Welk is most likely to be classified as a luxury good. A luxury good is a good for which demand increases more than proportionally with income.
This means that as people's incomes increase, they are more likely to spend a larger proportion of their income on luxury goods.
The income elasticity of demand for a good is a measure of how responsive demand is to changes in income. A positive income elasticity of demand indicates that demand increases as income increases, while a negative income elasticity of demand indicates that demand decreases as income increases.
The income elasticity of demand for Welk is 4.667, which is much higher than the income elasticities of demand for Kang (-3) and Lafgar (1.667). This indicates that demand for Welk is much more responsive to changes in income than demand for Kang or Lafgar.
Therefore, Welk is most likely to be classified as a luxury good.
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Provide an appropriate response. Determine the critical value, z o. to test the claim about the population proportion p *0.325 given n-42 and p-0247 Use a 0.05. a O 11.96 +2.33 O +1.645 O +2.575
Based on the information given, it should be noted that the critical value is +2.33.
How to explain the valueWe are given that the sample size is 42, the sample proportion is 0.247, and the significance level is 0.05. We want to test the claim that the population proportion is 0.325.
The critical value is the z-score that separates the rejection region from the non-rejection region. The rejection region is the area under the standard normal curve where we would reject the null hypothesis. The non-rejection region is the area under the standard normal curve where we would fail to reject the null hypothesis.
The z-table shows that the critical value for a two-tailed test with a significance level of 0.05 is +2.33. This means that if the z-score is greater than or equal to +2.33, we would reject the null hypothesis.
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I NEED HELP ASAP PLEASEEEEE!!!!! ONLY ON 9a ND 9b!!!!!!
9A:
The problem says 35% of the 3560 applications are from boys who lived in other states.
This can be expressed as:
3560*35% = 3560*0.35 = 1246 applications.
9B:
The problem says applications to the university (3560 applications) represented 40% of all applications.
This can be expressed as:
3560 = 40% * A, where A = number of applications received in all.
To find how many applications received in all, just solve for A.
A = 3560 / 0.4 = 8900 applications.
The key to these types of problems is that "of" signals multiplication. For example, 40% of all applications is 40% * all applications.
I need help with this we my class skipped this section
Answer:
v^ 3 = 1000
so v = cube root of 1000
v = 10
Define two binary operations + and on the set Z of integers by x + y = max(x, y) and x - y = min(x, y). a. Show that the commutative, associative, and distributive properties of a Boolean algebra hold for these two operations on Z. b. Show that no matter what element of Z is chosen to be the property x + 0 = x of a Boolean alge- bra fails to hold?
The given binary operations are defined as below:
For the integers x and y, the binary operations are defined as: x + y = max (x, y) and x – y = min (x, y)
a) Commutative Property: The commutative property holds for both binary operations, + and – , because: For any x, y ∈ Z,x + y = y + x and x – y = -(y – x)Therefore, both + and – are commutative.
Associative Property: Associativity can also be shown for both binary operations, + and –, as follows: For any x, y and z ∈ Z,x + (y + z) = max (x, max(y, z)) = max (max(x, y), z) = (x + y) + z(x – y) – z = min (x, min (y, z)) = min (min(x, y), z) = (x – y) – z
Therefore, both + and – are associative.
Distributive Property: The distributive property can be shown for these binary operations, + and –, as follows: For any x, y, and z ∈ Z,x + (y – z) = max (x, min (y, z)) = min (max(x, y), max(x, z)) = (x + y) – (x + z)Therefore, both + and – are distributive.
b) For the given operations, the element that violates the property x + 0 = x is:0If x = 2, then x + 0 = max (2, 0) = 2If x = 0, then x + 0 = max (0, 0) = 0So, the property x + 0 = x holds for all integers except for 0.
For this particular element, the value of x + 0 is always 0.
Therefore, the given property fails to hold.
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If a person was reading a 528 page book and they read 22 pages every day how many days would it take them
Answer:
24 days
Step-by-step explanation:
Given
[tex]Pages = 528[/tex]
[tex]Rate = 22[/tex]
Required
The number of days
The number of days (d) is calculated as:
[tex]d = \frac{Pages}{Rate}[/tex]
[tex]d = \frac{528}{22}[/tex]
[tex]d = 24[/tex]
-0.5f - 5 < -1
help please asap!!
Step-by-step explanation:
-.5f -5 < -1
-.5f < 4
f > -8
hope this helps
The circumference of a circle is 106.76 yards. What is the circle's radius?
Answer:
The answer is 16.99
Answer:
It's 17!
Step-by-step explanation:
please help i’ll give brainliest
Michael is making scale drawings of rectangular rooms using a scale of 1 inch : 1 and 1/2 feet. He wants to use paper that has a width of 8 and 1/2 inch and a length of 11 in for the drawing. Determine whether the scale drawings for each of these rooms will fit on one piece of paper.
Choose yes or no for each set of dimensions:
16 feet by 16 feet: yes or no?
10 ft by 15 ft: yes or no?
15 ft by 20 ft: yes or no
12 ft by 16 ft: yes or no?
Answer:
16ft by 16 ft: NO
10ft by 15ft: YES
15ft by 20ft: NO
12ft by 16ft: YES
Step-by-step explanation:
First, we know that the scale used is:
1 in = (1 + 1/2) ft.
This means that each inch on the drawing is equivalent to (1 + 1/2) ft.
We know that Michael uses a paper that has the measures:
width = (8 + 1/2) in
length = 11in
Then the maximum dimensions that can be represented with this paper are:
WIDTH = (8 + 1/2)*(1 + 1/2) ft. = (8 + 8/2 + 1/2 + 1/4) ft
= (8 + 4 + 2/4 + 1/4) ft
= (12 + 3/4) ft
LENGTH = 11*(1 + 1/2) ft = (11 + 11/2)ft = (11 + 10/2 + 1/2)ft
= (11 + 5 + 1/2)ft = (16 + 1/2) ft.
Now let's analyze the options, we can only draw the rooms in the paper if the measures are equal or smaller than the ones we found above:
Where the measures are written as: "width by length".
a) 16ft by 16 ft.
width = 16ft
length = 16ft
We can not draw this, because the maximum width that we can draw is (12 + 3/4) ft, which is smaller than 16ft.
b) 10 ft by 15 ft
width = 10ft
length = 15ft
Both are smaller than the maximum measures we found, then yes, we can draw this room.
c) 15 ft by 20 ft
width = 15ft
length = 20ft
Both are larger than the maximum measures, so no, we can not draw this.
d) 12ft by 16ft
width = 12ft < (12 + 3/4) ft = maximum width
lenth = 16ft < (16 + 1/2) ft = maximum length.
Both measures are smaller than the maximum ones, then we can draw this one
Un estudiante reparte el tiempo de un día de la siguiente forma:
1/4 del día duerme,
1/12 del día lo usa para desplazarse caminando hasta el colegio, 5/12
del día lo usa para estudiar. ¿Qué parte del día le queda para compartir con su familia?
Answer:
4 horas para compatir com su familia
Let f: X → R be a linear function, where X is a topological vector space. (a) Suppose that f is bounded above on a neighborhood V of the origin. That means to 7>0 such that f(x) ≤ y for all x € V. Prove that there exists a neighborhood W of the origin such that f(x)| ≤ y for all x € W. (b) Suppose that f is bounded above on a neighborhood V of the origin. Prove that f is co (c) Prove that if f is bounded above on a set 2 with int(2) Ø, then f is continuous.
(a) To prove that there exists a neighborhood W of the origin such that f(x) ≤ y for all x ∈ W, given that f is bounded above on a neighborhood V of the origin, we can use the linearity of f.
Since f is a linear function, it satisfies the following properties:
f(0) = 0
f(rx) = rf(x) for any scalar r and vector x
f(x + y) = f(x) + f(y) for any vectors x and y
Given that f is bounded above on V, there exists a positive number M such that f(x) ≤ M for all x ∈ V. Now, let's consider the neighborhood W defined as follows:
W = {x ∈ X | ||x|| < M}
We claim that for any x ∈ W, f(x) ≤ y.
Let x ∈ W. Since x is in the neighborhood W, we have ||x|| < M. By linearity, we can express x as x = rx' for some scalar r and vector x' with ||x'|| = 1.
Now, consider f(x):
f(x) = f(rx') = rf(x')
Since ||x'|| = 1, we have ||rx'|| = |r| ||x'|| = |r|.
Therefore, ||rx'|| < M implies |r| < M.
Using the fact that f is bounded above on V, we have f(x') ≤ M.
Combining these results, we get:
|f(x)| = |rf(x')| = |r| |f(x')| ≤ M
Since this inequality holds for any x ∈ W and |r| < M, we have shown that f(x) ≤ y for all x ∈ W, where W is a neighborhood of the origin.
(b) To prove that f is continuous, we can show that f is bounded above on any compact set in X. Let K be a compact set in X.
Since K is compact, it is also closed and bounded. By the linearity of f, we have:
f(K) = {f(x) | x ∈ K}
Since K is bounded, there exists a positive number M such that ||x|| ≤ M for all x ∈ K. By the linearity of f, we have:
f(K) = {f(x) | x ∈ K} ⊆ {f(x) | ||x|| ≤ M}
Thus, f(K) is bounded above by M.
By the previous result in part (a), if f is bounded above on a neighborhood of the origin, then it is bounded above on any neighborhood of the origin. Therefore, f is bounded above on the neighborhood V of the origin.
Since K is compact, it can be covered by finitely many neighborhoods of the origin, say V1, V2, ..., Vk. Thus, f is bounded above on each Vi, i = 1, 2, ..., k.
Now, consider the open cover {V1, V2, ..., Vk} of K. By compactness, there exists a finite subcover {V1, V2, ..., Vm}. Therefore, f is bounded above on K.
Since f is bounded above on any compact set K, it follows that f is continuous.
(c) The previous part (b) already proves that if f is bounded above on any compact set, it is continuous. Therefore, if f is bounded above on a set 2 with int(2) ≠ Ø (i.e., the interior of 2 is not empty), then f is continuous.
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find the mean of the table below.
Answer: 12.5
Step-by-step explanation:
So, basically there is 5 minutes and the number of times occuring is 5 so its 5x1 which is 5 and then there is 10 minutes and the number of times occuring is twice so its 2x10 which is 20. Next, is 15 minutes and the times occuring is 2 so, 2x15 which is 30. And finally theres 20 minutes and the amount of times occuring is once so 20x1 which is 20. Then you add them all up which is 5+20+30+20=75. And the formula for mean/average is The sum of all number/ The amount of numbers there are so 75/6 is 12.5.
Amy makes the following statement:
"There is a 60% chance of snow tomorrow and a 10% chance I will be late for school."
What is the probability that it will snow and Amy will be late for school? (1 point)
a
3%
b
6%
c
50%
d
70%
Answer:
B
Step-by-step explanation:
If y is 19 times x, which equation shows the relationship between x and 7?
A. X = 19y
B. Y = 19x
C. Y = 19 + x
D. X = 19 + y
Answer:
b. y=19x
Step-by-step explanation:
Example 6.7. Find the largest two digit integer a which satisfies the following congruence 3.x = 4(mod 7).
The largest two-digit integer a = 99 satisfies the congruence 3x ≡ 4
To obtain the largest two-digit integer that satisfies the congruence 3x ≡ 4 (mod 7), we need to find an integer value for x that satisfies the congruence equation.
First, we can rewrite the congruence as an equation:
3x = 4 + 7k, where k is an integer.
Next, we can iterate through two-digit integers in descending order to find the largest value of a that satisfies the equation.
Starting with a = 99, we substitute it into the equation:
3(99) = 4 + 7k
297 = 4 + 7k
By trying different values of k, we can see that k = 42 satisfies the equation:
297 = 4 + 7(42)
Therefore, x = 99 is a solution to the congruence equation 3x ≡ 4 (mod 7).
However, we need to find the largest two-digit integer, so we continue the iteration.
Next, we try a = 98:
3(98) = 4 + 7k
294 = 4 + 7k
By trying different values of k, we can see that k = 42 also satisfies the equation:
294 = 4 + 7(42)
Therefore, x = 98 is another solution to the congruence equation 3x ≡ 4 (mod 7).
Since we have found the largest two-digit integer a = 99 that satisfies the congruence, we can conclude that a = 99 is the answer to the problem.
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To cheracterize the relationship between the response variable y and i covariates of interest, the following multiple linear regression model is used to fit the observed data: y=X3+ hyl Ble, B₁-, Bc. where y denotes the nx 1 response vector, 3 represents the px 1 parameter vector consisting of p=k+1 regression coefficients Bo, 31, 32,k, and e denotes the nx 1 vector of error terms. Assume that the model matrix X is an n x p full-column-rank matrix, and the entries of the first column of X are all equal to 1. In addition, assume that the error terms are independent and identically distributed normal random variables, that is, €12N (0,0³). Letz, denote the ith column of X. Suppose that a, sa for every i, where a represents a positive constant. Show that Var and the equality would be attained if X¹X aI, where 8, represents the ith entry of the least square estimator 3.
Multiple linear regression model is used to fit the observed data, in order to characterize the relationship between the response variable y and i covariates of interest.
The following regression model is used:
[tex]y=Xβ+ ε[/tex]
where:y denotes the n × 1 response vector.
[tex]β[/tex]represents the [tex]p × 1[/tex]parameter vector consisting of [tex]p = k + 1[/tex] regression coefficients.
[tex]X[/tex]denotes the n × p model matrix.
[tex]ε[/tex] denotes the [tex]n × 1[/tex] vector of error terms.
The entries of the first column of X are all equal to 1.
The entries of other columns of X correspond to the i covariates of interest.
It is given that the model matrix X is an n × p full-column-rank matrix.
The least squares estimator of [tex]β[/tex] is given by:
[tex]β^ = (X'X)^-1X'y[/tex]
The error terms are assumed to be independent and identically distributed normal random variables.
The variance-covariance matrix of the least squares estimator is given by:
[tex]Var(β^) = σ^2(X'X)^-1[/tex]
It is given that all covariates have the same variance-covariance structure.
Hence,
[tex]σ^2 = σ0^2[/tex] for every i.
It is also given that a, σ0 for every i, where a represents a positive constant.
Hence,
[tex]σ^2 = σ0^2[/tex]
= [tex]a^2[/tex]
Show that Var and the equality would be attained if
[tex]X'X = a^2I[/tex],
where [tex]β^[/tex] represents the ith entry of the least square estimator [tex]β[/tex].
From the given data, the variance-covariance matrix of the least squares estimator is given by:
[tex]Var(β^) = σ^2(X'X)^-1[/tex]
[tex]= (a^2/n)(X'X)^-1[/tex]
It is given that all covariates have the same variance-covariance structure.
Hence,
[tex]σ^2 = σ0^2[/tex] for every i.
It is also given that a, σ0 for every i, where a represents a positive constant.
Hence,
[tex]σ^2 = σ0^2[/tex]
[tex]= a^23[/tex]
Hence,
[tex]Var(β^) = (a^2/n)(X'X)^-1[/tex]
Now, let the diagonal entries of (X'X) be d1, d2, ..., dp.
Hence,
(X'X) = [dij]
i=1,2,...,p;
X¹ = [0, 0, ..., 1, ..., 0]'
Let X¹ denote the ith column of X.
Hence, X¹ is given by:
[tex]X¹ = [0, 0, ..., 1, ..., 0]'[/tex]
where 1 is in the ith position.
Hence, the ith diagonal entry of X'X is given by:
[tex](X'X)ii = Σj(Xj¹)^2[/tex]
where the sum is over all i.
From the given data, the entries of the first column of X are all equal to 1.
Hence, [tex]X1¹ = [1, 1, ..., 1]'.[/tex]
Hence, [tex](X'X)ij = nai[/tex] and
[tex](X'X)ij = nai[/tex] for [tex]i ≠ j.[/tex]
Hence,[tex](X'X) = a^2I + n11'[/tex]
The inverse of (X'X) is given by:
[tex](X'X)^-1 = (1/n)(I - (1/n)a^-2(1 1'))[/tex]
Hence,
[tex]Var(β^) = (a^2/n)(X'X)^-1[/tex]
=[tex]a^2[(1/n)(I - (1/n)a^-2(1 1'))][/tex]
The variance of the ith entry of the least square estimator is given by:
[tex]Var(β^i) = ai^2[(1/n)(I - (1/n)a^-2(1 1'))]ii[/tex]
Hence,
[tex]Var(β^i)[/tex]= [tex]ai^2[(1/n)(1 - (1/n)a^-2)][/tex]
= [tex]a^2/n[/tex]
Therefore, the variance of the ith entry of the least square estimator is given by:
[tex]Var(β^i) = a^2/n[/tex]
The equality would be attained if
[tex]X'X = a^2I,[/tex]
where β^i represents the ith entry of the least square estimator β^. Therefore, the required result has been obtained.
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A local hamburger shop sold a combined total of 711 hamburgers and cheeseburgers on Monday. There were 61 more cheeseburgers sold than hamburgers. How many hamburgers were sold on Monday?
Answer:
772 hamburgers sold on monday
Step-by-step explanation:
add the problem
Calculate the curvature ofy = x3 at x=1. Graph the curve and the osculating circle using GeoGebra.
The curvature of the function y = x³ at x = 1 is 6
How to calculate the curvature of the functionFrom the question, we have the following parameters that can be used in our computation:
y = x³
To start with, we need to differentiate the function
So, we have
y' = 3x²
Next, we differentiate the function for the second time to calculate the curvature of the function
So, we have
y'' = 6x
The value of x is 1
So, we have
y'' = 6(1)
Evaluate
y'' = 6
Hence, the curvature of the function y = x³ at x = 1 is 6
See attachment for the graph
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Evaluate -14 - 6 - 12 =??
Answer: -32
Step-by-step explanation: -14 - 6 - 12 = -32
Two numbers have a sum of 49, and 4 TIMES the first number MINUS the second number is EQUAL to 141. What are the two numbers?
Answer:
I got 38 and 11
Step-by-step explanation:
What is the value of x?
r \ 130°
Answer:
Step-by-step explanation:
Answer:
I don't see an "x" or an equation-
Alyssa is enrolled in a public-speaking class. Each week she is required to give a speech of grater length than the speech she gave the week before. The table shows the lengths of several of her speeches.
The week se will give a 12-minute speech is week 22 (option A)
Which week will she give a 12 - minute speech?The table is a linear table. This is because the variables change by a fixed amount.
Rate of change = change in length of speech / change in week
(180 - 150) / (4 - 3)
= 30 / 1 = 30 seconds
The next step is to convert minutes to seconds
1 minute = 60 seconds
60 x 12 = 720 seconds
Length of speech in week 2 = 150 - 30 = 120 seconds
Length of speech in week 1 = 120 - 30 = 90
Week the speech would have a length of 720 seconds = 90 +[ 30 x (week number - 1)]
720 = 90 + [30 x (x -1)
720 = 90 + 30x - 30
720 = 60 + 30x
720 - 60 = 30x
660 = 30x
x = 660 / 30
x = 22
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Alicia would like to know if there is a difference in the average price between two brands of shoes. She selected and analyzed a random sample of 40 different types of Brand A shoes and 33 different types of Brand B shoes. Alicia observes that the boxplot of the sample of Brand A shoe prices shows two outliers. Alicia wants to construct a confidence interval to estimate the difference in population means.
This question is incomplete, the complete question is;
Alicia would like to know if there is a difference in the average price between two brands of shoes. She selected and analyzed a random sample of 40 different types of Brand A shoes and 33 different types of Brand B shoes. Alicia observes that the boxplot of the sample of Brand A shoe prices shows two outliers. Alicia wants to construct a confidence interval to estimate the difference in population means.
Is the sampling distribution of the difference in sample means approximately normal?
a) Yes, because Alicia selected a random sample
b) Yes, because for each brand it is reasonable to assume that the population size is greater than ten times its sample size.
c) Yes, because the size of each sample is at least 30
d) No, because the distribution of Brand A shoes has outliers
e) No, because the shape of the population distribution is unknown.
Answer:
the sampling distribution of the difference in sample means is approximately normal because the size of each sample is al least 30
Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.
Step-by-step explanation:
Given the data in the question;
sample size of brand A shoes [tex]n_A[/tex] = 40
sample size of brand B shoes [tex]n_B[/tex] = 33
As we can notice,
[tex]n_A[/tex] = 40 > 30
[tex]n_B[/tex] = 33 > 30
Hence, we consider both samples to be large.
Therefore, the sampling distribution of the difference in sample means is approximately normal because the size of each sample is al least 30.
Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.
A researcher studies alcohol's effect on reaction time and finds no difference between people who consumed 1 versus 2 beverages. She included 40 participants randomly selected from the dining hall on campus, with a range in age from 18 to 38 years and a range in weight from 100 to 325 pounds. She decides to rerun her study with 30 women from 18 to 22 years of age and who are all of normal body weight and finds there is a statistical difference in reaction times between those who consumed 1 versus 2 beverages. Why might her results have changed
Answer:
Her results changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.
Step-by-step explanation:
She is trying to find if there is difference between people who consumed 1 versus 2 beverages.
Now initially she surveyed 40 people between ages 18 to 38 years and of huge weights and found out there was no difference. However, she decided to run the survey again on only women aged 18 to 22 years with normal body weight.
Since she used a lesser sample size and surveyed only women, it means there will be lesser variability in the result as the sample is smaller and streamlined to only women.
Thus, her result changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.
Court Casuals has the following beginning balances in its stockholders'equity accounts on January 1.2021:Common Stock,$90.000 Additional Paid-in Capital,$4.100.000:and Retained Earnings,$3.000,000.Net income for the year ended December 31,2021,is $900.000.Court Casuals has the following transactions affecting stockholders'equity in 2021: May 18 Issues 26,000 additional shares of $1 par value common stock for $50 per share. May 31 Purchases 4,500 shares of treasury stock for $40 per share. Julyl Declares a cash dividend of $2 per share to ail stockholders of record on July 15. Hint: Dividends are not paid on treasury stock. July 31 Pays the cash dividend declared on July 1. August 18 Resells 2,500 shares of treasury stock purchased on May 31 for $52 per share Taking into consideration all the entries described above,prepare the statement of stockholders'equity for the year ended December 31,2021,using the format provided.(Amounts to be deducted should be indicated with a minus sign.) COURT CASUALS Stelement of Stockholdara'Equity For the Yoar Ended December31.2021 Additional Common Ratained Pald-in Stock Earmings Capltal 90,000 $4,100,000 $3,000,000 Treasury Stock Total Stockholders Equlty $7,190,000 Balance,January 1 issue common stock Purchase treasury stock Cash dividends Resell treasury stock Net income Balance,December 31 90,000$4.100,000$3,000,000$ 0$7.190.000
The preparation of the stockholders' equity statement for the year ended December 31, 2021, is as follows:
Court Casuals
Statement of stockholers' equity
December 31, 2021
Common Stock $116,000
Additional Paid-in Capital $5,326,000
Retained Earnings $3,677,000
Treasury Stock $-2,000
Total stockholders' equity $9,117,000
How the stockholders' equity statement is prepared:The stockholders' equity statement includes the common stock, additional paid-in capital, retained earnings, and the subtraction of the treasury stock.
Court Casuals
Stockholers' equity on January 1, 2021
Common Stock $90,000
Additional Paid-in Capital $4,100,000
Retained Earnings $3,000,000
Net income for the year, 2021, = $900,000
Transactions Analysis:May 18: Cash $1,300,000 Common Stock $26,000 Additional Paid-in Capital $1,274,000 (26,000 x $50 - $26,000)
May 31: Treasury Stock $4,500 Additional Paid-in Capital $175,500 (4,500 x $40 - $4,500) Cash $180,000
Jul 1: Cash Dividend $223,000 (90,000 + 26,000 - 4,500) x $2 Dividends Payable $223,000
July 31: Dividends Payable $223,000 Cash $223,000
August 18: Cash $130,000 Treasury Stock $2,500 Additional Paid-in Capital $127,500 (2,500 x $52 - $2,500)
Statement of Retained Earnings, December 31, 2021:
Beginning balance $3,000,000
Net income $900,000
Dividends -223,000
Ending balance $3,677,000
Common Stock Account:Beginning balance $90,000
May 18: Cash 26,000
Ending balance $116,000
Additional Paid-in Capital Account:Beginning balance $4,100,000
May 18: Cash 1,274,000
May 31: Cash -175,500
August 18: Cash 127,500
Ending balance $5,326,000
Treasury Stock Account:May 31: Cash $4,500
August 18: Cash -2,500
Ending balance $2,000
Thus, we can summarize from the stockholders' equity that Court Casuals has outstanding shares of 114,000 (116,000 - 2,000) at $1 par, which is the difference between the ending balances of the common stock and the treasury stock accounts, after reflecting the equity transactions for the year.
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А
5 50.1
Find the measurement of the missing side
indicated
с
B
х
Answer:
Step-by-step explanation:
B
x is approximately equal to 6.
What are Trignometric ratios ?The ratios of the sides of a right triangle are called trigonometric ratios.
Three common trigonometric ratios are the sine (sin), cosine (cos), and tangent (tan).
sin = Perpendicular/ Hypotenuse
Cos = Base / Hypotenuse
tan = Perpendicular/Base
In the figure attached with the answer we can see that in Triangle ABC ,
tan 50.1 = x / 5
5(tan 50.1) = x
Therefore x is approximately equal to 6.
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yeah i need help anyone?
Answer:
A
Step-by-step explanation:
A secant is a line that goes through a circle at two points.
Answer:
A
Step-by-step explanation:
a straight line that cuts a curve in two or more parts.
Please answer correctly! I will mark you Brainliest!
Answer:
37
Step-by-step explanation:
There is one dot at 3.45. What does that value
represent?
In one simulated SRS of 100 students, the average
GPA was ; = 3.45.
In one simulated SRS of 100 students, the
population mean GPA was = 3.45.
There is a 1 out of 100 chance that the population
mean GPA is u = 3.45.
A majority of the sample of 100 students had a GPA
of 1 = 3.45.
Answer:
A- In one simulated SRS of 100 students, the average GPA was = 3.45.
Step-by-step explanation:
Just took the assignment myself, but here's a bit of insight to help anyone later on!
The x-bar means sample statistic. Since it was one sample taken out of the dot plot, 3.45 is a sample statistic.
A person P, starting at the origin, moves in the direction of the positive x-axis, pulling a weight along the curve C, called a tractrix, as shown in the figure. The weight, initially located on the y-axis at (0, s), is pulled by a rope of constant length s, which is kept taut throughout the motion. Assuming that the rope is always tangent to C solve the differential equation dy y dx 2 - y2 of the tractrix.
The required differential equation of the tractrix is dy/dx = (2y - y²)/s. This is obtained by substituting s = x/cos x into the differential equation 2y dy/dx - y² = 0 and simplifying. The differential equation dy/dx = (2y - y^2)/s is solved as shown below:
Solving the given differential equation In the given figure, let (x, y) be the coordinates of the point P. The weight is at (0, s) and the length of the rope is s. At a point P(x, y) on the tractrix, the tangent to the curve is parallel to the x-axis.The slope of the tangent is given by the differential coefficient dy/dx. We can determine dy/dx in terms of x and y by differentiating y^2 + x^2 = s^2 using implicit differentiation to get 2y dy/dx + 2x = 0.Differentiating again, we have d²y/dx² + y = 0.This differential equation is a second-order linear differential equation, with characteristic equation r² + 1 = 0. This yields r = ±i. Therefore, the general solution is y = c1cos x + c2 sin x, where c1 and c2 are constants. To find c1 and c2, we use the given initial condition that y = s when x = 0. This gives c1 = s and c2 = 0. Therefore, the solution to the differential equation is: y = s cos x. We can now eliminate x from the equation x² + y² = s² by substituting y = s cos x to obtain x² + s² cos² x = s². Solving for s, we have:s = x/cos x. Substituting this expression into the differential equation 2y dy/dx - y² = 0 yields the following equation:dy/dx = (2y - y²)/s This is the required differential equation of the tractrix.
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