15:60 simlplfed a. 3:4 b 1:4 c 2:5

In a hypothesis testing context, before examining the data, one should typically decide whether the alternative hypothesis is one-sided or two-sided. This decision is based on the specific research question and the expected direction of the effect being tested.

It helps determine the appropriate statistical test and the formulation of the null and alternative hypotheses.

The computation of the p-value and the decision of whether or not to reject the null hypothesis are made after examining the data and conducting the statistical analysis. The p-value is a measure of the strength of the evidence against the null hypothesis, and it is compared to a predetermined significance level to make a decision. If the p-value is below the significance level, the null hypothesis is typically rejected in favor of the alternative hypothesis.

Therefore, the correct answer is (c) decided whether the alternative hypothesis is one-sided or two-sided.

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The probability of rolling a sum greater than 5 when rolling a pair of standard six-sided dice is 5/6 or approximately 0.833.

To determine the probability of rolling a sum greater than 5, we can first find the total number of possible outcomes when rolling two dice, which is 36 (6 possible outcomes for each of the 6 sides on the first die). We can then count the number of outcomes where the sum of the two dice is greater than 5, which includes the outcomes (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), and (6,6). There are 21 such outcomes, so the probability of rolling a sum greater than 5 is 21/36, which simplifies to 5/6 or approximately 0.833.

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If c and d are positive, then -c/2>d/2  is an example of a positive fraction c/d that satisfies the inequality -50 (c/d) > - 50

We know that c/d is a positive fraction,

so c>0 and d>0.

Multiplying both sides of the inequality by -d (which is negative since d>0), we get:

-50(c/d)>-50(-d)

-50c>50d

Dividing both sides by 50 (which is positive), we get:

-c>d

-c/2>d/2

Since c and d are positive, this is an example of a positive fraction c/d that satisfies the inequality -c/2>d/2

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Answer:

Step-by-step explanation:

Given two of three numbers correct, you want to find the correct value for the third number of 8 faces, 7 vertices, and 10 edges.

The relation between faces, vertices, and edges is ...

  F + V = E + 2

The given numbers are off by 3:

  8 + 7 = 15 ≠ 12 = 10 + 2

We can decrease the numbers of Faces or Vertices by 3, or we can increase the number of Edges by 3.

The numbers will be correct if we change to ...

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The exact length of the curve is 8/3 (5sqrt(26) - 1).

To find the length of the curve, we can use the arc length formula:

L = ∫[tex][a,b]sqrt(dx/dt)^2 + (dy/dt)^2 dt[/tex]

where a and b are the starting and ending values of the parameter t, and dx/dt and dy/dt are the derivatives of x and y with respect to t, respectively.

Plugging in the given equations, we get:

[tex]dx/dt = 24t[/tex]

[tex]dy/dt = 24t^2[/tex]

Therefore,

[tex](sqrt(dx/dt)^2 + (dy/dt)^2) = sqrt((24t)^2 + (24t^2)^2) = sqrt(576t^2 + 576t^4)[/tex]

Substituting these expressions into the arc length formula, we get:

L = ∫[tex][0,2]sqrt(576t^2 + 576t^4) dt[/tex]

We can factor out 576t^2 from the square root:

L = ∫[tex][0,2]sqrt(576t^2(1 + t^2)) dt[/tex]

And then simplify the expression inside the square root:

L = ∫[tex][0,2]24t sqrt(1 + t^2) dt[/tex]

This integral can be evaluated using the substitution[tex]u = 1 + t^2, du/dt = 2t, dt = du/2t:[/tex]

L = ∫[tex][1,5]12 sqrt(u) du[/tex]

Now we can use the power rule of integration to evaluate this integral:

[tex]L = [8/3 u^(3/2)]_1^5 = 8/3 (5sqrt(26) - 1)[/tex]

Therefore, the exact length of the curve is 8/3 (5sqrt(26) - 1).

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There are 10 boys in Mr. Johnson's class, and the total number of Students in the class is:10 + 15 = 25

If the ratio of boys to girls in Mr. Johnson's class is 2 to 3, this means that for every 2 boys, there are 3 girls. Let's represent the number of boys in the class as "b". Then we can set up the following proportion:2/3 = b/15

To solve for "b", we can cross-multiply:2 x 15 = 3 x b

30 = 3b

b = 10

Therefore, there are 10 boys in Mr. Johnson's class, and the total number of students in the class is:10 + 15 = 25

It's worth noting that we could have also found the number of girls in the class by using the ratio. Since the ratio of boys to girls is 2 to 3, this means that the total number of parts in the ratio is 2 + 3 = 5. To find the number of girls, we can divide the total number of students (25) by the total number of parts (5) and then multiply by the number of parts representing girls (3):(25/5) x 3 = 15

So we can see that there are indeed 15 girls in the class.

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Answer:

Step-by-step explanation:

a= 3(x+11)

B=9(x+8)

C=5(x+5)

D=3(3x+2)

The probability density function on [0 1] is 36, the correct option is C.

We are given that;

Function f(x) = ke-^(3x+2)

Now,

The function f(x) = ke^(-3x+2) is a probability density function on [0 1] if and only if the integral of f(x) from 0 to 1 is equal to 1.

∫(0 to 1) ke^(-3x+2) dx = -1/3ke^(-3x+2) from 0 to 1 = -1/3ke^-1 + 1/3ke^2

For f(x) to be a probability density function, the above expression must be equal to 1. Therefore,

-1/3ke^-1 + 1/3ke^2 = 1

Solving for k gives us k = 36

Therefore, by probability the answer will be 36.

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The probability that it is not raining and the flight leaves on time is 0.83 rounded to the nearest thousandth.

Let's use the formula for the probability of the intersection of two events to find the probability that it will rain and the flight will be delayed:

P(rain and delay) = 0.12

The probability of either rain or delay or both by using the formula for the probability of the union of two events:

P(rain or delay) = P(rain) + P(delay) - P(rain and delay)

Substituting the given probabilities, we get:

P(rain or delay) = 0.15 + 0.14 - 0.12

P(rain or delay) = 0.17

The probabilities of rain or delay or both add up to 0.17 can find the probability that it is not raining and the flight leaves on time by subtracting this value from 1:

P(not rain and on time) = 1 - P(rain or delay)

P(not rain and on time) = 1 - 0.17

P(not rain and on time) = 0.83

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Answer:

.125 = 3.25/l

.125l = 3.25, so l = 26 feet

.125 = 2.5/w

.125w = 2.5, so w = 20 feet

A = lw = (26 feet)(20 feet)

= 520 square feet

The correct answer is A.

All the correct statements are,

⇒ the absolute value a number is from 0 on a number line

⇒ The additive inverse of a number

⇒ Horizontal left and right

⇒ An integer number or its opposite

⇒ Straight vertical and down

⇒ A set of two opposite numbers that equal zero when combined.

We have to given that;

To fill the blanks in all the statement.

Hence, We get;

We know that;

⇒ The absolute value a number is from 0 on a number line

⇒ The additive inverse of a number

⇒ Horizontal left and right

⇒ An integer number or its opposite

⇒ Straight vertical and down

⇒ A set of two opposite numbers that equal zero when combined.

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The center and shape of the distribution are respectively:

9.6

Skewed left

The center of a distribution is a found using a statistics such as mean, median, or mode, and then provides a single value that is representative of the data. The spread is one that describes how close the data values are to each other using the range or standard deviation. The shape is one that describes how the data looks on a graph.

From the bar graph, we see that the center of the distribution is at 9.6.

Similarly, we also see that it is skewed left because it is longer on the left side of its peak than on its right.

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The factored function is given as follows:

[tex]x^4 + 2x^3 - 20x^2 + 64x - 32 = (x^2 + 6x - 4)(x^2 - 4x + 8)[/tex]

The function for this problem is defined as follows:

[tex]y = x^4 + 2x^3 - 20x^2 + 64x - 32[/tex]

The zeros are given as follows:

Hence the function is factored as follows:

[tex]x^4 + 2x^3 - 20x^2 + 64x - 32 = (ax^2 + bx + c)(x - 2 + 2i)(x - 2 - 2i)[/tex]

(we have the multiplication of two second degree polynomials resulting in a fourth degree polynomial, we must obtain the other second degree polynomial).

[tex]x^4 + 2x^3 - 20x^2 + 64x - 32 = (ax^2 + bx + c)(x^2 - 4x + 8)[/tex]

[tex]x^4 + 2x^3 - 20x^2 + 64x - 32 = ax^4 + (-4 + b)x^3 + \cdots + 8c[/tex]

(it is not necessary to make the calculations in the middle of the function as they are not needed to obtain the constants).

Hence the value of a is given as follows:

a = 1.

The value of b is given as follows:

-4 + b = 2

b = 6.

The value of c is given as follows:

8c = -32

c = -4.

Hence the factored expression is of:

[tex]x^4 + 2x^3 - 20x^2 + 64x - 32 = (x^2 + 6x - 4)(x^2 - 4x + 8)[/tex]

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The function r(t) for the line passing through the points p(8,3,3) and q(6,8,7) is r(t) = <8-2t, 3+5t, 3+4t>

We can use the vector form of the equation of a line to find a function r(t) for the line passing through the points p(8,3,3) and q(6,8,7).

Let's first find the direction vector of the line by subtracting the coordinates of the two points:

q - p = <6-8, 8-3, 7-3> = <-2, 5, 4>

Now, we can write the vector equation of the line in terms of a parameter t as:

r(t) = p + t(q - p)

Substituting the values of p and q, we get:

r(t) = <8, 3, 3> + t<-2, 5, 4>

Expanding, we get:

r(t) = <8-2t, 3+5t, 3+4t>

Therefore, the function r(t) for the line passing through the points p(8,3,3) and q(6,8,7) is:

r(t) = <8-2t, 3+5t, 3+4t>

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According to the statement the total probability is approximately 0.026, which is the second option in the list provided.

The probability that Brandon will earn a free homework pass can be found using the binomial probability formula. In this case, the number of trials (n) is 10, the probability of success (p) is 1/5 (since there are five answer choices per problem), and the probability of failure (q) is 4/5. We need to calculate the probability of getting at least 5 correct answers, meaning we will consider 5, 6, 7, 8, 9, and 10 correct answers.
The binomial probability formula is:
P(x) = C(n, x) * (p^x) * (q^(n-x))
Where P(x) is the probability of x successes, C(n, x) is the number of combinations of n items taken x at a time, and p and q are the probabilities of success and failure, respectively.
Calculating the probabilities for each of the desired outcomes (5 to 10 correct answers) and summing them up will give us the probability of Brandon earning a free homework pass. After doing the calculations, the total probability is approximately 0.026, which is the second option in the list provided.

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No, if A and B are disjoint, they cannot be independent.

Yes, A and B can be independent and disjoint.

Yes, A and B can be independent even if A is a subset of B.

No, if A and B are independent, A and A ⊂ B (A is a proper subset of B) cannot be independent.

Let's address each question separately:

1. If A and B are disjoint (mutually exclusive), meaning they cannot occur simultaneously, can they be independent?

No, if A and B are disjoint, they cannot be independent. The definition of independence states that the probability of the intersection of two independent events is equal to the product of their individual probabilities. Since A and B are disjoint, their intersection is empty, and the probability of an empty set is zero. Therefore, the condition for independence does not hold.

2. If A and B are independent, can they be disjoint?

Yes, A and B can be independent and disjoint. Disjoint events mean they have no common outcomes, while independent events mean that the occurrence of one event does not affect the probability of the other. Therefore, if A and B are independent, it is possible for them to be disjoint.

3. If A ⊂ B (A is a subset of B), can A and B be independent?

Yes, A and B can be independent even if A is a subset of B. The independence of events is determined by the conditional probabilities. In this case, if A ⊂ B, then the occurrence of A provides information about B. However, if the conditional probability of B given A is equal to the probability of B (P(B|A) = P(B)), then A and B can still be considered independent.

4. If A and B are independent, can A and A ⊂ B be independent?

No, if A and B are independent, A and A ⊂ B (A is a proper subset of B) cannot be independent. When A is a proper subset of B, the occurrence of A provides information about B. As a result, the probability of B given A, denoted as P(B|A), is affected by the knowledge that A has occurred. Therefore, A and A ⊂ B are not independent in this scenario.

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Answer:

£20.99

Step-by-step explanation:

£7 in U.S dollars is $8.80 meaning that each book was $4.40, so 4.40 x 6 is $26.4 so once it's converted it will be £20.99. Hope that helped:)

Answer:

To simplify the given expression -2wx² (7w^5-3w³x² +8x^4), we can use the distributive property of multiplication over addition/subtraction. First, we can distribute -2wx² to each term inside the parentheses:-2wx² * 7w^5 = -14w^6x²-2wx² * (-3w³x²) = 6w³x^4-2wx² * 8x^4 = -16wx^6\

Now, we can combine these simplified terms by adding or subtracting them based on their exponents:-14w^6x² + 6w³x^4 - 16wx^6. This is the final simplified form of the given expression.

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A quartic polynomials is more complicated than quadratic or cubic polynomials. There are no obvious linear or quadratic factors. It is possible that the polynomial x^4 + x^3 + 2x^2 + x + 2 is irreducible over F3.

The polynomial 2x³ + 2x² + 2 over the field F₃, we first notice that we can factor out a 2 from all three terms to obtain:

2(x³ + x² + 1)

Now we need to factor the polynomial x³ + x² + 1 over F₃. One way to do this is to simply plug in all possible values for x (which are 0, 1, and 2 in F₃) and see if any of them result in a zero polynomial. We find that none of them do, so we know that x³ + x² + 1 is irreducible over F₃.

Therefore, our final factorization of 2x³ + 2x² + 2 over F₃ is:

2(x³ + x² + 1)

(b) To factor the polynomial x⁴ + x³ + 2x² + x + 2 over F₃, we can start by plugging in all possible values for x and checking if any of them result in a zero polynomial. Doing so, we find that x = 1 is a root of the polynomial, which means that x - 1 is a factor. Using polynomial long division or synthetic division, we can divide x⁴ + x³ + 2x² + x + 2 by x - 1 to obtain:

x⁴ + x³ + 2x² + x + 2 = (x - 1)(x³ + 2x² + 4x + 2)

Now we need to factor the cubic polynomial x³ + 2x² + 4x + 2 over F₃. Again, we can plug in all possible values for x and check for roots, but we won't find any in this case. However, we can use the fact that the sum of the coefficients of the polynomial is zero (1 + 2 + 4 + 2 = 0 in F₃) to infer that x = 1 is a root mod 3, and therefore x - 1 is a factor. Dividing x³ + 2x² + 4x + 2 by x - 1 using polynomial long division or synthetic division, we obtain:

x³ + 2x² + 4x + 2 = (x - 1)(x² + 3x + 2)

Now we need to factor the quadratic polynomial x² + 3x + 2 over F₃. We can do this by factoring it as (x + 1)(x + 2), since (x + 1)(x + 2) = x² + 3x + 2 mod 3.

Therefore, our final factorization of x⁴ + x³ + 2x² + x + 2 over F₃ is:

(x - 1)(x + 1)(x + 2)²

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A matrix with orthonormal columns satisfies this condition and is therefore orthogonal. The statement is false. A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.

An orthogonal matrix is a square matrix whose columns and rows are orthonormal, which means they are orthogonal (perpendicular) to each other and have a magnitude of 1. If a matrix has orthonormal columns but is not square, it cannot be considered an orthogonal matrix.

The statement "A matrix with orthonormal columns is an orthogonal matrix" is false, and the correct answer is (B) - A matrix with orthonormal columns is an orthogonal matrix if the matrix is also square.

In summary, a matrix with orthonormal columns is not necessarily an orthogonal matrix. The statement is only true if the matrix is also square.

To explain, an orthogonal matrix is a square matrix where all columns (and rows) are orthonormal, meaning they are of unit length and orthogonal to each other. However, a matrix with orthonormal columns does not necessarily meet the requirements of being square and having orthonormal rows. In fact, a rectangular matrix with orthonormal columns cannot have orthonormal rows.

Therefore, the only way for a matrix with orthonormal columns to be an orthogonal matrix is if it is also square. This is because a square matrix has an equal number of rows and columns, which ensures that its columns and rows are orthonormal to each other, and hence it is an orthogonal matrix.

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There are different questions that can be asked regarding this scenario, but one common question is: what is the probability that all four chocolates have coconut filling?

To answer this question, we can use the hypergeometric distribution, which describes the probability of obtaining a certain number of "successes" (in this case, chocolates with coconut filling) in a sample of a given size (in this case, four) taken from a population of a given size (in this case, 16 chocolates with four of them having coconut filling), without replacement. The probability of getting four chocolates with coconut filling is then:

P(X = 4) = (4 choose 4) * (16 - 4 choose 0) / (16 choose 4) = 1/182

where "n choose k" denotes the number of ways of choosing k items from a set of n items, and the probability of each possible sample is the same. Therefore, the probability of all four chocolates having coconut filling is very low, only about 0.55%.

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The experimental probability that Sue will hit the bullseye on her next toss is 2/7.

We have,

The proportion of outcomes where a specific event occurs in all trials, not in a hypothetical sample space but in a real experiment, is known as the empirical probability, relative frequency, or experimental probability of an event.

Here, we have

Given: Sue is playing darts. So far, she has hit the bullseye 4 times and missed the bullseye 10 times.

We have to find the experimental probability that Sue will hit the bullseye on her next toss.

experimental probability = 4/(4+10) = 4/14 = 2/7

The next toss = P = 2/7

Hence,  the experimental probability that Sue will hit the bullseye on her next toss is 2/7.

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complete question:

Sue is playing darts. So far, she has hit the bullseye 4 times and missed the bullseye 10 times. What is the experimental probability that Sue will hit the bullseye on her next toss?

Answer:

113.1 cm³

Step-by-step explanation:

diameter = 2 X radius

Volume of sphere = (4/3) X π X r ³

= (4/3) π (3)³

= 36π

= 113.1 cm³ to nearest tenth

Answer:

x −4 −2 0

y 0 2 4

Step-by-step explanation:

:)

The equation 2cos(2x) + 1 = 0 has no solutions in the interval [0, π/2].

We can start by rearranging the equation:

2cos(2x) = -1

cos(2x) = -1/2

Since the cosine function has a maximum value of 1 and a minimum value of -1, the equation cos(2x) = -1/2 has solutions in the interval [0, π/2] if and only if -1/2 is between -1 and 1/2. However, this is not the case, so the equation has no solutions in the given interval.

To see this more clearly, we can use the inverse cosine function (also known as arccosine) to find the angles whose cosine is -1/2. Using a calculator or a table, we find that the two angles in the interval [0, π] whose cosine is -1/2 are π/3 and 5π/3. Since π/3 is less than π/2 and 5π/3 is greater than π/2, neither of these angles is in the interval [0, π/2]. Therefore, the equation 2cos(2x) + 1 = 0 has no solutions in this interval.

In summary, the equation 2cos(2x) + 1 = 0 has no solutions in the interval [0, π/2] because the cosine of any angle in this interval is greater than or equal to -1/2, which is not a solution to the equation.

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The association in this graph can best be described as C. Negative linear.

A negative linear association is one that moves from the left to the right. In this kind of association, the predictor increases while the response decreases. The linear nature of this association is seen in the straight line formed from the plot.

A positive linear association would fall from the right towards the left side and a non-linear association will form a curve. So, the association in the table is negative linear.

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The probabilities of hitting the next bat are  Mitchell = 5/12 and Travis= 9/20

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

Mitchell hits 5 out of 12 times

Travis hits 9 out of 20 times

The probabilities of hitting the next bat is calculated as

P(Hit) = Hit/Total number of times

using the above as a guide, we have the following:

P(Mitchell Hit) = 5/12

P(Travis Hit) = 9/20

Hence, the probabilities of hitting the next bat are 5/12 and 9/20

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if "e(θˆ)" shows the expected value of the unbiased estimator (θˆ), and b(θˆ) = 5, that shows that E[θˆ] = θ + 5, as the estimator (θˆ) consistently overestimates the accurate value of the parameter (θ) by 5 units.

An unbiased estimator is a statistical tool used to estimate a population parameter (like θ) by minimizing the difference between the estimator (θˆ) and the true value of the parameter.

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[tex]6\sqrt{5} \ cis(\frac{11\pi}{6}) \div 3\sqrt{6} \ cis (\frac{\pi}{2} )[/tex] can be expressed in polar form as[tex]6\sqrt{5} \ cis(\frac{11\pi}{6}) \div 3\sqrt{6} \ cis (\frac{\pi}{2} )=\underline{\frac{\sqrt{30} }{3} } \ cis\ (\underline{\frac{4\pi}{3}})[/tex] . Therefore the values to be dragged in the box are [tex]\frac{\sqrt{30} }{3}[/tex]  and [tex]\frac{4\pi}{3}[/tex].

We have to express in polar form, polar form of complex number:

[tex]r(cos\theta+isin\theta) \rightarrow rcis\theta[/tex]

where, r = modulus of complex number

[tex]\theta[/tex] = argument of complex number

The division of two complex number, [tex]z=[/tex] [tex]r_{1} cis \theta_{1}[/tex] and [tex]x=[/tex] [tex]r_{2} cis \theta_{2}[/tex]

[tex]\frac{z}{x} = \frac{r_{1} }{r_{2} }\ cis(\theta_{1}- \theta_{2})[/tex]

Similarly, let a = [tex]6\sqrt{5} \ cis(\frac{11\pi}{6})[/tex]

                    b = [tex]3\sqrt{6} \ cis(\frac{pi}{2})[/tex]

[tex]\frac{a}{b}= \frac{6\sqrt{5} }{3\sqrt{6} } \ cis (\frac{11\pi }{6}- \frac{\pi}{2} )[/tex]

 [tex]= \frac{{\sqrt{2}}\times\sqrt{2}\times\sqrt{5} }{\sqrt{2} \times\sqrt{3} } \ cis(\frac{11\pi-3\pi}{6} )[/tex]

 [tex]=\sqrt{\frac{10}{3} }\ cis\ \frac{8\pi}{6}[/tex]

[tex]\frac{a}{b}= \sqrt{\frac{10}{3} } \ cis\ \frac{4\pi}{3}[/tex]

It can also be written as, [tex]\frac{a}{b}= \frac{\sqrt{30} }{3} \ cis\ \frac{4\pi}{3}[/tex]

⇒ [tex]6\sqrt{5} \ cis(\frac{11\pi}{6}) \div 3\sqrt{6} \ cis (\frac{\pi}{2} )= \frac{\sqrt{30} }{3} \ cis\ \frac{4\pi}{3}[/tex]

Comparing it with the question we get:

[tex]6\sqrt{5} \ cis(\frac{11\pi}{6}) \div 3\sqrt{6} \ cis (\frac{\pi}{2} )=\underline{\frac{\sqrt{30} }{3} } \ cis\ (\underline{\frac{4\pi}{3}})[/tex]

Therefore, the first blank is [tex]\frac{\sqrt{30} }{3}[/tex] and the second blank is [tex]\frac{4\pi}{3}[/tex].

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