what percentage of the total sum of squares can be accounted for by the estimated regression equation (to decimal)?

Answer:

A) 7

B) (47.5,54.5)

Step-by-step explanation:

a) The margin of error is half the length of the confidence interval, so the length of the confidence interval is 2 times the margin of error:

Length of confidence interval = 2 x 3.5 = 7.

b) If the sample mean is 51 and the margin of error is 3.5, then the confidence interval can be calculated as follows:

Lower bound = sample mean - margin of error = 51 - 3.5 = 47.5

Upper bound = sample mean + margin of error = 51 + 3.5 = 54.5

Therefore, the 95% confidence interval for the population mean is (47.5, 54.5).

L<1, the series is absolutely convergent by the Ratio Test.

What is ratio test?

The ratio test is a convergence test used to determine whether an infinite series converges or diverges. It is based on the idea that if the ratio of successive terms in a series approaches a limit L as n approaches infinity, then the series converges absolutely if L < 1, diverges if L > 1, and inconclusive if L = 1.

More formally, given a series [tex]\sum_{n=1}^\infty a_n[/tex], we consider the limit

[tex]L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|[/tex]

If L < 1, then the series converges absolutely. If L > 1, then the series diverges. If L = 1, then the ratio test is inconclusive and we may need to use other tests to determine convergence or divergence.

Using the ratio test, we have:

[tex]\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{(-1)^{n+1}\frac{(n+1)^3 3^{n+1}}{(n+1)!}}{(-1)^n\frac{n^3 3^n}{n!}}\right|\\\&=\lim_{n\to\infty} \left|\frac{(n+1)^3 3}{n^3}\right|\&=\lim_{n\to\infty} \left|\frac{n^3+3n^2+3n+1}{n^3}\cdot 3\right|\\&=\lim_{n\to\infty} \left(3+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\right)\\&=3[/tex]

Since L<1, the series is absolutely convergent by the Ratio Test.

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Answer:

64

Hope this helps!

Step-by-step explanation:

1, 4 ( 1 × 4 ), 16 ( 4 × 4 ), 64 ( 16 × 4 ), 256 ( 64 × 4 ), etc.

The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant rate. The probability density function of the exponential distribution is given by f(x) = λe^(-λx), where λ is the rate parameter.

a. To show that EX" = EXn-1, we need to use the memoryless property of the exponential distribution. This property states that the conditional probability of X > t+s given that X > s is equal to the unconditional probability of X > t, for any s,t > 0. Using this property, we can write:

EX" = E(X|X > n-1) + (n-1) = E(X) + n-1 = 1/λ + n-1

EXn-1 = E(X|X > 1) + (n-2) = E(X) + n-2 = 1/λ + n-2

Since EX" = EXn-1, we have shown that the memoryless property holds.

b. To find EX" = n!, we can use the moment generating function (MGF) of the exponential distribution, which is given by M(t) = λ/(λ-t). The nth moment of X is defined as E(X^n) = (-1)^n d^n M(t)/dt^n at t=0. Differentiating the MGF n times, we get:

E(X^n) = n!λ^n/(λ-t)^n+1 at t=0

Setting n=1, we get E(X) = 1/λ, which is the mean of the exponential distribution. Setting n=2, we get E(X^2) = 2/λ^2, which is the variance of the exponential distribution.

For n>2, we can use the formula above to find the nth moment:

E(X^n) = n!λ^n/(-1)^{n+1} = n!/(λ^n)

Therefore, for n = 1,2,3,..., we have:

EX" = E(X^n|X > n-1) = (n-1)!/(λ^n-1) = n!/λ^n

Thus, we have shown that EX" = n! for n = 1,2,3,...

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To calculate the weighted mean of student 11's exam scores when exam 4 is weighted three times that of the other 3 exams, we need to first multiply the score of exam 4 for student 11 by 3, and then add up all four exam scores for student 11, multiplied by their respective weights.

Let's denote the exam scores for student 11 as follows: E1, E2, E3, E4 (where E4 is the score for exam 4). The weights for each exam are given in the second sheet of the spreadsheet. Let's denote these weights as W1, W2, W3, W
The weighted mean for student 11 can be calculated as follows:

Weighted Mean = (W1*E1 + W2*E2 + W3*E3 + 3*W4*E4) / (W1 + W2 + W3 + 3*W4)
We plug in the values or student 11 from the spreadsheet to get:

Weighted Mean = (0.15*86 + 0.2*93 + 0.25*78 + 3*0.4*89) / (0.15 + 0.2 + 0.25 + 3*0.4)
Weighted Mean = 243.9 / 1.45
Weighted Mean = 168.28

Therefore, the weighted mean of student 11's exam scores when exam 4 is weighted three times that of the other 3 exams is 168.28.
To calculate the weighted mean of student 11's exam scores with exam 4 weighted three times that of the other 3 exams, you should follow these steps:

1. Locate student 11's scores for exams 1, 2, 3, and 4 in the second sheet of the spreadsheet.
2. Assign the weights: 1 for exams 1, 2, and 3, and 3 for exam 4.
3. Multiply each exam score by its corresponding weight.
4. Add the weighted scores together.
5. Divide the sum by the total sum of weights (1 + 1 + 1 + 3 = 6).

Weighted mean = (Exam1 * 1 + Exam2 * 1 + Exam3 * 1 + Exam4 * 3) / 6
Please note that without the actual spreadsheet and data, I cannot provide you with a specific answer. Once you have the data, follow these steps to calculate the weighted mean for student 11.

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The highest free throw of 82% has an average bowling score of 108

This ordered pair represents the ordered pair of the highest x value in the graph

From the graph, the ordered pair is (82, 108)

It means that the student with the highest free throw of 82% has an average bowling score of 108

This ordered pair represents the ordered pair of the highest y value in the graph

From the graph, the ordered pair is (80, 112)

It means that the student with the highest free throw of 80% has an average bowling score of 112

From the graph, we can see that the association is a positive linear association

This is because as the free throw percents increases, the average bowling score is also expected to increase

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Complete question

The members of a school basketball team went bowling for a season ending party. They made this scatter plot to compare their free throw percents for the season and their average bowling score for 3 games.

Part A:

What ordered pair represents the student with the highest free throw percent? Explain the meaning of each coordinate in the ordered pair.

Part B:

What ordered pair represents the student with the highest bowling average? Explain the meaning of each coordinate in the ordered pair.

Part C:

Is the association between free throw percent and bowling average linear or nonlinear? If it is linear, is the relationship positive, negative, or neither? State the association, if any, in terms of the variables.

The percentage of students who travel are :

Walk = 50%, Cycle = 8.33%, Car = 25% and bus = 16.67%.

Given that,

Total number of students surveyed = 60

Number of students who travel by walk = 30

Percentage of students who walk = 30/60 × 100 = 50%

Number of students who travel by cycle = 5

Percentage of students who cycle = 5/60 × 100 = 8.33%

Number of students who travel by car = 15

Percentage of students who car = 15/60 × 100 = 25%

Number of students who travel by bus = 10

Percentage of students who bus = 10/60 × 100 = 16.67%

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The complete question is given below.

60 students are asked the following question in a survey :

How do you travel to school?

Here are the results:

Travel method   No of students

Walk                   30

Cycle                   5

Car                      15

Bus                      10

Complete the pie chart to show this information.

The recurrence relation for the number of ways to form postage of n cents with 4-cent, 6-cent, and 10-cent stamps, with order mattering, is P(n) = P(n-4) + P(n-6) + P(n-10), with initial conditions P(0) = 1 and P(n) = 0 for n < 0.

To form postage of n cents, we can use either a 4-cent stamp, a 6-cent stamp, or a 10-cent stamp.

Therefore, the number of ways to form postage of n cents can be calculated by considering the number of ways to form postage of (n-4) cents, (n-6) cents, and (n-10) cents.

Let P(n) denote the number of ways to form postage of n cents with these stamps.

Then we have:

P(n) = P(n-4) + P(n-6) + P(n-10)

This is a recurrence relation for P(n).

The initial conditions for this recurrence relation are:

P(0) = 1 (There is one way to form postage of 0 cents, by using no stamps.)

P(n) = 0 for n < 0 (There are no ways to form negative postage.)

We can also find P(4), P(6), and P(10) directly:

P(4) = 1 (We can use one 4-cent stamp.)

P(6) = 2 (We can use one 6-cent stamp, or two 4-cent stamps.)

P(10) = 4 (We can use one 10-cent stamp, or one 6-cent stamp and one 4-cent stamp, or two 4-cent stamps and one 2-cent stamp, or four 2-cent stamps.)

Using these initial conditions and the recurrence relation, we can calculate P(n) for any positive integer n.

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Different function model used in to model different situations of real life.l, for example Linear, quadratic and polynomial function model. The slope of a hill, roller coaster designers are real life example of polynomial model.

Various functions can be used to test real-world situations. We have a linear business model associated with the product or the main features of the business that makes them ascending or descending. A quadratic function simulates an increase followed by a decrease.

Polynomial functions simulate many changes in direction. Multinomial models can now be used to investigate situations where the relationship between variable and estimator is curvilinear. Sometimes nonlinear relationships at the small scale of the description can also be modeled with polynomials. For example, roller coaster designers may use polynomial model to describe the bends of their rides. Other examples include the continuation of slopes, curved bridges or mountains which are based on polynomial function modelling.

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To evaluate the principal value of the integral 00 dx / (x^2 + 2), we need to find the limit of the integral as the limits of integration approach 0 from both the positive and negative sides. The Principal value of the integral = arctan(x / sqrt(2)) + C.

Evaluate the principal value of the integral ∫(1 / (x^2 + 2)) dx
Here is a step-by-step explanation to solve this integral:
Step 1: Identify the function to integrate
The given function is f(x) = 1 / (x^2 + 2).
Step 2: Perform substitution
We can perform a trigonometric substitution to make integration easier. Let x = sqrt(2) * tan(u), so dx = sqrt(2) * sec^2(u) du.
Step 3: Rewrite the integral with substitution
The integral becomes ∫(sqrt(2) * sec^2(u) du / (2*tan^2(u) + 2)).
Step 4: Simplify the integrand
Simplify the expression inside the integral to get ∫(sec^2(u) du / (sec^2(u))).
Step 5: Integrate the simplified expression
Since the numerator and denominator are the same, the integrand simplifies to 1. Now, we just need to integrate ∫1 du. The result is the integral of 1 with respect to u, which is simply u + C, where C is the constant of integration.
Step 6: Replace u with the original variable
Recall that we set x = sqrt(2) * tan(u), so u = arctan(x / sqrt(2)). Therefore, the final answer is:
Principal value of the integral = arctan(x / sqrt(2)) + C

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The volume of the prism is 96 cm³.

To calculate the volume of the prism, we need to multiply the cross-sectional area by the length of the prism. Given that the cross-sectional area is 24 cm² and the length is 4 cm, we can use the formula:

Volume = Cross-sectional area * Length

Volume = 24 cm² * 4 cm

Volume = 96 cm³

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The derivative of the natural logarithm function, ln(x), is simply the reciprocal of x i.e. d/dx ln(x) = 1/x.

The derivative of ln(x) with respect to x, denoted as d/dx ln(x), can be found using the logarithmic differentiation technique. First, we express ln(x) as the natural logarithm of e raised to the power of ln(x):
ln(x) = ln(e^(ln(x)))
Using the chain rule, we can then find the derivative of ln(x) as:
d/dx ln(x) = d/dx ln(e^(ln(x)))
          = 1/x

Therefore, the derivative of ln(x) with respect to x is simply 1/x. This result can be useful in solving problems involving exponential and logarithmic functions, particularly when finding the slopes of tangent lines or rates of change.

In the context of exponentials and logs, "d/dx ln(x)" refers to finding the derivative of the natural logarithm function, ln(x), with respect to x. The natural logarithm is the inverse of the exponential function with base e (approximately equal to 2.718).
The derivative of ln(x) with respect to x is given by:
d/dx ln(x) = 1/x

So, the derivative of the natural logarithm function, ln(x), is simply the reciprocal of x.

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The estimated temperature at the point (2.04, 0.98) on the unevenly heated metal plate is approximately 138.66 degrees Celsius.

To estimate the temperature at the point (2.04, 0.98), we can use the first-order Taylor approximation, which includes the partial derivatives of the temperature T(x,y) with respect to x and y.

Given the information T(2,1) = 138, Tx(2,1) = 12, and Ty(2,1) = -9, we can estimate T(2.04, 0.98) as follows:

T(2.04, 0.98) ≈ T(2,1) + Tx(2,1) * (2.04 - 2) + Ty(2,1) * (0.98 - 1) T(2.04, 0.98) ≈ 138 + 12 * (0.04) - 9 * (-0.02) T(2.04, 0.98) ≈ 138 + 0.48 + 0.18 T(2.04, 0.98) ≈ 138.66.

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Kylie has to try at most 3,024 different combinations in order to call Katerina if she does not dial repetitive phone numbers.

Kylie only recalls the first six numbers of Katerina's phone number, so she must guess the last four. Because she cannot repeat any of the numbers she has already successfully predicted, the first remaining digit has just nine viable alternatives (all digits from 0 to 9 except the one she already knows). Similarly, the second remaining digit has just eight options, the third has seven, and the fourth has six.

Therefore, the total number of possible combinations is:

9 x 8 x 7 x 6 = 3,024

This implies Kylie will have to try at most 3,024 different combinations to reach Katerina, provided she gets it right on the last try.

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To write the iterated integral for the shaded region r in exercises 21-26, we need to use the concept of double integrals. A double integral is a type of iterated integral that allows us to integrate over a two-dimensional region.

The iterated integral for the shaded region r can be written as:

∫∫r f(x,y) dA

where f(x,y) is the function we are integrating and dA represents the infinitesimal area element.

To evaluate the double integral, we can use either the row-first or column-first method. The row-first method involves fixing the value of y and integrating with respect to x first, while the column-first method involves fixing the value of x and integrating with respect to y first.

For example, in exercise 21, the shaded region r is the rectangle with vertices (0,0), (0,2), (3,2), and (3,0). If we want to integrate the function f(x,y) over this region, we can write the iterated integral as:

∫0^3 ∫0^2 f(x,y) dy dx

This means we first integrate f(x,y) with respect to y from y=0 to y=2, and then integrate the resulting expression with respect to x from x=0 to x=3.

Similarly, we can write the iterated integral for the shaded region r in exercises 22-26 using the same concept of double integrals.

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Answer:

Step-by-step explanation:

Answer:

[tex] \sqrt{72 {m}^{5} {n}^{2} } [/tex]

[tex] \sqrt{2 \times 36 \times {m}^{4} \times m \times {n}^{2} } [/tex]

[tex]6 {m}^{2} n \sqrt{2m} [/tex]

C is the correct answer.

Vivian paid $17.36 for 3.5 pounds.

Given that, one pound of sirloin steak costs $4.95, Vivian bought 3.5 pounds of the sirloin steak for a church cookout,

We need to find the total she paid for 3.5 pounds,

So,

if 1 pound = 4.96

so, 3.5 = 4.96×3.5

= 17.36

Hence, Vivian paid $17.36 for 3.5 pounds.

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Since all three conditions are satisfied, we can conclude that S is a subspace of M2(R). To prove that s is a subspace of (m2(r), , .), we need to show that s satisfies the three axioms of a subspace:

1. s contains the zero vector:
The zero vector in m2(r) is the 2x2 matrix with all entries equal to zero. We can verify that this matrix satisfies ab = ba for any matrix b, so the zero vector is in s.

2. s is closed under vector addition:
Let b1 and b2 be matrices in s. We need to show that their sum, b1 + b2, is also in s.

(ab1 + ab2) = a(b1 + b2) = ab1 + ab2 (using the distributive property of matrix multiplication)

Similarly,

(b1a + b2a) = (b1 + b2)a = b1a + b2a

So b1 + b2 satisfies the condition ab = ba and is therefore in s.

3. s is closed under scalar multiplication:
Let b be a matrix in s, and let c be a scalar. We need to show that the product cb is also in s.

(acb) = a(cb) = a(bc) = (ab)c = (ba)c = b(ac)

So cb satisfies the condition ab = ba and is therefore in s.

Since s satisfies all three axioms of a subspace, we can conclude that s is indeed a subspace of (m2(r), , .).
To determine if the set S is a subspace of the vector space M2(R), we need to check if it satisfies three conditions: closure under addition, closure under scalar multiplication, and the existence of the zero vector.

1. Closure under addition:
Let B1 and B2 be two matrices in S such that AB1 = B1A and AB2 = B2A. We need to check if the sum B1 + B2 is also in S.
A(B1 + B2) = AB1 + AB2 = B1A + B2A = (B1 + B2)A, which shows that the sum B1 + B2 is in S.

2. Closure under scalar multiplication:
Let B be a matrix in S such that AB = BA, and let c be a scalar in R. We need to check if cB is also in S.
A(cB) = c(AB) = c(BA) = (cB)A, which shows that the product cB is in S.

3. Existence of the zero vector:
The zero matrix 0 satisfies A0 = 0A = 0, so the zero matrix is in S.

Since all three conditions are satisfied, we can conclude that S is a subspace of M2(R).

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For i) we get: a = -1/3 and b = -20/3 and for ii) we get the polynomial satisfying all the given properties is: P(x) = x³ - (1/3)x² - (20/3) and for iii) we get c=0

Explanation:

To satisfy property iii) P(0)=0, we know that c must be equal to 0.

Let's now use the first two properties to find the values of a and b.

i) At x = -3, P'(x) = 0 and P''(x) < 0 for a local maximum.

P'(x) = 3x² + 2ax + b

P''(x) = 6x + 2a

Substituting x = -3 in the above equations, we get:

9a - 9 + b = 0

-18 + 2a < 0

Solving the above two equations simultaneously, we get:

a = -1/3 and b = -20/3

ii) At x = 7, P'(x) = 0 and P''(x) > 0 for a local minimum.

Using the same approach as above, we get:

a = -2/3 and b = 532/9

Therefore, the polynomial satisfying all the given properties is:

P(x) = x³ - (1/3)x² - (20/3)x

Note that property iii) is satisfied because we set c=0 earlier.

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The latest time Jane and her family can start hiking is 3:19 p.m.

The latest time Jane and her family can start hiking is as follows:

1. First, find the total time needed for hiking and setting up camp: 1 hour and 36 minutes for hiking, plus 57 minutes for setting up camp.

2. Convert the hours and minutes to minutes only: 1 hour = 60 minutes, so 1 hour and 36 minutes = 60 + 36 = 96 minutes.

3. Add the hiking time (96 minutes) to the camp setup time (57 minutes): 96 + 57 = 153 minutes.

4. Convert 153 minutes back to hours and minutes: 153 minutes = 2 hours and 33 minutes.

5. Subtract the total time needed (2 hours and 33 minutes) from the sunset time (5:52 p.m.): 5:52 p.m. - 2 hours and 33 minutes.

6. The latest time Jane and her family can start hiking is 3:19 p.m.

Your answer: The latest time Jane and her family can start hiking is 3:19 p.m.

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To find aw/as and aw/at using the Chain Rule:

(a) Using the

Chain Rule

, we have:

aw/as = (dw/ds) * (ds/as) = ((dw/dx) * (dx/ds) + (dw/dy) * (dy/ds) + (dw/dz) * (dz/ds)) * (ds/as)

Substituting

the given expressions for x, y, and z, we get:

x = 2t, y = s - 2t, z = st

dx/ds = 0, dy/ds = 1, dz/ds = t

dx/dt = 2, dy/dt = -2, dz/dt = s

Therefore:

aw/as = ((z/x) * 2 + (z/y) * (-2) + (x*y)) * (1/s) = (2st/x - 2st/y + 2st) * (1/s)

= 2st * (y/x - y/y + 1) * (1/s) = 2st * (1 - 2/s) * (1/s)

= 2t * (1 - 2/s)

Similarly, we can find

aw/at

:

aw/at = (dw/dx) * (dx/dt) + (dw/dy) * (dy/dt) + (dw/dz) * (dz/dt)

= z * 2 + (-z) * 2 + xy

= 2st - 2st + 2ts

= 2ts

Therefore, aw/as = 2t * (1 - 2/s) and aw/at = 2ts.

(b) To find aw/as and aw/at by converting w to a function of s, we substitute the expressions for x and y in terms of s:

x = 2t, y = s - 2t

into the expression for w:

w = xyz = (2t)(s - 2t)(st) = 2st^2 - 4t^2s

Then we differentiate w with respect to s and t to get:

dw/ds = 4t^2 - 4t

dw/dt = 4st - 8ts

Using the Chain Rule, we can find aw/as and aw/at:

aw/as = dw/ds = 4t^2 - 4t

aw/at = dw/dt = 4st - 8ts

Therefore,

aw/as = 4t^2 - 4t

and aw/at = 4st - 8ts.

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The measure of the angle x will be 57°.

The angle of an arc in a circle is defined as the angle subtended by the arc at the centre of the circle. The angle of the arc is measured in degrees or radians.

There is a theorem related to the angle of an arc in a circle called the central angle theorem. This theorem states that the angle subtended by an arc at the centre of a circle is equal to double the angle subtended by the same arc at any point on the circumference of the circle.

The measure of the angle x is,

x = ( 360 - 123 - 123 ) / 2

x = 57°

Hence, the value of an angle x is 57°.

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The total number of ways is the product of these combinations: C(100, 10) * C(90, 10) * C(80, 10) * ... * C(20, 10). To divide one hundred people into ten discussion groups with ten people in each group, we can use the concept of combinations.

A combination represents the number of ways to choose items from a larger set, without considering the order of the items. In this case, we can use the formula:

C(n, r) = n! / (r!(n-r)!)

where C(n, r) represents the number of combinations, n is the total number of items, r is the number of items to be chosen, and ! represents the factorial.

For your problem, we'll divide the people into groups sequentially. First, we choose 10 people out of 100 for the first group, then 10 out of the remaining 90 for the second group, and so on. So the total number of ways is the product of these combinations:

C(100, 10) * C(90, 10) * C(80, 10) * ... * C(20, 10)

Calculating these combinations and multiplying them together, we get the total number of ways to divide one hundred people into ten discussion groups of ten people each.

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The data collection method in this scenario can best be described as "random sampling." The student randomly draws 200 marbles from a bag of 800 total marbles, which helps ensure an unbiased representation of the overall population.

The data collection method used by the student can be best described as a random sampling. This is because the student selected a sample of 200 marbles from the bag randomly, without any predetermined pattern or bias. This sample was then used to make an inference about the entire bag of marbles, without having to count all 800 marbles individually. The statistics obtained from the sample (73 red marbles out of 200) can then be used to estimate the proportion of red marbles in the entire bag (which would be 300/800 or 0.375). By analyzing the proportion of red marbles in the sample (73 out of 200), they can estimate the number of red marbles in the entire bag using statistics.

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To graph the function d = 8 tan 2xt on the interval (0,2), we can plot several points by selecting different values of t and calculating the corresponding value of d. The function d = 8 tan 2xt is undefined whenever the tangent function is undefined, which occurs when the angle of the tangent is equal to (2k + 1)π/2, where k is an integer.

a) For example, when t = 0.5 seconds, we have d = 8 tan(2(0.5)x) = 8 tan(x), and we can evaluate this expression for different values of x to get points on the graph. Repeat this process for different values of t to obtain more points and connect them to form the graph.

b) Since tan(x) is undefined when x = (2k + 1)π/2, we have 2xt = (2k + 1)π/2, which implies t = (2k + 1)π/4x for k ∈ Z. Therefore, the function is undefined at times t = (2k + 1)π/4x for k = -4, -3, -2, -1, 0, 1, 2, 3, 4. Plugging in x = 10.2, we get t = ±0.3877, ±1.162, ±1.936, ±2.71. Therefore, the function is undefined at these values of t.

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If 1.4 g of chemical A is used, the amount of chemical B needed to balance the chemical reaction is 1.6 g.

The mass of chemical B needed is calculated by reading off their corresponding values from the graph as shown below;

If 1.4 g of chemical A is used, the amount of chemical B needed to balance the chemical reaction must be taken from the graphed values by tracing the value of chemical A from the horizontal axis to the corresponding value of chemical B on vertical axis.

Chemical A = 1.4 g

Chemical B = 1.6 g (this value is obtained by tracing the intersection of 1.4 g on the curve to the y-axis).

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Using the concept of sine rule, the missing side of the triangle is: 12.05 units

Sine rule is one in geometry that is used to show the relationship between the sides and angles of a triangle. It is given as:

a/sinA = b/sinB = c/sinC

Where:

A, B, C are the angles and a, b and c are the opposite sides to the angles.

The sum of angles in a triangle is 180 degrees.

Thus:

Missing angle of triangle = 180 - (70 + 61)

Missing angle = 49°

Using sine rule, we can easily say that let the missing side be x and as such we have:

x/sin 49 = 15/sin 70

x = (15 * sin 49)/sin 70

x = 12.05 units

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To solve the equation 2y^(4) +11y^(3) + 18y" + 4ť' - 8y=0, we can first factor out a common factor of 2y: 2y(y^3 + 5y^2 + 9y + 2ť' - 4) = 0 Next, we can focus on the expression inside the parentheses, which can be written as:

y^3 + 5y^2 + 9y + 2ť' - 4

We can recognize this expression as the polynomial 2r^4 +11r^3 + 18r^2 + 4r – 8 evaluated at r=y-1/2.

So, 2r^4 +11r^3 + 18r^2 + 4r – 8 = (2r - 1)(r + 2)^3.

Finally, we can factor out a common factor of 2 and use the factored form of the polynomial:

2(y-1)(y+2)^3(2y+1) = 0

Therefore, the solutions to the original equation are:

y = 1, -2, -1/2 (multiplicity 3)
To solve the given polynomial equation, we will first rewrite it in a more accurate and coherent form by removing the irrelevant terms:

2y^4 + 11y^3 + 18y^2 + 4y - 8 = 0

We are given a hint that the equation can be factored as:

(2r - 1)(r + 2)^3

Now, we will replace r with y to match the variable in the given equation:

(2y - 1)(y + 2)^3 = 0

Now that we have the factored form of the equation, we can set each factor equal to zero and solve for y:

1) 2y - 1 = 0
2y = 1
y = 1/2

2) (y + 2)^3 = 0
y + 2 = 0
y = -2

So the solutions to the given equation are y = 1/2 and y = -2.

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