Nae Maria Zaragoza 11 Practice Anment You took independent random samples of 20 students at City College and 25 sett SF State. You cach student how many sodas they drank over the course of you. The complemenn at City College was the sample standard deviation was 10. Al Suate the sample en was 90 and the sample standard deviation was is Use script of e for City College and subscriptors for State 1. Calculate a point estimate of the difference between the two population man = 20 n = 25 XI = 80 X = 90 N-12= XT-X2 = 80-90=-10 61 = 10 SI = 15 2.

Answer:

g=4

Step-by-step explanation:

this is a 30 60 90 triangle. the hypotenuse is 2x while the shortest side is x. if 8=2x then x must be 4.

Answer:

yes

Step-by-step explanation:

We are to determine if 6 yards is enough t to go round the perimeter of the window

The length is not given, so we have to determine the length from the area

Area of a rectangle = length x breadth

19.25 = 3.5 x length

length = 5.5 feet

Perimeter = 2 x ( length + breadth )

2 x (5.5 + 3.5) = 18 feet

We need to convert the string to foot

1 yard = 3 foot

6 x 3 = 18 foot

the string and the perimeter are equal, so it is enough

Answer:

4

Step-by-step explanation:

Line up the numbers at the decimal point and then find the number the same number of spaces away from the decimal point.

12,345.00

00067.89

Answer:

9.8 units

Step-by-step explanation:

distance = sqrt (x2 - x1)^2 + ( y2 - y1)^2

sqrt (-3 - (-6))^2 + (5 - 1)^2

sqrt (9)^2 + (4)^2

sqrt 81 + 16

sqrt 97

9.848857802

Answer:

See Explanation

Step-by-step explanation:

I will answer this question with the attached square pyramid

From the attached pyramid, we have:

[tex]Base\ Length = 20m[/tex]

So, the base area is:

[tex]Area = Length * Length[/tex]

[tex]A_1= 20m*20m[/tex]

[tex]A_1= 400m^2[/tex]

The dimension of each of the 4 triangles is:

[tex]Height = 16.4m[/tex]

[tex]Base = 20m[/tex]

So, the area of 4 triangles is:

[tex]Area = 4 * 0.5 * Base * Height[/tex]

[tex]A_2 = 4 * 0.5 * 20m * 16.4m[/tex]

[tex]A_2 = 656m^2[/tex]

So, the surface area is:

[tex]Area = A_1 + A_2[/tex]

[tex]Area = 400m^2 + 656m^2[/tex]

[tex]Area = 1056m^2[/tex]

Answer:

10 inches

Step-by-step explanation:

5/6*12

5*2 (since 12/6=2)

10 inches long!

hope it helps you!

Answer:

It would be 10 inches

Step-by-step explanation:

The 5/6 of 12 is 10 since(or you can simply say that we just subtract 2, I don't really know how to explain my work)

Answer:

a

Step-by-step explanation:

the 5y and the negative one cancel each other out. add the rest together you end up with 5x=-15. and divide each side by 5. you'll end up with x=-3

Answer: there is no picture

The computed values are:

u + v = (-3, -2)

v + u = (-3, -2)

5u = (10, -15)

2u + 3v = (-11, -3)

2u + 4w = (0, 2, 0)

u - v + 2w = (5, 0, 0)

|v + w| = 7.95

Vector addition is the operation of adding two vectors together to obtain a new vector. It is performed by adding the corresponding components of the vectors. For example, if we have two vectors u = [tex](u_1, u_2, u_3)[/tex] and v = [tex](v_1, v_2, v_3)[/tex], their sum u + v is given by [tex](u_1 + v_1, u_2 + v_2, u_3 + v_3)[/tex].

Scalar multiplication is the operation of multiplying a vector by a scalar (a real number). It is performed by multiplying each component of the vector by the scalar. For example, if we have a vector u = [tex](u_1, u_2, u_3)[/tex] and a scalar k, their product k * u is given by [tex](k * u_1, k * u_2, k * u_3[/tex]).

Both vector addition and scalar multiplication are fundamental operations in linear algebra and are used to manipulate and combine vectors in various applications.

To compute the given expressions, we perform vector addition and scalar multiplication as follows:

u + v =

[tex]= (2, -3) + (-5, 1) \\= (2 - 5, -3 + 1) \\= (-3, -2)[/tex]

v + u =

[tex]=(-5, 1) + (2, -3) \\= (-5 + 2, 1 - 3) \\= (-3, -2)[/tex]

5u =

[tex]= 5 * (2, -3) \\= (5 * 2, 5 * -3)\\ = (10, -15)[/tex]

2u + 3v =

[tex]=2 * (2, -3) + 3 * (-5, 1) \\= (4, -6) + (-15, 3)\\ = (4 - 15, -6 + 3) \\= (-11, -3)[/tex]

2u + 4w =

[tex]= 2 * (2, -3) + 4 * (-1, 2, 3/2) \\= (4, -6) + (-4, 8, 6)\\ = (4 - 4, -6 + 8, -6 + 6)\\ = (0, 2, 0)[/tex]

u - v + 2w =

[tex]= (2, -3) - (-5, 1) + 2 * (-1, 2, 3/2) \\= (2, -3) + (5, -1) + (-2, 4, 3) \\= (2 + 5 - 2, -3 - 1 + 4, 0 - 3 + 3) \\= (5, 0, 0)[/tex]

|v + w| =

[tex]= |(-5, 1) + (-1, 2, 3/2)| \\= |(-5 - 1, 1 + 2, 0 + 3/2)| \\= |(-6, 3, 3/2)| \\= \sqrt{((-6)^2 + 3^2 + (3/2)^2)} \\= \sqrt{(36 + 9 + 9/4)} \\= \sqrt{(63.25)} \\= 7.95[/tex]

Therefore, the computed values are:

u + v = (-3, -2)

v + u = (-3, -2)

5u = (10, -15)

2u + 3v = (-11, -3)

2u + 4w = (0, 2, 0)

u - v + 2w = (5, 0, 0)

|v + w| = 7.95

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The given metric is as follows: $$ ds^2 = e^{2A(r)} dt^2 - e^{2B(r)} dr^2 - 2(r^2 +\sin^2\theta) (d\phi^2 + \sin^2\theta d\phi^2) $$

The Riemann tensor is given as: $$ R^a_{bcd} = \partial_c \Gamma^a_{bd} - \partial_d \Gamma^a_{bc} + \Gamma^a_{ce}\Gamma^e_{bd} - \Gamma^a_{de}\Gamma^e_{bc} $$

Here, $\Gamma^a_{bc}$ is the Christoffel symbol of the second kind defined as:

$$ \Gamma^a_{bc} = \frac{1}{2} g^{ad}(\partial_b g_{cd} + \partial_c g_{bd} - \partial_d g_{bc}) $$

In this problem, we need to calculate the 20 independent components of the Riemann tensor. First, let's calculate the Christoffel symbols of the second kind.

Here, $g_ {00} = e^{2A(r)}$, $g_ {11} = -e^{2B(r)} $, $g_ {22} = -(r^2 + \sin^2\theta) $, and $g_{33} = -(r^2 + \sin^2\theta) \sin^2\theta$.

Using these, we get:$$ \Gamma^0_{00} = A'(r)e^{2A(r)}$$$$ \Gamma^0_{11} = B'(r)e^{2B(r)}$$$$ \Gamma^1_{01} = A'(r)e^{2A(r)}$$$$ \Gamma^1_{11} = -B'(r)e^{2B(r)}$$$$ \Gamma^2_{22} = -r(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^3_{33} = -\sin^2\theta(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^2_{33} = \cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ \Gamma^3_{32} = \Gamma^3_{23} = \cot\theta $$

Using these Christoffel symbols, we can now calculate the components of the Riemann tensor. There are a total of $4^4 = 256$ components of the Riemann tensor, but due to symmetry, only 20 of these are independent. Using the formula for the Riemann tensor, we get the following non-zero components:

$$ R^0_{101} = -A''(r)e^{2A(r)}$$$$ R^0_{202} = R^0_{303} = (r^2 + \sin^2\theta)(\sin^2\theta A'(r) + rA'(r))e^{2(A-B)}$$$$ R^1_{010} = -A''(r)e^{2A(r)}$$$$ R^1_{121} = -B''(r)e^{2B(r)}$$$$ R^2_{232} = r(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{323} = \sin^2\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^2_{323} = -\cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{322} = -\cos\theta\sin\theta(r^2 + \sin^2\theta)^{-1}$$$$ R^0_{121} = A'(r)B'(r)e^{2(A-B)}$$$$ R^1_{020} = A'(r)B'(r)e^{2(A-B)}$$$$ R^2_{303} = -\sin^2\theta A'(r)e^{2(A-B)}$$$$ R^3_{202} = -rA'(r)e^{2(A-B)}$$$$ R^0_{202} = (r^2 + \sin^2\theta)\sin^2\theta A'(r)e^{2(A-B)}$$$$ R^0_{303} = (r^2 + \sin^2\theta)A'(r)e^{2(A-B)}$$$$ R^1_{010} = A''(r)e^{2(A-B)}$$$$ R^1_{121} = B''(r)e^{2(A-B)}$$$$ R^2_{232} = r(r^2 + \sin^2\theta)^{-1}$$$$ R^3_{323} = \sin^2\theta(r^2 + \sin^2\theta)^{-1}$$

Therefore, these are the 20 independent components of the Riemann tensor.

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The expression 10/3 can be expressed as a percent by multiplying it by 100. The result is approximately 333.33%.

To express a fraction as a percent, we need to convert it into a decimal and then multiply by 100 to get the percentage representation. In this case, we have 10/3 as the fraction.

To convert the fraction 10/3 to a decimal, we divide 10 by 3, which gives us approximately 3.3333. To express this decimal as a percentage, we multiply it by 100. Thus, 3.3333 * 100 = 333.33%.

Therefore, the expression 10/3 can be expressed as approximately 333.33% when converted to a percentage.

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Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12

The solution to the initial value problem is :

y(t) = 12e³ᵗ + 3.

We have 3t y'' - 9y' + 18y = 6t e

Taking Laplace transform on both sides, we get

3L(ty'') - 9L(y') + 18L(y) = 6L(te)

Using Laplace transform formulas, we get:

3[s²Y(s) - sy(0) - y'(0)] - 9[sY(s) - y(0)] + 18Y(s) = 6/s²L(e)

⇒ 3s²Y(s) - 3s(5) + 6 - 9sY(s) + 45 + 18Y(s) = 6/s² * 1/sY(s)[3s² - 9s + 18] = 6/s² * 1/s - 3s + 12Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12

Now, we need to find inverse Laplace transform of Y(s) to obtain the solution y(t).

Let's solve for the first term by Partial Fraction Expansion.

6/s * 1/(s * (s - 3))= A/s + B/(s - 3)6 = A(s - 3) + Bs

Therefore, A = -2 and B = 2y(t) = L⁻¹[Y(s)] = L⁻¹[6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12]= -2L⁻¹[1/s] + 2L⁻¹[1/(s - 3)] + 5L⁻¹[1/s] + 12L⁻¹[1/(s - 3)]= -2 + 2e³ᵗ + 5 + 12e³ᵗ= 12e³ᵗ + 3

Therefore, Y(s) = 6/s * 1/(s * (s - 3)) + 1/s * 5 + 1/(s - 3) * 12 and the solution to the initial value problem is y(t) = 12e³ᵗ + 3.

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Answer:

(Thermometer B reading - Thermometer A reading)

Step-by-step explanation:

The thermometer reading aren't given in the question.

However, hypothetically.

The difference between two temperature values (morning and evening values) would be :

Temperature in the evening - morning temperature

Therefore,

If ;

Thermometer A reading = morning temperature

Thermometer B reading = evening temperature

Difference in the temperature :

(Thermometer B reading - Thermometer A reading)

The least square straight line y = -2x + 3 to fit the data points (0, 3), (2, 1), (3, 1) is found. The minimum square error is 61.

Given data points are (0, 3), (2, 1), (3, 1).

To find the least square straight line, y = mx + b.

The line that fits these points will have the minimum square error.(0,3)      y = mx + b;  3 = 0 + b;  b = 3(2,1)        

y = mx + b;  1 = 2m + b;  b = 1 - 2m(3,1)        

y = mx + b;  1 = 3m + b;  b = 1 - 3m

Substitute the value of b in (2) and (3)1 - 2m = 3 - 3m;  m = -2y = mx + b;  

y = -2x + 3

The least square straight line y = -2x + 3 to fit the data points (0, 3), (2, 1), (3, 1) is found.

Now, we need to compute the minimum square error.

Square error of each point: Point 1 (0, 3):  Square error = (3 - 3)² = 0

Point 2 (2, 1):  Square error = (1 - (-4))² = 25

Point 3 (3, 1):  Square error = (1 - (-5))² = 36

The minimum square error is the sum of the square error of all the points, Minimum square error = 0 + 25 + 36 = 61

Therefore, the least square straight line y = -2x + 3 to fit the data points (0, 3), (2, 1), (3, 1) is found. The minimum square error is 61.

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In regression models, there are different methods that can be used to evaluate the accuracy of the model and the relationship between the variables. One of the most commonly used methods for evaluating the accuracy of the model is by calculating the R-squared value.

R-squared value represents the proportion of variation in the dependent variable that is explained by the independent variable(s). It ranges from 0 to 1, with a higher value indicating a better fit. To evaluate the accuracy of the model is to use residual plots. Residual plots can be used to identify patterns or trends in the errors or residuals, which can help to identify potential problems with the model and suggest ways to improve it. Additionally, the residuals can be tested for normality and homoscedasticity. Normality can be checked using a normal probability plot, and homoscedasticity can be checked using a scatter plot of residuals versus fitted values.

If the residuals are normally distributed and have a constant variance, then the assumptions of the regression model are met. Another way to evaluate the relationship between the variables is to use correlation analysis. Correlation analysis is a statistical technique that measures the strength and direction of the linear relationship between two variables. The correlation coefficient can range from -1 to +1, with a value of 0 indicating no correlation and a value of -1 or +1 indicating a perfect negative or positive correlation, respectively.

However, correlation analysis only measures the strength and direction of the linear relationship and does not take into account other factors that may affect the relationship, such as outliers or nonlinearities.

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Answer:

(5,354 + x)

or

536.4*x

Step-by-step explanation:

We know that x = 10.

Now we want to write an expression (in terms of x) for the number 5,364.

This could be really trivial, remember that x = 10.

Then:  (x - 10) = 0

And if we add zero to a number, the result is the same number, then if we add this to 5,364 the number does not change.

5,364 = 5,364 + (x - 10) = 5,364 + x - 10

5,364 = 5,354 + x

So (5,354 + x) is a expression for the number 5,364 in terms of x.

Of course, this is a really simple example, we could do a more complex case if we know that:

x/10 = 1

And the product between any real number and 1 is the same number.

Then:

(5,364)*(x/10) = 5,364

(5,364/10)*x = 5,364

536.4*x = 5,364

So we just found another expression for the number 5,364 in terms of x.

Answer:  = -8

Step-by-step explanation: Your welcome!

Answer:

y intercept (0;4)

Step-by-step explanation:

let x = 0 because the graph will intersect the y-axis at the value of 0 for the x-axis

Step-by-step explanation:

This is an odd question  (do we have all of the info??)....I had to make an assumption...

Well..... you will not get a numerical answer...it is a quadratic equation

area = (x+5) ft  (3x+6) ft         (I assumed one was length and one was width)

area =   (3x^2 +21x + 30)     ft^2

The solution to the initial-value problem using the Laplace transform is y(t) = (1/25)(1 - [tex]e^{(-5t)[/tex]) - (1/25)t + (1/125)[tex]e^{(-5t)[/tex].

To solve the given initial-value problem using Laplace transform, we will first take the Laplace transform of the given differential equation and apply the initial condition.

Take the Laplace transform of the differential equation:

Applying the Laplace transform to the equation y' + 5y = f(t), we get:

sY(s) - y(0) + 5Y(s) = F(s),

where Y(s) represents the Laplace transform of y(t) and F(s) represents the Laplace transform of f(t).

Apply the initial condition:

Using the initial condition y(0) = 0, we substitute the value into the transformed equation:

sY(s) - 0 + 5Y(s) = F(s).

Substitute the given function f(t):

The given function f(t) is defined as:

f(t) = t, 0 ≤ t < 1

f(t) = 0, t ≥ 1

Taking the Laplace transform of f(t), we have:

F(s) = L{t} = 1/s²,

Solve for Y(s):

Substituting F(s) and solving for Y(s) in the transformed equation:

sY(s) + 5Y(s) = 1/s²,

(Y(s)(s + 5) = 1/s²,

Y(s) = 1/(s²(s + 5)).

Inverse Laplace transform:

To find y(t), we need to take the inverse Laplace transform of Y(s). Using partial fraction decomposition, we can write Y(s) as:

Y(s) = A/s + B/s² + C/(s + 5),

Multiplying both sides by s(s + 5), we have:

1 = A(s + 5) + Bs + Cs².

Expanding and comparing coefficients, we get:

A = 1/25, B = -1/25, C = 1/125.

Therefore, the inverse Laplace transform of Y(s) is:

y(t) = (1/25)(1 - [tex]e^{(-5t)[/tex]) - (1/25)t + (1/125)[tex]e^{(-5t)[/tex].

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Answer:

$247.50

Step-by-step explanation:

Answer:

48

Step-by-step explanation:

The nth term of an AP is expressed as;

Tn = a+(n-1)d

Id 12th term is 32, hence;

T12 = a+11d

32 = a+11d ...1

If the 5th term is 18, then;

T5 = a+4d

18 = a + 4d ....2

Subtract 1 from 2;

32 - 18 = 11d - 4d

14 = 7d

d = 14/7

d = 2

From 1; 32 = a+11d

32 = a+ 11(2)

32 = a + 22

a = 32-22

a = 10

Get the 20th term

T20 = a+19d

T20 = 10 + 19(2)

T20 = 10 + 38

T20 = 48

Hence the 20th term is 48

Answer:

90 = π/2

45 = π/4

0 and 360 = 0 and 2π

135 = 3π/4

180 = π

225 = 5π/4

270 = 2π/3

315 = 7π/4

315 =

Step-by-step explanation:

Answer:

(-2, -4)

Step-by-step explanation:

The Wronskian formula is given by:$$W(y_1,y_2)=\begin {vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$$To prove the Wronskian formula of two functions, let $y_1$ and $y_2$ be two non-zero solutions of the differential equation $y'' + p(x)y' + q(x)y = 0$.

Then the Wronskian of these two functions is given by: $W(y_1,y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=Ce^{-\int p(x)dx}$ where $C$ is a constant that depends on $y_1$ and $y_2$ but not on $x$.

Part (a) of the given Wronskian formulas is: $$W(2\sin v(x), J_v(x))=\begin{vmatrix} 2\sin v(x) & J_v(x) \\ 2v\cos v(x) & J_v'(x) \end{vmatrix}=2\sin v(x)J_v'(x)-2v\cos v(x)J_v(x)$$

Note that this formula is almost the same as the standard Wronskian formula, but with the constant $C$ replaced by $2\sin v(x)$.

We can verify that this is indeed a valid Wronskian by taking the derivative with respect to $x$:$$\frac{d}{dx}[2\sin v(x)J_v'(x)-2v\cos v(x)J_v(x)]=2\cos v(x)J_v'(x)-2\sin v(x)[vJ_v(x)+J_v'(x)]=0$$

The last step follows from the differential equation satisfied by the Bessel functions: $x^2y''+xy'+(x^2-v^2)y=0$

Part (b) of the given Wronskian formulas is: $$W(Y_\nu(x),Y_{\nu+1}(x))=\begin{vmatrix} Y_\nu(x) & Y_{\nu+1}(x) \\ Y_\nu'(x) & Y_{\nu+1}'(x) \end{vmatrix}=W_0Y_{\nu+1}(x)-W_1Y_\nu(x)$$where $W_0$ and $W_1$ are constants that depend on $\nu$ but not on $x$. This formula is also a valid Wronskian, since we can verify that its derivative with respect to $x$ is zero:

$$\frac{d}{dx}[W_0Y_{\nu+1}(x)-W_1Y_\nu(x)]=W_0Y_{\nu+1}'(x)-W_1Y_\nu'(x)=0$$

This follows from the recurrence relations satisfied by the Bessel functions:$Y_{\nu-1}'(x)-\frac{\nu}{x}Y_{\nu-1}(x)+\frac{\nu+1}{x}Y_{\nu+1}(x)=0$ $Y_{\nu+1}'(x)-\frac{\nu+1}{x}Y_{\nu+1}(x)+\frac{\nu+2}{x}Y_{\nu+2}(x)=0$

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Answer:

45/49

decimal form:

0.91836734

Step-by-step explanation:

The area of the surface formed by the part of the sphere [tex]x^2 + y^2 + z^2 = 4z[/tex] that lies inside the paraboloid [tex]z = x^2 + y^2[/tex] is π/6 square units.

To find the area of the surface, we need to calculate the double integral over the region that lies inside both the sphere and the paraboloid.

The given sphere equation can be rewritten as [tex]x^2 + y^2 + (z - 2)^2 = 4[/tex]. This represents a sphere centered at (0, 0, 2) with a radius of 2.

The paraboloid equation [tex]z = x^2 + y^2[/tex] represents an upward-opening paraboloid centered at the origin.

To find the region of intersection, we set the sphere equation equal to the paraboloid equation:

[tex]x^2 + y^2 + (x^2 + y^2 - 2)^2 = 4[/tex]

Simplifying, we get [tex]x^4 + y^4 - 4x^2 - 4y^2 + 4 = 0[/tex].

This equation represents the boundary curve of the region of intersection.

By evaluating the double integral over this region, we find the area of the surface to be π/6 square units.

Therefore, the area of the surface formed by the given part of the sphere lying inside the paraboloid is π/6 square units.

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