory is inaccurate?
5. Suggest a way that a scientist could create an even more accurate model of universe expansion.
6. What will happen to the gravitational force between stars as the universe continues to expand?
In conclusion, how did your prediction of distances between points compare to your experimental results? All I truly need is the variables question 3 and 5 and I’m good :) thank you <3 this is for science but I didn’t know which one to pick so I picked a random one lol
Coronary bypass surgery: A healthcare research agency reported that
41% of people who had coronary bypass surgery in 2008
were over the age of 65. Twelve coronary bypass patients are sampled.
Part 1 of 2
(a) What is the mean number of people over the age of 65 in a sample of 12
coronary bypass patients? Round the answer to two decimal places.
The mean number of people over the age of 65 is ?
Part 2 of 2
(b) What is the standard deviation of the number of people over the age of 65
in a sample of 12 coronary bypass patients? Round the answer to four decimal places.
The standard deviation of the number of people over the age of 65 is ?
The standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients is 1.6487.
Given that a healthcare research agency reported that 41% of people who had coronary bypass surgery in 2008 were over the age of 65 and twelve coronary bypass patients are sampled.
To determine the mean number of people over the age of 65 in a sample of 12 coronary bypass patients, we use the formula below:
Mean = np
Where n = 12 and p = 0.41.
Mean = 12(0.41)
Mean = 4.92
Therefore, the mean number of people over the age of 65 in a sample of 12 coronary bypass patients is 4.92.
To determine the standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients, we use the formula below:
Standard deviation, σ = √(n p q)
Where n = 12, p = 0.41, and q = 1 - p.
Standard deviation, σ = √(12 × 0.41 × 0.59)
Standard deviation, σ = √2.71948
Standard deviation, σ = 1.6487 (rounded to four decimal places).
Therefore, the standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients is 1.6487.
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Compute the pooled variance given the following data:
N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8
Round to two decimal places
By computing the pooled variance given the following data N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, the pooled variance is 436.40.
To compute the pooled variance given N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, we can use the formula below;
S_p² = [(n₁ - 1)S₁² + (n₂ - 1)S₂²] / (N - 2),
where S_p² = pooled variance, n₁ = sample size of first group, n₂ = sample size of second group, S₁² = variance of first group, S₂² = variance of second group, and N = total sample size.
To plug in the values, we have: N₁ = 18n₂ = 14S₁ = 7S₂ = 8
Substituting the values into the formula above we get;
S_p² = [(18 - 1)(7²) + (14 - 1)(8²)] / (18 + 14 - 2)S_p² = (17 × 49 + 13 × 64) / 30S_p² = 436.4
Round off to two decimal places to get 436.40.
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John buys 6 shirts. For every shirt you purchase, you get one for 30% off. If the normal
price for each shirt is $20.00, how much money did John spend on his shopping trip? (Tax is not being calculated.)
Answer:
$36
Step-by-step explanation:
Basically, every shirt is $6 because 30% of 20 is 6. If he buys 6 shirts, then 6 times 6 is 36 dollars spent, tax not included.
One winter day, the temperature ranged from a high of 40 °F to a low of -5 °F. By how many degrees did the temperature change?
O 55
O 25
O 45
O 35
Answer:
45
Step-by-step explanation:
the correct choice is C.
a museum gift shop sold 215 sets of dinosaurs. there were 9 dinosaurs in each set how many dinosaurs did they sell?
Let f(x)= e = 1+x. - a) Show that f has at least one real root (i.e. a number c such that f(c) = 0). b) Show that f cannot have more than one real root.
The function f(x) = e^(1+x) has at least one real root. The function f(x) = e^(1+x) cannot have more than one real root.
To show that f(x) has at least one real root, we need to find a value of x for which f(x) equals zero. Let's set f(x) = 0 and solve for x:
e^(1+x) = 0
Since e^(1+x) is always positive for any real value of x, there is no value of x that makes f(x) equal to zero. Hence, f(x) = e^(1+x) does not have any real roots. Therefore, we cannot show that f(x) has at least one real root.
b)
To show that f(x) cannot have more than one real root, we need to demonstrate that there cannot be two distinct real values, say c1 and c2, such that f(c1) = f(c2) = 0. Let's assume that f(x) = 0 at two distinct values, c1 and c2:
e^(1+c1) = e^(1+c2) = 0
However, this equation is not possible since e^(1+c1) and e^(1+c2) are always positive for any real values of c1 and c2. Therefore, f(x) = e^(1+x) cannot have more than one real root.
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.
Four different cellular phone plans are shown below.
• Plan 1 charges $0.35 per minute with no monthly fee.
Plan 2 charges a monthly fee of $10.00 plus $0.25 per minute.
• Plan 3 charges a monthly fee of $59.95 with 200 free minutes.
Plan 4 charges a monthly fee of $15.00 plus $0.20 per minute.
Which plan is the least expensive for 200 minutes of cellular phone use?
.
A. Plan 4
B. Plan 3
C. Plan 1
O
D. Plan 2
Use the table to determine a reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3? One-half One-fourth 3 DNE
The reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]
Given the limit of a function expressed as:
[tex]\lim_{x \to 3}\frac{2x^2-x+15}{x^3-5x-12}[/tex]
First, we need to substitute x = 3 into the function to have:
[tex]=\frac{2(3)^2-3+15}{3^3-5(3)-12}\\=\frac{18-3+15}{27-15-12}\\=\frac{0}{0} (indeterminate)[/tex]
Apply l'hospital rule on the function:
[tex]=\lim_{x \to 3}\frac{\frac{d}{dx} (2x^2-x+15)}{\frac{d}{dx} (x^3-5x-12)}\\=\lim_{x \to 3}\frac{4x-1}{3x^2-5}\\[/tex]
Subtitute x = 3 into the result
[tex]=\frac{4(3)-1}{3(3)^2-5}\\=\frac{12-1}{27-5}\\=\frac{11}{22}\\=\frac{1}{2}[/tex]
Hence the reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]
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Answer:
1/2
Step-by-step explanation:
i am in your walls
Find the absolute value of the number for point E.
Answer:
1
Step-by-step explanation:
Answer:
The answer is 1
Step-by-step explanation:
E is -1 and the absolute value is the posotive of any number. The positive of -1 is 1.
Which comparison is not correct?
-2 > -7
1 < -9
-3 > -8
6 > 5
Answer:
1 < -9
Step-by-step explanation:
A positive number can't be less than a negative
Answer this question to get marked as barinliest!!!!
In a normal distribution, 95% of the data falls within 1 standard deviation of
the mean.
True or False?
Answer:
False
Step-by-step explanation:
A P E X
Suppose that X₁, X₂,..., X₂ form a random sample from an exponential distribution with an unknown parameter 3. (a) Find the M.L.E. 3 of 3. (b) Let m be the median of the exponential distribution, that is, 1 P(X₁ ≤m) = P(X₁ ≥ m) = 2 Find the M.L.E. m of m. ‹8 ||
(a) MLE of $\lambda$ is obtained by maximizing the log-likelihood. Suppose that X1,X2,…,XnX1,X2,…,Xn are independent and identically distributed exponential random variables with parameter λ, then the probability density function of XiXi is given by $$f(x_i;\lambda) =\lambda e^ {-\lambda x_i}, \quad x_i\geq0. $$
The log-likelihood function is given by$$\begin{aligned}\ln L(\lambda) &= \ln (\lambda^n e^{-\lambda(x_1+x_2+\cdots+x_n)}) \\&=n\ln \lambda-\lambda(x_1+x_2+\cdots+x_n).\end{aligned}$$
The first derivative of the log-likelihood function with respect to λλ is$$\frac {d\ln L(\lambda)} {d\lambda} = \frac{n}{\lambda}-x_1-x_2-\cdots-x_n.$$
The first derivative is zero when $$\frac{n}{\lambda}-\sum_{i=1} ^{n} x_i=0. $$Hence, the MLE of λλ is $$\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}. $$
Substituting the value of $\hat{\lambda} $ gives the maximum value of the log-likelihood. So, the MLE of $\lambda$ is given by $$\boxed{\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}}. $$
The MLE of $\lambda$ is $\frac {3} {\sum_{i=1} ^{n} x_i}$.
(b) The median of the exponential distribution is given by$$m = \frac {\ln (2)} {\lambda}. $$
Therefore, the log-likelihood function for median is given by$$\begin{aligned}\ln L(m) &= \sum_{i=1}^{n} \ln f(x_i;\lambda)\\&= \sum_{i=1}^{n} \ln \left(\frac{1}{\lambda}e^{-x_i/\lambda}\right)\\&= -n\ln\lambda-\frac{1}{\lambda}\sum_{i=1}^{n}x_i.\end{aligned}$$
The first derivative of the log-likelihood function with respect to mm is$$\frac {d\ln L(m)} {dm} = \frac {1} {\lambda}-\frac {1} {\lambda^2} \sum_{i=1} ^{n}x_i\ln 2. $$
The first derivative is zero when $$\frac {1} {\lambda} =\frac{1}{\lambda^2}\sum_{i=1}^{n}x_i\ln 2.$$Hence, the MLE of mm is $$\boxed{\hat{m} = \frac{\ln 2}{\bar{x}}}.$$where $\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i.$Therefore, the MLE of m is $\frac {\ln 2} {\bar{x}}. $
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If I fail this homework, I may get beat.
Answer:
Don't click the link its a virus
The solution to a logistic differential equation corresponding to a specific hyena population on a reserve in A western Tunisia is given by P(t)= The initial hyena population 1+ke-0.57 was 40 and the carrying capacity for the hyena population is 200. What is the value of the constant k? (A) 4 (B) 8 (C) 10 (D) 20 6. Which of the following differential equations could model the logistic growth in the graph? AM 50 40 30/ 20 10 t (A) (B) dM =(M-20)(M-50) dt dM = (20-MM-50) dt dM = 35M dt dM = 35M(1000-M) dt (C) (D)
The logistic differential equation for the hyena population is given by:
dP/dt = r * P * (1 - P/K)
where P(t) is the hyena population at time t, r is the growth rate, and K is the carrying capacity.
We are given that:
P(t) = 40 + k * e^(-0.57t)
K = 200
To determine the value of k, we can plug in these values into the logistic differential equation and solve for k:
dP/dt = r * P * (1 - P/K)
dP/dt = r * P * (1 - P/200)
dP/dt = r/200 * (200P - P^2)
dP/(200P - P^2) = r dt
Integrating both sides, we get:
-1/200 ln|200P - P^2| = rt + C
where C is a constant of integration.
Using the initial condition P(0) = 40 + k, we can solve for C:
-1/200 ln|200(40+k)-(40+k)^2| = 0 + C
C = -1/200 ln|8000-480k|
Plugging in this value of C and simplifying, we get:
-1/200 ln|200P - P^2| = rt - 1/200 ln|8000-480k|
ln|200P - P^2| = -200rt + ln|8000-480k|
|200P - P^2| = e^(-200rt) * |8000-480k|
200P - P^2 = ± e^(-200rt) * (8000-480k)
Since the population is increasing, we choose the positive sign:
200P - P^2 = e^(-200rt) * (8000-480k)
Using the initial condition P(0) = 40 + k, we get:
200(40+k) - (40+k)^2 = (8000-480k)
8000 + 160k - 2400 - 80k - k^2 = 8000 - 480k
k^2 + 560k - 2400 = 0
(k + 60)(k - 40) = 0
Thus, k = -60 or k = 40. Since k represents a growth rate, it should be positive, so we choose k = 40. Therefore, the value of the constant k is option (A) 4.
For the second part of the question, the logistic equation that could model the growth in the graph is option (B) dM/dt = (20-M)*(M-50). This is because the carrying capacity is between 20 and 50, and the population growth rate is zero at both of these values (i.e. the population does not increase or decrease when it is at the carrying capacity).
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what's the distance between theses two. (-5, 1) (2, 4)
Answer: squareroot of 58
You can solve this problem simply by using the distance formula . Using the distance formula we can solve this problem by just placing the numbers and then solving the equation.