The double dot plot shows the values in two data sets.

Double dot plot shows the values of two data sets. It shows the number line ranges from 10 to 70 showing the following values that appear as dots above the line. For data set A. Two dots above 10. Three dots above 20. Four dots above 30. Two dots above 40. One dot above 50. For data set B. One dot above 30. Three dots above 40. Four dots above 50. Two dots above 60. Two dots above 70.

Compare the data sets using measures of center and variation.

Data set A Data set B
Median $=$
Median $=$
IQR $=$
IQR $=$

Question: "Identify the correct description for the formula g'(x) ≈ g(x)/h – g(x – h)/h from the following options: FFD1: forward finite difference with stepsize h for the first derivative of g at a BFD1: backward finite difference with stepsize h for the first derivative of g at a CFD1: central finite difference with stepsize h for the first derivative of g at x CFD2: central finite difference with stepsize h for the second derivative of g at x None of the Above"

The correct description for the formula g'(x) ≈ g(x)/h – g(x – h)/h is BFD1: backward finite difference with stepsize h for the first derivative of g at a.

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1. y = 2
dy/dx = 0 (Constant terms have a derivative of 0)

2. y = 2x^2 + 2
dy/dx = 4x (Apply power rule: d(ax^n)/dx = a * n * x^(n-1))

3. y = 2x
dy/dx = 2 (Linear terms have a derivative equal to their coefficient)

4. y = 4x^3 - 4
dy/dx = 12x^2 (Apply power rule and constant term has derivative 0)

5. y = 3x^6
dy/dx = 18x^5 (Apply power rule)

6. y = 2(x-5)^2
dy/dx = 4(x-5) (Apply chain rule: d(u^2)/dx = 2u * du/dx)

7. y = 1 - 3x
dy/dx = -3 (Linear terms have a derivative equal to their coefficient)

8. y = 2/x^3
dy/dx = -6/x^4 (Rewrite as 2x^(-3) and apply power rule)

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Answer:

47 ÷ 3681 is approximately 0.0128

Answer:

The nswer is 0.0127682694919858

Ok I’ll tell you how to do it.

1. Base x Hight x length
I can’t solve it because I can’t see all the numbers.

The angle measure to the nearest degree of tan B = .5543 is 29°.

Given that tan B = 0 .5543, we need to find the measure to the nearest degree of tan B,

Since, we need to find the measurement of the angle, so we will use the concept of inverse of trigonometric functions,

tan B = 0 .5543

B = tan⁻¹ (0.5543)

B = 28.99 ≈ 29°

Hence, the angle measure to the nearest degree of tan B = .5543 is 29°.

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The solution of the linear equation:

7x – 13 = ─2x + 5

is x = 2

Here we want to find the value of x that is a solution of:

7x - 13 = -2x + 5

To solve it, we need to isolate x in one of the sides of the equation.

7x - 13 = -2x + 5

7x + 2x = 5 + 13

9x = 18

x = 18/9

x = 2

The value x = 2 makes the given linear equation true.

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The required dimensions are as follows

coordinates of point M (1, 2.5)

PQ = 4

QR = 5.4

PM = 3.9

OM = 2.7

The perimeter of the parallelogram = 18.8

angle QMP = 60 degrees

Angle ROP =  100 degrees

The dimensions are calculated by plotting the coordinates and measuring the dimensions from the graph.

From the graph we can see that

PQ = 4

QR = 5.4

PM = 3.9

OM = 2.7

The perimeter of the parallelogram

= 2(4 + 5.4)

= 18.8

angle QMP = 180 - angle QMR = 180 - 120 = 60 degrees

Angle ROP = 180 - angle QRO = 180 - 80 = 100 degrees

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In problem 2, we are comparing the Lagrange form of interpolation polynomial and the Newton form of the interpolating polynomial of degree 3. To solve this problem, we first need to understand the concepts of interpolation, polynomial, and Lagrange.

A set of basis polynomials are used to create the interpolating polynomial in the Lagrange method of polynomial interpolation.

Returning to issue 2, we are given the degree 3 interpolating polynomial, which is a degree 3 polynomial that traverses a specified set of data points.

We are asked to contrast this polynomial with the interpolation polynomial in the Lagrange form.

Another approach to creating a polynomial that traverses a given set of data points is to use the Lagrange form of interpolation polynomials.

We must assess the degree 3 interpolating polynomial and the Lagrange form of the interpolation polynomial at the specified point p(0) in order to compare the two findings.

In conclusion, we can say that to compare the Lagrange form of interpolation polynomial and the Newton form of the interpolating polynomial of degree 3, we need to evaluate both polynomials at the given point and choose the one that gives the same value as the data point.

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Answer:

9. Associative property

10. Distributive property

Name the property shown.

9. 2 x= x 2

td

11. 7(z+ y) = 7z+ 7y

10. 231= 23

12. a + (2b + 3c) = (a + 2b) + 3c

Answer:

Answer:

  (a)  S = {1, 2, 3, 4, 5, 6, 7}

Step-by-step explanation:

You want the sample space for rolling a 7-sided die.

The sample space is the list of all possible outcomes.

Possible outcomes from rolling a 7-sided die are any of the numbers 1 through 7.

The sample space is ...

  S = {1, 2, 3, 4, 5, 6, 7} . . . . . choice A

<95141404393>

a.  The probability distribution of the test statistic is approximately a standard normal distribution.

b. The p-value for the test of  factors that affect the prevalence of tuberculosis among intravenous drug users is 0.0202.  

c. We can conclude that there is a statistically significant difference between the two groups in terms of their proportions of positive skin test results.

d. The 95% confidence interval does not contain zero, so there is a statistically significant difference between the two groups in terms of their proportions of positive skin test results.

To estimate the common value of p assuming that the population proportions of positive skin test results are equal, we can compute the pooled proportion:

p-hat = (x1 + x2) / (n1 + n2)

= (24.7 + 17.4) / (97 + 161)

= 0.195

where x1 and x2 are the number of individuals with positive skin test results in the two groups, and n1 and n2 are the sample sizes.

a. To test the null hypothesis that the proportions of intravenous drug users who have positive tuberculin skin test results are identical for those who share needles and those who do not, we can use a two-sample z-test for proportions. The test statistic is:

z = (p1 - p2) / sqrt(phat * (1 - phat) * (1/n1 + 1/n2))

where p1 and p2 are the sample proportions, phat is the pooled proportion, and n1 and n2 are the sample sizes.

Plugging in the values, we get:

z = (0.247 - 0.174) / sqrt(0.195 * (1 - 0.195) * (1/97 + 1/161))

= 2.05

The probability distribution of the test statistic is approximately a standard normal distribution, since the sample sizes are large enough (both n1 and n2 are greater than 30).

b. The p-value for the test is the probability of observing a z-value of 2.05 or more extreme under the null hypothesis. From a standard normal distribution table or calculator, we find that the p-value is approximately 0.0202 (or 0.0404 for a two-tailed test).

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the proportions of intravenous drug users who have positive tuberculin skin test results are not identical for those who share needles and those who do not.

c. To construct a 95% confidence interval for the true difference in proportions, we can use the formula:

(p1 - p2) ± z* sqrt(phat * (1 - phat) * (1/n1 + 1/n2))

where z is the critical value for a 95% confidence interval from a standard normal distribution (z = 1.96).

Plugging in the values, we get:

(0.247 - 0.174) ± 1.96 * sqrt(0.195 * (1 - 0.195) * (1/97 + 1/161))

= 0.073 ± 0.090

Therefore, we can be 95% confident that the true difference in proportions of intravenous drug users who have positive tuberculin skin test results between those who share needles and those who do not is between 0.073 and -0.073 (which can be written as an absolute value of 0.073).

d.  We can infer that there is a statistically significant difference between the two groups in terms of the proportions of positive skin test results because the interval does not contain zero.

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a) The coordinates of K' after the translation are given as follows: K'(-2,3).

b) The coordinates of M' after the translation are given as follows: M'(-4, 1).

The four translation rules are defined as follows:

From the vector, the composite translation rule in this problem is given as follows:

(x,y) -> (x - 6, y + 2).

The coordinates of K and M are given as follows:

K(4,1), M(2,-1).

Hence the coordinates of the translated vertices are obtained applying the operation as follows:

K'(-2, 3) and M'(-4, 1).

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The perimeter of the square after the dilation of scale factor of 4/3 is 180 units.

Given that,

Perimeter of the square = 135 units = 4a, where 'a' is the length of a side.

Scale factor = 4/3

We have to find the perimeter of the square if the square is dilated by a scale factor of 4/3.

If the square is dilated by a scale factor of 4/3,

length of each side = 4/3 a

Perimeter of the new square = 4 × 4/3 a

                                                = 4/3 × 4a

                                                = 4/3 × 135

                                                = 180 units

Hence the new perimeter of the square is 180 units.

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Louisa would need to invest $32,172.75 semiannually to have $44000 in four years if the fund she has selected pays 8.1% annually, compounded semiannually.

To have $44000 in four years, Louisa would need to invest a total of $x in the fund that pays 8.1% annually, compounded semiannually.

Using the formula for compound interest, we can solve for x:

A = P(1 + r/n)^(nt)

Where:
A = the total amount (in this case, $44000)
P = the principal amount (the amount Louisa needs to invest)
r = the annual interest rate (8.1%)
n = the number of times the interest is compounded per year (semiannually = 2)
t = the number of years (4)

Plugging in these values:

44000 = P(1 + 0.081/2)^(2*4)

44000 = P(1.0405)^8

44000 = P(1.366)

P = 32172.75

So Louisa would need to invest $32,172.75 semiannually to have $44000 in four years if the fund she has selected pays 8.1% annually, compounded semiannually.

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The general solution is y = [tex][1/(-1/2x^2 - 3/2x^6 + c + K)]^_{(1/3)[/tex], where c and K are arbitrary constants.

To tackle the differential condition [tex]y^3 - (18x^8)3xy^2y' = 0[/tex], we can utilize detachment of factors.

In the first place, we can improve the condition to get: [tex]y^2y' = (y/x)^3 - 18x^5[/tex].

Then, we can isolate the factors by duplicating the two sides by dx and partitioning the two sides by [tex](y^2(y/x)^3 - 18x^5)[/tex] to get:

[tex](y^2/y^3)dy = [(1/x)^3 - 18x^3]dx[/tex]

Incorporating the two sides, we get:

[tex]-1/y + c = (- 1/2x^2) - (3/2)x^6 + K[/tex]

Where K is an erratic steady of coordination.

At last, we can settle for y to get:

[tex]y = [1/(- 1/2x^2 - 3/2x^6 + c + K)]^_{(1/3)[/tex]

where c + K is the erratic steady.

Accordingly, the overall answer for the differential condition is:

[tex]y^3 - (18x^8)3xy^2y' = 0[/tex] is [tex]y = [1/(- 1/2x^2 - 3/2x^6 + c + K)]^(1/3)[/tex], where c and K are inconsistent constants.

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The probability that a customer will use plus gas and fill the tank is 0.21 or 21%.

Let's use the following notation:

R: the event that a customer uses regular gas

P: the event that a customer uses plus gas

M: the event that a customer uses premium gas

F: the event that a customer fills their tank

We are given:

P(R) = 0.4, P(P) = 0.35, P(M) = 0.25

P(F|R) = 0.3, P(F|P) = 0.6, P(F|M) = 0.5

We want to find P(P and F), the probability that a customer uses plus gas and fills their tank. We can use the following formula:

P(P and F) = P(F and P) = P(F|P) * P(P)

Substituting the values, we get:

P(P and F) = P(F|P) * P(P) = 0.6 * 0.35 = 0.21

Therefore, the probability that a customer will use plus gas and fill the tank is 0.21 or 21%.

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Given the function h(x) = 2x - 3 and g(x) = -2x + 2, g(h(x)) = -4x + 8

A function is a mathematical relationship that shows the relationship between two variables.

Given the functions h(x) = 2x - 3 and g(x) = -2x + 2

We desire to find (goh)(x), we proceed as follows.

Since we have the functions h(x) = 2x - 3 and g(x) = -2x + 2

We notice that  (goh)(x) = g(h(x))

So, substituting the values of the variables into the equation, we have that

(goh)(x) = g(h(x))

= -2x + 2

Substituting h(x) = 2x - 3 into the equation, we have that

g(h(x)) = -2x + 2

= -2(2x - 3) + 2

= -4x + 6 + 2

= -4x + 8

So, g(h(x)) = -4x + 8

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Since the posterior distribution is normal, the conditional expectation E[θ|X] is also a linear function of X.

Therefore, if 8(X)

(a)

If 8(X) is an unbiased estimate of 0, then we have E[8(X)] = 0, which means that ∫ 8(x)Pe(x)dx = 0 for all possible values of 0.

Now, the Bayes estimate with respect to quadratic loss is given by

δ(X) = argmin (E[(δ(X) - θ)^2|X]) = E[θ|X]

It can be shown that the Bayes estimate with respect to quadratic loss is the conditional expectation of θ given X.

Now, if δ(X) = 8(X), then we have

E[(δ(X) - θ)^2|X] = E[(8(X) - θ)^2|X]

= E[(8(X) - E[θ|X] + E[θ|X] - θ)^2|X]

= E[(8(X) - E[θ|X])^2|X] + E[(E[θ|X] - θ)^2|X] + 2E[(8(X) - E[θ|X])(E[θ|X] - θ)|X]

= Var[θ|X] + (E[θ|X] - θ)^2

where the last equality follows from the fact that 8(X) is an unbiased estimate of θ, and hence, E[8(X) - θ|X] = 0.

Since we are using quadratic loss, the above expression needs to be minimized with respect to δ(X), which is equivalent to minimizing Var[θ|X] + (E[θ|X] - θ)^2.

It can be shown that the minimum is achieved when δ(X) = E[θ|X].

Therefore, if 8(X) is the Bayes estimate with respect to quadratic loss, then we must have 8(X) = E[θ|X] for all possible values of X.

This means that the posterior distribution of θ given X is degenerate, i.e., P[δ(X) = θ|X] = 1 for all possible values of X.

Conversely, if P[δ(X) = θ|X] = 1 for all possible values of X, then δ(X) = E[θ|X] for all possible values of X.

This means that 8(X) is the Bayes estimate with respect to quadratic loss, and it is also an unbiased estimate of θ.

(b)

Suppose Pe = N(0,02%). Then, we have

E[θ^2] = Var[θ] + E[θ]^2 = 0.02

Since E[θ^2] < [infinity], we can conclude that Var[θ] < [infinity].

Now, suppose there exists a prior distribution π such that X is a Bayes estimate with respect to quadratic loss. Then, we must have

8(X) = E[θ|X]

It can be shown that if Pe = N(0,02%), then the posterior distribution of θ given X is also normal with mean

μ = (0.02/(0.02 + nσ^2))x

and variance

σ^2 = (0.02σ^2)/(0.02 + nσ^2)

where n is the sample size and σ^2 is the variance of Pe.

Since the posterior distribution is normal, the conditional expectation E[θ|X] is also a linear function of X.

Therefore, if 8(X)

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Answer:

[tex] {9}^{x} \div {81}^{ \frac{2}{x} } = 1[/tex]

[tex]{9}^{x} \times {81}^{ \frac{ - 2}{x} } = 1[/tex]

[tex]{9}^{x} \times { ({9}^{2}) }^{ \frac{ - 2}{x} } = 1[/tex]

[tex]{9}^{x} \times {9}^{ \frac{ - 4}{x} } = 1[/tex]

[tex] {9}^{(x - \frac{4}{x}) } = 1[/tex]

[tex]{9}^{0 } = 1 \: \: \: \: \: (known)[/tex]

Equating the exponents

[tex]x - \frac{4}{x} = 0[/tex]

[tex] \frac{ {x}^{2} - 4 }{x} = 0[/tex]

[tex] {x}^{2} = 4 \\ x = + 2 \: or \: 2[/tex]

The function f(x) = 5.1 log(x) is a dilation of the parent function by a factor of 5.1

From the question, we have the following parameters that can be used in our computation:

f(x) = log x

f'(x) = 5.1 log x

When the above functions are compared, we have

f'(x) = 5.1 log(x)

This means that the function f(x) is dilated by 5.1 to get the function f'(x)

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Two angles are co-terminal, meaning they differ by a multiple of 360°.

a) We know that sin A = opposite/hypotenuse = 5/12. Therefore, the adjacent side of angle A must be negative, since it is located in the second quadrant. We can use the Pythagorean theorem to find the hypotenuse:

(5/12)^2 + (adjacent)^2 = hypotenuse^2

25/144 + (adjacent)^2 = hypotenuse^2

(adjacent)^2 = hypotenuse^2 - 25/144

(adjacent)^2 = (hypotenuse^2 * 144 - 25)/144

We also know that cosine is adjacent/hypotenuse and tangent is opposite/adjacent, so:

cos A = adjacent/hypotenuse = sqrt(hypotenuse^2 - 25/144)/hypotenuse

tan A = opposite/adjacent = 5/sqrt(hypotenuse^2 - 25/144)

b) To find angle A, we can use the inverse sine function:

A = sin^-1(5/12)

A ≈ 24.02°

We know that cosecant is the reciprocal of sine, so:

csc A = 1/sin A

We want to find angles that have a cosecant of V3, so:

1/sin A = V3

sin A = 1/V3

We can use the unit circle to find angles whose sine is 1/V3. One such angle is 60°, since sin 60° = V3/2. Another angle is 300°, since sin 300° = -V3/2. These two angles are co-terminal, meaning they differ by a multiple of 360°.

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The frog's probability of being on lily pad B after an hour (or in the long run) is around 0.2975 or 29.75%.

To calculate the probability that the frog will be on lily pad B after an hour, we can start by calculating the long-run distribution or steady-state probabilities of the frog's movement between the two lily pads.

Let p represent the probability that the frog is on lily pad B at any given time. We may develop two equations based on the probabilities given in the problem:

p = 0.85(1-p) (the frog moves from point A to point B with a probability of 0.85 and stays on point B with a probability of 1-p).

1-p = 0.65p (the frog moves from B to A with the probability of 0.65 and stays on A with the probability of 1-p).

Simplifying the equation, we get:

p = 0.85 - 0.85p + 0.65p

p = 0.85(1 - 0.65) = 0.2975

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When x = 2, f(g(x)) is approximately equal to 187/9.

To solve for f(g(x)) when x = 2, we need to substitute the value of x into the function g(x) and then substitute the result into the function f(x). Let's calculate it step by step:

Step 1: Calculate g(x) when x = 2:

g(x) = x + 4/3

g(2) = 2 + 4/3

g(2) = 2 + 4/3

g(2) = 10/3

Step 2: Substitute the result from step 1 into f(x):

f(x) =[tex]x^2[/tex] + 2x + 3

f(g(x)) = f(10/3)

f(g(2)) = f(10/3)

Step 3: Calculate f(g(2)):

f(10/3) = (10/3[tex])^2[/tex] + 2(10/3) + 3

f(10/3) = 100/9 + 20/3 + 3

f(10/3) = 100/9 + 60/9 + 27/9

f(10/3) = 187/9

Therefore, when x = 2, f(g(x)) is approximately equal to 187/9.

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The equations which satisfy (6, -3) are: y = -5x + 27 and y = 2x - 15 (Option C and D)

To know which equation will result in (6, -3), we shall determine the value of y in each equation since we know that x = 6. Details below:

For A

y = -3x + 6

y = -3(6) + 6

y = -18 + 6

y = 12

For B

y = 2x - 9

y = 2(6) - 9

y = 12 - 9

y = 3

For C

y = -5x + 27

y = -5(6) + 27

y = -30 + 27

y = -3

For D

y = 2x - 15

y = 2(6) - 15

y = 12 - 15

y = -3

For E

y = -4x + 27

y = -4(6) + 27

y = -24 + 27

y = 3

From the above, the equation that satisfy (6, -3) are:

Thus, the correct answer to the question is Option C and D

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The general solution to the original differential equation is:  

y(t) = [tex]C1 e^t cos t + C2 e^t sin t + (1/2)ex sin t + (1/4)ex sin(2t) + (1/4)ln|[/tex]

We first solve the associated homogeneous differential equation:

[tex]2y'' - 4y' + 4y[/tex] = 0

The characteristic equation is[tex]r^2[/tex] - 2r + 2 = 0, which has roots r = 1 ± i. Therefore, the general solution to the homogeneous equation is:

[tex]y_h(t) = e^t([/tex]C1 cos t + C2 sin t)

To use the method of variation of parameters to find the particular solution to the original equation, we assume that the solution has the form:

[tex]y_p(t) = u(t)e^t cos t + v(t)e^t sin t[/tex]

where u(t) and v(t) are functions to be determined.

[tex]y_p''(t) \\\\2u'(t)e^t cos t + 2v'(t)e^t sin t + 2u(t)e^t cos t - 2v(t)e^t sin t - 2u(t)e^t sin t - 2v(t)e^t cos t[/tex]

[tex]y_p'(t) = u'(t)e^t cos t + v'(t)e^t sin t + u(t)e^t cos t + v(t)e^t sin t[/tex]

Substituting these into the original equation and simplifying, we get:

[tex]2u'(t)e^t cos t + 2v'(t)e^t sin t = ex sec x[/tex]

We need to find u'(t) and v'(t) such that this equation holds for all t. To do this, we take the derivative of the assumed solution with respect to t and equate coefficients of cos t and sin t separately:

[tex]u'(t)e^t cos t + v'(t)e^t sin t + u(t)e^t cos t + v(t)e^t sin t = 0 (1)\\v'(t)e^t cos t - u'(t)e^t sin t + u(t)e^t sin t - v(t)e^t cos t = ex sec x (2)[/tex]

Solving equation (1) for u'(t) and v'(t) and substituting into equation (2), we get:

[tex]v(t) = ∫ [ex sec x / (e^(2t))] dt\\u(t) = -∫ [ex sec x / (e^(2t))] tan t dt[/tex]

Evaluating the integrals, we get:

[tex]v(t) = (1/2)ex tan x - (1/2)ln|cos x| + C1\\u(t) = (1/4)ex [sin(2t) - 2cos(2t)] + (1/4)ln|cos x| tan x + C2[/tex]

where C1 and C2 are arbitrary constants.

The general solution to the original differential equation is:  

y(t) = [tex]C1 e^t cos t + C2 e^t sin t + (1/2)ex sin t + (1/4)ex sin(2t) + (1/4)ln|[/tex]

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Here is the function to show how much money Milo saves each week:

f(x) = 0.5(x - 15)

A mathematical function is a rule that relates each element of a set called the domain to exactly one element of a set called the range.

The domain is the set of all possible input values for the function, and the range is the set of all possible output values.

Taking Milo for example,

Milo saves each week, given his weekly allowance x, can be expressed as:

f(x) = 0.5(x - 15)

where

x = amount of money Milo saves each week.

if Milo's weekly allowance is $50, then his savings can be calculated using the function as follows:

f(50) = 0.5(50 - 15) = 0.5(35) = 17.5

Therefore, Milo saves $17.5 each week when his weekly allowance is $50.

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There are 720 different ways to position six individuals in a line for a photo.

From the information, the number of ways six people can be placed in a line for a photo can be determined using the expression 6!.

6! is the factorial of 6, which is the sum of all positive numbers ranging from 1 to 6. So,

6! = 6 x 5 x 4 x 3 x 2 x 1

When we simplify this expression, we get:

6! = 720

As a result, there are 720 different ways to position six individuals in a line for a photo.

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(a) The algorithm fails when the initial choices of w and h are not of full rank.

(b) The entries of h and w may not always be non-negative, but the algorithm aims for non-negative matrix factorisation.

(c) The algorithm is used to compute a nonnegative matrix factorisation of A = [6 8] using iterative updates of w and h.

(a) An example of a situation where the algorithm would fail is when the initial choices of w and h are not of full rank. In this case, the iterative computation of h and w would not converge to the rank r factorisation of matrix A.

(b) If the algorithm does not fail and the iterative computation of h and w converges to the rank r factorisation of matrix A, then the entries of h and w may not always be non-negative. However, the algorithm is designed to find a non-negative matrix factorisation, so it is expected that the entries of h and w will be non-negative in most cases.

(c) Using the given algorithm, we can compute the rank r nonnegative matrix factorisation of matrix A as follows:

- Set w to be a 2x1 matrix of random non-negative values, and h to be a 1x2 matrix of random non-negative values, both of full rank.
- Compute h = h .* (w' * A) ./ (w' * w * h) and w = w .* (A * h') ./ (w * h * h'), where .* denotes element-wise multiplication, and ' denotes matrix transpose.
- Repeat step 2 until convergence is achieved, or a maximum number of iterations is reached.

Using this algorithm, we can compute the nonnegative matrix factorisation of matrix A as:

w = [0.1829; 0.9119]
h = [3.6953 4.9237]

where w and h are non-negative matrices of rank 1 that satisfy A = w * h.

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For a metal rod of length 31 cm is placed in a magnetic field of strength 2. 3 T, the induced emf, in volts, between the ends of the rod when the rod is not moving is equals to zero.

When a conducting rod is moving in magnetic field perpendicular to its velocity, electro motive force( EMF ) between the ends of the rod is generated due to the Lorentz force exerted on free charges of the rod. The value of [tex]EMF = BvLsin⁡θ[/tex], where B is magnetic field, L is the length of the rod, v is the rod speed, θ is the angle between the rod and velocity vector. We have a metal rod, with length of metal rod, L = 31 cm

The strength of magnetic field, B = 2.3 T, oriented perpendicular to the field, θ

= 90°

Now, the rod is not moving,so v = 0 m/s, then EMF = BvLsin⁡θ = 31× 2.3 × 0

=> EMF = 0 V.

So, The induced emf between the ends of the rod when the rod is not moving is zero.

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Complete question:

A metal rod of length 31 cm is placed in a magnetic field of strength 2. 3 t, oriented perpendicular to the field. Determine the induced emf, in volts, between the ends of the rod when the rod is not moving.