a) The value of k is 1.
b) The variance of X is 1/12.
c) Pr(X² < 2) = Fx(√2) = (√2) - 1
e) The density function of Y is fY(y) = 1 / (2√(y + 1)), for 0 ≤ y ≤ 3.
(a) We need to integrate the probability density function (pdf) over its entire range and set it equal to 1.
∫[1,2] k dx = 1
Integrating, we get:
k[x] from 1 to 2 = 1
k(2 - 1) = 1
k = 1
So, the value of k is 1.
(b) The expectation (mean) of a continuous random variable can be calculated using the following formula:
E(X) = ∫[−∞,∞] x f(x) dx
In our case, since the pdf is zero outside the range [1, 2], we can simplify the calculation:
E(X) = ∫[1,2] x f(x) dx = ∫[1,2] x dx
E(X) = [x²/2] from 1 to 2
E(X) = (2²/2) - (1²/2) = 3/2
So, the expectation of X is 3/2.
The variance of a continuous random variable can be calculated using the formula:
Var(X) = E(X²) - [E(X)]²
E(X²) = ∫[−∞,∞] x² f(x) dx
In our case, since the pdf is zero outside the range [1, 2]:
E(X²) = ∫[1,2] x² f(x) dx = ∫[1,2] x² dx
E(X²) = [x³/3] from 1 to 2
E(X²) = (2³/3) - (1³/3) = 7/3
Now, we can calculate the variance:
Var(X) = E(X²)- [E(X)]²
Var(X) = (7/3) - (3/2)²
Var(X) = 7/3 - 9/4
Var(X) = 28/12 - 27/12
Var(X) = 1/12
So, the variance of X is 1/12.
(c) The cumulative distribution function (CDF) F(x) is the integral of the pdf from negative infinity to x:
Fx(z) = ∫[−∞,z] f(x) dx
Since the pdf is zero outside the range [1, 2], the CDF is:
Fx(z) = ∫[1,z] f(x) dx = ∫[1,z] dx
Fx(z) = [x] from 1 to z
Fx(z) = z - 1
To calculate probabilities, we can substitute the given values into the CDF:
Pr(X < 4/3) = Fx(4/3) = (4/3) - 1 = 1/3
Pr(X² < 2) = Fx(√2) = (√2) - 1
(e) Let Y = X² - 1. To find the density function of Y, we can use the transformation technique.
First, we need to find the cumulative distribution function (CDF) of Y.
To do this, we express Y in terms of X:
Y = X² - 1
Now, we can solve for X:
X = √(Y + 1)
To find the density function of Y, we differentiate the CDF of Y with respect to Y:
fY(y) = d/dy [FX(√(y + 1))]
Using the chain rule, we have:
fY(y) = fX(√(y + 1)) (1 / (2√(y + 1)))
Substituting the given pdf of X (fx(x) = 1, 1 ≤ x ≤ 2), we have:
fY(y) = 1 (1 / (2√(y + 1)))
fY(y) = 1 / (2√(y + 1)), for 0 ≤ y ≤ 3
So, the density function of Y is fY(y) = 1 / (2√(y + 1)), for 0 ≤ y ≤ 3.
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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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ratio of the surface area to volume? 280 in² / 300 in³
The ratio of the surface area to the volume is 14/15 in²/in³.
To find the ratio of surface area to volume for a given object, we divide the surface area by the volume. In this case, we have a ratio of 280 in² to 300 in³.
The surface area represents the total area of all the exposed surfaces of the object, while the volume represents the amount of space occupied by the object.
To calculate the ratio, we divide the surface area by the volume:
Ratio = Surface Area / Volume
Ratio = 280 in² / 300 in³
Simplifying the ratio:
Ratio = (280/300) in²/in³
Ratio = (28/30) in²/in³
Ratio = (14/15) in²/in³
Therefore, the ratio of the surface area to the volume is 14/15 in²/in³.
This ratio represents the relationship between the amount of surface area and the amount of volume for the given object. It indicates that, for every 15 cubic inches of volume, there are 14 square inches of surface area.
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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]
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.
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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Find the Laplace transform of the function f(t) = t sin(4t) +1.
The Laplace transform of the function f(t) = t sin(4t) +1 is
L[f(t)] = (4 / (s^2 + 16) + 1) / s
To find the Laplace transform of the function f(t) = t sin(4t) + 1, we can use the linearity property of the Laplace transform.
The Laplace transform of t sin(4t) can be found using the derivative property and the Laplace transform of sin(4t). The derivative property states that if F(s) is the Laplace transform of f(t), then sF(s) is the Laplace transform of f'(t).
Taking the derivative of t sin(4t) with respect to t, we get:
f'(t) = 1⋅sin(4t) + t⋅(4⋅cos(4t))
Now, we can find the Laplace transform of f'(t) using the derivative property:
L[f'(t)] = sL[f(t)] - f(0)
Since f(0) = 0, the Laplace transform becomes:
L[t sin(4t)] = sL[t sin(4t) + 1]
Next, we need to find the Laplace transform of sin(4t). The Laplace transform of sin(at) is [tex]a / (s^2 + a^2)[/tex]. Therefore, the Laplace transform of sin(4t) is [tex]4 / (s^2 + 16).[/tex]
Now, substituting these values into the equation, we have:
sL[t sin(4t)] - 0 = sL[t sin(4t) + 1]
Simplifying, we get:
sL[t sin(4t)] = sL[t sin(4t) + 1]
Dividing both sides by s, we have:
L[t sin(4t)] = L[t sin(4t) + 1] / s
Now, we can substitute the Laplace transform of sin(4t) and simplify further:
[tex]L[t sin(4t)] = (4 / (s^2 + 16) + 1) / s[/tex]
Therefore, the Laplace transform is: [tex]L[f(t)] = (4 / (s^2 + 16) + 1) / s[/tex]
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PLEASE HELP WILL MARK BRAINLY
.In the zombie apocalypse, the ratio of children to adults was 3 to 4. The ratio
of adults to zombies was 4 to 5. If there were 200 zombies, then how many
children were there?
2. (4^5)(4^−7) = 4^?
3. . What is the circumference and area of the circle? Use 22/7
for π. Radius=42cm
Step-by-step explanation:
diameter is 84 thereare 2 time 42 .84.
follow me nice study bye .
PLS HELP ILL GIVE BRAINLEST!!!!!!!!
Answer:
pay back is so good!..
Step-by-step explanation:
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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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.
Please help.
Is algebra.
Answer:
question 3= c question 4= a
Step-by-step explanation:
I need help with this we my class skipped this section
Answer:
v^ 3 = 1000
so v = cube root of 1000
v = 10
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:
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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On a coordinate plane, what is the distance between the point at (8, 7) and the point at (-9, 7)?
Answer:
They are 17 units away from each other!
Step-by-step explanation:
Because they have the same y coordinate, this make us not have to do the formula. To find how far they are away, add the absolute value of each x coordinate. The answer is 17 units.
give me brainliest
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
Please answer correctly! I will mark you Brainliest!
Answer:
37
Step-by-step explanation:
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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Determine if the relation below represents a function. Explain your reasoning
Given:
The graph of a relations.
To find:
Whether the relation is a function or not.
Solution:
A relation is a function if there exist unique outputs for each input. If a graph passes the vertical line test then is a function.
Vertical line test: Each vertical line intersect the curve at most one.
In the given figure draw a vertical at [tex]x=1[/tex] as shown in the below figure.
From the below graph, it is clear that the vertical line intersects the curve at two points. So, it does not pass the vertical line test.
Therefore, the given relation is not a function.
The Royal Fruit Company produces two types of fruit drinks. The first type is 30% pure fruit juice, and the second type is 80% pure fruit juice. The company is attempting to produce a fruit drink that contains 35% pure fruit juice. How many pints of each of the two existing types of drink must be used to make 90 pints of a mixture that is 35% pure fruit juice?
Answer: I- 81 pints, II-9 pints
Step-by-step explanation:
Given
The first type of juice has 30% pure Juice
The second type of juice has 80% pure Juice
The final mixture has 95 pints of 35% pure juice
Suppose we take x pints from the first Juice
So, the second Juice contributes 90-x
for 35% content
[tex]\Rightarrow 35=\dfrac{x\times 30+(90-x)80}{90}\\\\\Rightarrow 3150=30x+7200-80x\\\\\Rightarrow 50x=4050\\\Rightarrow x=81\ \text{pints}[/tex]
First contributes 81 pints. second contributes 9 pints
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:
Evaluate -14 - 6 - 12 =??
Answer: -32
Step-by-step explanation: -14 - 6 - 12 = -32
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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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:
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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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:
6=1x+5
Find Y . . . . . . . .
Answer:
If you are referring to the y-intercept, then the answer is 5. Otherwise, the answer is probably 6.
Step-by-step explanation:
Answer:
i got 1
Step-by-step explanation:
What is the area of this cross section of this rectangular prism? Enter your answer in the box.
10 in.
16 in.
16 in.
Answer:
1152in²
Step-by-step explanation:
Given the following
Length = 10in
Width = 16in
Height = 16in
surface area of the prism = 2(LW + WH + LH)
surface area of the prism = 2(10*16+ 16*16 + 10*16)
surface area of the prism = 2(160+256+160)
surface area of the prism = 2(576)
surface area of the prism = 1152in²
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
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.