A. Write true or false after each sentence. If the sentence is
false, change the Capitalization word or words to make it true.

1. In the expression 7x + 15, 15 is a COEFFICIENT .

2. 3x + 7 means (3x + 7) DIVIDED BY 2

3. You can rewrite 2(4 + 8) as (2)(4) + (2)(8) using the DISTRIBUTIVE PROPERTY.

Thee factory is able to provide 94.38 boxes per hour to the warehouse.

Rate = (2000 - 1245) / 8 = 94.38 boxes per hour

Therefore, the factory is able to provide 94.38 boxes per hour to the warehouse.

Boxes added in 40 hours = rate * time = 94.38 * 40 = 3,775.2

Therefore, the inventory at the end of a 40-hour work week would be:

1245 + 3775.2 = 5020.2 boxes

rate = (50000 - 1245) / time

Simplifying this equation, we get:

time = (50000 - 1245) / rate = 511.64 hours (rounded to two decimal places).

Therefore, it will take approximately 511.64 hours to fill the warehouse to its 50,000 box capacity,

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There are 55,440 different ways for the coach to select and order the 5 players for penalty kicks.

A soccer team has 11 players on the field at the end of a scoreless game. According to league rules, the coach must select 5 of the players and designate an order in which they will take penalty kicks. To determine the number of different ways the coach can do this, you need to calculate the number of permutations of 11 players taken 5 at a time. This can be calculated using the formula:

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

Where n = 11 (total players) and r = 5 (players to be selected).

P(11, 5) = 11! / (11-5)!

P(11, 5) = 11! / 6!

P(11, 5) = 39,916,800 / 720

P(11, 5) = 55,440

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Yes, it would be unusual if less than 52% of the sampled teenagers owned smartphones.

We are given the mean (μp) as 0.55 and the standard deviation (σp) as 0.0397. We need to find the probability of having less than 52% (0.52) of teenagers owning smartphones.

1) Calculate the z-score.
z = (x - μp) / σp
z = (0.52 - 0.55) / 0.0397
z ≈ -0.76

2) Find the probability associated with the z-score.
Using a z-table or a calculator, we find that the probability of having a z-score less than -0.76 is approximately 0.224. This means there is a 22.4% chance that less than 52% of the sampled teenagers would own smartphones.

Since the probability of having less than 52% of the sampled teenagers owning smartphones is 22.4%, it would be considered unusual, as the probability is relatively low.

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Huffman coding is a lossless data compression algorithm that assigns variable-length codes to characters in a message based on their frequency of occurrence. It was invented by David A. Huffman in 1952.

The algorithm works by creating a binary tree of nodes, where each node represents a character and its frequency of occurrence. The two nodes with the lowest frequencies are then combined into a single node, with a weight equal to the sum of the two frequencies. This process is repeated until all the nodes have been combined into a single tree.

The resulting tree is then traversed to assign unique binary codes to each character. The left branches of the tree are assigned the binary value 0, and the right branches are assigned the binary value 1. The binary code for a character is obtained by concatenating the binary values assigned to the branches on the path from the root to the node representing that character.

The advantage of Huffman coding is that it produces variable-length codes that are more efficient than fixed-length codes, since frequently occurring characters are assigned shorter codes. This leads to significant compression of data, especially in cases where certain characters or symbols occur much more frequently than others.

Let's address each part step-by-step:

1. Show that Huffman coding is uniquely decipherable:
Huffman coding is uniquely decipherable because it is a prefix code. A prefix code is a type of variable-length code in which no codeword is a prefix of another codeword. This means that, when reading a message encoded with a prefix code, you can always identify the correct symbol as soon as you read the corresponding codeword. Since Huffman coding constructs a prefix code, it is uniquely decipherable.

2. Show that Huffman coding is instantaneous:
A code is considered instantaneous if it can be decoded without having to look at future symbols in the message. Since Huffman coding is a prefix code, it is also instantaneous. As mentioned earlier, with a prefix code, you can always identify the correct symbol as soon as you read the corresponding codeword, meaning you don't need to wait for future symbols to decode the message. Therefore, Huffman coding is instantaneous.

3. Show that Huffman coding is not unique:
Huffman coding is not unique because the order in which the nodes are merged during the construction of the Huffman tree can be different, leading to different codes. When constructing a Huffman tree, the algorithm starts by creating a node for each symbol and assigning it a frequency. It then iteratively merges the two nodes with the lowest frequencies until only one node, the root of the tree, remains. However, if two or more nodes have the same frequency, the algorithm can choose to merge them in any order. This can result in different Huffman trees and thus different codes, which demonstrates that Huffman coding is not unique.

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A survey does not help capture:

(b) Everything a population knows.

A survey is a research method used to collect data from a sample of individuals or population through a series of standardized questions or measures, typically conducted through a questionnaire, interview, or online form. Surveys are commonly used in social science, marketing research, and other fields to gather information on a range of topics such as attitudes, opinions, behaviors, and demographics.

While surveys can provide information on knowledge, behaviors, and attitudes, they may not be able to capture the full perspective of individuals or their experiences. Surveys are limited by the questions asked and the way in which they are designed, so they may not always capture the nuances and complexities of a population's beliefs and experiences.

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In any planar triangulation, there are always fewer edges than three times the number of vertices, so the average degree of a vertex must be less than 6.

Let V stand for the triangulation's collection of vertices and E for its set of edges. As each edge adds two degrees to the total degree count, the triangulation's total degree count is equal to twice the number of edges. Thus,

Σdeg(v) = 2|E| where deg(v) is the degree of vertex v and |E| is the number of edges in the triangulation.

|E| = (3/2) |T|, number of triangles in the triangulation is |T| .

Furthermore, we know that the sum of the degrees of the vertices is equal to 3 times the number of triangles, since each triangle contributes 3 to the total degree count:

Σdeg(v) = 3|T|

Putting these equations together, we have:

Σdeg(v) = 3|T| = (3/2) * 2|E| = 3|E|

Dividing both sides by the number of vertices, n, we obtain:

(1/n) Σdeg(v) = 3/ n * |E|

Thus, the average degree of a vertex in the triangulation is strictly less than 6, since the average degree of a vertex in the corresponding graph is at most 2 (since each triangle is incident to at most 3 other triangles).

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The solution to the initial boundary value problem is u(x,t) = 1/π.

To solve the initial boundary value problem ut = 2uxx for x ∈ (-π, π], t ∈ [0, + [infinity]), ux(0, t) = ux,(1,t) = 0, for t ∈ [0, + [infinity]), u(x,0) = π^2 – π^2 for r ∈ [ -π, π], we can use the method of separation of variables.

Assume u(x,t) = X(x)T(t), then we have:

X''(x) + λX(x) = 0, T'(t) + 2λT(t) = 0

where λ is a separation constant. The general solution for the spatial equation is X(x) = A sin(nx) + B cos(nx), where n = sqrt(λ) and A, B are constants. Since u(0,t) = u(1,t) = 0, we have A = 0 and B cos(nπ) = 0, which implies n = kπ for k = 1, 2, 3, ... Thus, the spatial eigenfunctions are X_k(x) = cos(kπx), and the corresponding eigenvalues are λ_k = -(kπ)^2.

The time equation can be solved as T(t) = Ce^(-2λ_k t), where C is a constant. Therefore, the general solution for the initial boundary value problem is:

u(x,t) = Σ C_k cos(kπx) e^(-2(kπ)^2 t)

where the sum is taken over all k = 1, 2, 3, .... To determine the constants C_k, we use the initial condition u(x,0) = π^2 – π^2 = 0. This gives:

Σ C_k cos(kπx) = 0

Since the eigenfunctions form an orthogonal set on [-π, π], we can multiply both sides by cos(mπx) and integrate over [-π, π] to obtain:

C_m = 0 for m = 1, 2, 3, ...

Thus, the only non-zero constant is C_0, which can be determined using the normalization condition:

1 = ∫_(-π)^π (u(x,t))^2 dx = C_0^2 π^2

Therefore, C_0 = 1/π. Thus, the solution to the initial boundary value problem is:

u(x,t) = (1/π) cos(0πx) e^(-2(0π)^2 t) = 1/π e^0 = 1/π

In conclusion, the solution to the initial boundary value problem is u(x,t) = 1/π.

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Answer:

The range of a set of data is the difference between the highest and lowest values. In this case, the highest value is 98 and the lowest value is 65, so the range is 98-65 = 33. This means that the scores on the test ranged from 65 to 98, a difference of 33 points.

The range is a measure of the spread of the data. In this case, the range is relatively large, which means that the scores were spread out over a wide range of values. This suggests that the test was challenging and that there was a wide range of student abilities.

For the first question, we need to assign truth values to q, r, and s such that the pair of statements are not equivalent. For (a) sv(sq) and svq, we can assign q = True, r = False, and s = False. This makes sv(sq) True and svq False, thus showing that the two statements are not equivalent. For (b) (s19)►r and (-84-9)vr, we can assign q = False, r = True, and s = False. This makes (s19)►r False and (-84-9)vr True, thus showing that the two statements are not equivalent.

For the second question, we can construct the compound proposition as follows: (a∧b)∨(a∧c)∨(a∧d)∨(b∧c)∨(b∧d)∨(c∧d). This proposition is true precisely when at least two of the variables a, b, c, and d are true. We can see that this is the case because for the proposition to be true, at least two of the terms in the disjunction need to be true, each of which represents the case where at least two variables are true. For example, (a∧b) represents the case where both a and b are true, and (a∧c) represents the case where both a and c are true, and so on. Therefore, the given compound proposition satisfies the condition of being true precisely when at least two of the variables are true.

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An inequality representing v, the number of visits Nayeli needs to make to get a free movie ticket is 40 + 14.5v ≥ 185.

Inequality describes a mathematical statement that states that two or more algebraic expressions are unequal.

Inequalities are represented as:

The rewards points Nayeli has just for signing up = 40

The points earned per visit to the movie theater = 14.5

The total number of points required for a free movie ticket ≥ 185

Let the number of visits Nayeli needs to make too get a free movie ticket = v

40 + 1.45v ≥ 185

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The temperature of the object at time t = 70 is approximately 93.26 degrees.

To solve the problem, we can use the solution to the differential equation given by equation 2.15:

u(t) = [tex]Ce^[/tex](-kt) + A,

where C is a constant that we need to determine from the initial condition u(0) = 120. Substituting t = 0 and u(0) = 120 into the equation, we get:

120 = Ce^(-k*0) + A

120 = C + A

Next, we need to determine the value of C using the information that at t = 50, the temperature of the object is 100 degrees:

100 = Ce^(-k*50) + 90

10 = Ce^(-2.5)

Solving for C, we get:

C = 10/e^(-2.5)

C ≈ 14.868

Now we can use the value of C and equation 2.15 to find the temperature of the object at t = 70:

u(70) = 14.868e^(-0.05*70) + 90

u(70) ≈ 93.26 degrees

Therefore, the temperature of the object at time t = 70 is approximately 93.26 degrees.

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a) The critical value for a two-tailed test with a significance level of 0.01 and 11 degrees of freedom is 3.11 (found using a t-distribution table).

b) The test statistic for comparing two standard deviations is the F-statistic. The formula for calculating it is [tex]F = \frac{s1^2}{s2^2}[/tex], where s1 is the sample standard deviation of the first group (base product), s2 is the sample standard deviation of the second group (modified product), and the larger standard deviation is always in the numerator. Using the sample data given, we find:
s1 = 7 cm (from the base product)
s2 = 6.5 cm (from the modified product)
[tex]F = \frac{(7)^{2} }{(6.5)^{2} }  = 1.223[/tex]

c) To determine if there is a difference between the standard deviations, we compare the calculated F-statistic to the critical value we found in part a. Since our calculated F-value (1.223) is less than the critical value (3.11), we fail to reject the null hypothesis. Therefore, we conclude that there is not enough evidence to suggest that there is a significant difference between the standard deviations of the two products.

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The assumption that P(Xi < 6) ≥ d for some 8 € (0, 1), we can show that Var(Xi) ≤ 6^2 - (6d)^

(a) To prove that P(ΣIXI0 for all t > 0, we can use Markov's inequality, which states that for any non-negative random variable Y and any positive constant a, we have:

P(Y ≥ a) ≤ E(Y)/a

Let Y = e^(tΣIXi) and a = e^t. Then we have:

P(ΣIXi ≥ t) = P(e^(tΣIXi) ≥ e^t) ≤ E(e^(tΣIXi))/e^t

Now, we need to show that E(e^(tΣIXi)) ≤ e^(t^2/2). To do this, we can use the fact that for any independent random variables Y1, Y2, ..., Yn, we have:

E(e^(t(Y1+Y2+...+Yn))) = E(e^(tY1)) E(e^(tY2)) ... E(e^(tYn))

Uszng this formula and the assumption that P(Xi < 6) ≤ 6 for all 6 € (0, 1), we get:

E(e^(tXi)) = ∫₀^₆ e^(tx) fXi(x) dx ≤ ∫₀^₆ e^(6t) fXi(x) dx = e^(6t) E(Xi)

Therefore, we have:

E(e^(tΣIXi)) = E(e^(tX1) e^(tX2) ... e^(tXn)) ≤ E(e^(6t)X1) E(e^(6t)X2) ... E(e^(6t)Xn) = (E(X1) e^(6t))^(n)

Since Xi is non-negative, we have E(Xi) = ∫₀^₆ fXi(x) dx ≤ 1, so we get:

E(e^(tΣIXi)) ≤ (e^(6t))^n = e^(6nt)

Finally, substituting this inequality into the earlier expression, we get:

P(ΣIXi ≥ t) ≤ E(e^(tΣIXi))/e^t ≤ (e^(6nt))/e^t = e^(6n-1)t

Since this inequality holds for all t > 0, we have:

P(ΣIXi ≥ 0) = lim t→0 P(ΣIXi ≥ t) ≤ lim t→0 e^(6n-1)t = 1

Therefore, we have shown that P(ΣIXi ≥ 0, as required.

(b) To prove that P(ΣIXi ≥ t) ≥ 1 - ne^(-2t^2/d^2) for all t > 0, we can use Chebyshev's inequality, which states that for any random variable Y with finite mean and variance, we have:

P(|Y - E(Y)| ≥ a) ≤ Var(Y)/a^2

Let Y = ΣIXi and a = t. Then we have:

P(|ΣIXi - E(ΣIXi)| ≥ t) ≤ Var(ΣIXi)/t^2

Now, we need to find an upper bound for Var(ΣIXi). Since the Xi are independent, we have:

Var(ΣIXi) = Var(X1) + Var(X2) + ... + Var(Xn)

Using the assumption that P(Xi < 6) ≥ d for some 8 € (0, 1), we can show that Var(Xi) ≤ 6^2 - (6d)^

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Answer: Use the table to identify values of p and q that can be used to factor x2 + x – 12 as (x + p)(x + q).A. –2 and 6. B. 3 and –4. C. –3 and 4. D. 2 and –6.

Step-by-step explanation:

Algebra is used to solve the mathematical problems, the total amount spent by Andy on lunches in this week is equals to $195.

Algebra is the branch of mathematics that use in the representation of problems or situations in the form of mathematical expressions. Mathematical ( arithmetic) operations say multiplication (×), division (÷), addition (+), and subtraction (−) are used to form a mathematical expression.

We have, a data of amount spent by Andy on lunches in a week. Let the total amount spent by him in this week be "x dollars". Using algebra of mathematics, we can written as x = sum of amounts spent by him in whole week so, x = $50 + $20 + $10 + $25 + $25 + $15 + $50 = $195

Hence, required total amount value is $195.

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Complete question:

Andy spent the following amounts on lunches this week,

day. amount

Sunday $50

Monday. $20

Tuesday $10

Wednesday $25

Thursday $25

Friday $15

Saturday $50

Calculate total amount he spent in this week.

X1 = U1 + U-1 = U1 (since U-1 = 0);
X2 = U2 + U0 = U2 + 1 (since U0 = y1 - y0 = -1);
X3 = U3 + U1 = U3 + U1;
X4 = U4 + U3 + U2.

Innovations algorithm is a method for computing the values of a time series at each time point using the lag operator and the innovation or error terms. The lag operator, denoted by the symbol B, is used to shift the time index of a time series by one unit. Specifically, B multiplied by the time series X is equivalent to the time series X shifted one unit into the past.

We are given the MA(2) process X, = (1+0,B²)W with W, - WN(0,1), this means that X is a moving average process of order 2, where the current value of X depends on the current and two past innovation terms. The innovation terms U1, U2, U3, ... are uncorrelated random variables with zero mean and unit variance.

i) To use the innovations algorithm to find X1, X2, and X3 in terms of U1, U2, and U3, we first need to express X in terms of B and the innovation terms. Using the definition of the MA(2) process, we have:

Xt = (1 + 0*B^2)Wt + B* (1 + 0*B^2)Wt-1 + B^2* (1 + 0*B^2)Wt-2
= (1 + 0*B^2)Wt + B* (1 + 0*B^2)Wt-1 + B^2* (1 + 0*B^2)Wt-2
= Wt + B*Wt-1 + B^2*Wt-2

Next, we can use the innovations algorithm to express Xt in terms of the innovation terms U1, U2, and U3. The algorithm works as follows:

- Compute the initial values of the time series using the given initial conditions. In this case, we have y(0)=1+0*2, y(1)=0, y(2)=0, and y(h)=0 for h>3.
- For each time point t > 2, compute the innovation term Ut as the difference between the observed value yt and the predicted value based on the past observations up to time t-1. Specifically, we have:
Ut = yt - (1 + 0*B + 0*B^2) yt-1
= yt - yt-1

- Compute the current value of Xt as a linear combination of the past innovation terms Ut-1, Ut-2, and Ut-3, using the coefficients from the MA(2) process. Specifically, we have:
Xt = Ut + 0*Ut-1 + Ut-2
= Ut + Ut-2

Using this algorithm, we can compute the values of X1, X2, and X3 as follows:
- X1 = U1 + U-1 = U1 (since U-1 = 0)
- X2 = U2 + U0 = U2 + 1 (since U0 = y1 - y0 = -1)
- X3 = U3 + U1 = U3 + U1

ii) To find the value of X4, we can use the result from part i to express X4 in terms of the innovation terms U1, U2, U3, and U4. Specifically, we have:

X4 = U4 + U2
= y4 - y3 + y2 - y1
= (1 + 0*B^2) W4 - (1 + 0*B + 0*B^2) W3 + (1 + 0*B^2) W2 - (1 + 0*B + 0*B^2) W1
- (1 + 0*B + 0*B^2) W0
= W4 - W3 + W2 - W1
= U4 + U3 + U2

Therefore, X4 = U4 + U3 + U2.

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The solutions to the equation 12x²-22x-5=-5x using the quadratic formula are x = 5/4 and x = -1/3.

The given equation is a quadratic equation in standard form, ax² + bx + c = 0, where a = 12, b = -22, and c = -5 + 5 = 0. We can use the quadratic formula to solve for x:

x = (-b ± √(b² - 4ac)) / 2a

Substituting the values of a, b, and c, we get:

x = (22 ± √(22² - 4(12)(0))) / 2(12)

Simplifying under the square root, we get:

x = (22 ± √484) / 24

x = (22 ± 22) / 24

Simplifying further, we get:

x = 44/24 or x = 0/24

Reducing the first fraction, we get:

x = 11/6 or x = 0

However, we need to check if x = 0 satisfies the given equation or not. Substituting x = 0, we get:

12(0)² - 22(0) - 5 = -5(0)

-5 = 0

This is not true, so x = 0 is an extraneous solution and should be discarded. Therefore, the solutions to the given equation are:

x = 5/4 and x = -1/3.

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To determine a condition which must be true if 4x, x, y is in V, we can first write 4x, x, y as a linear combination of the vectors in the given span.

Let's call the vectors in the span a and b, where:

a = [2, 3, 1]
b = [3, 3, 5]

Then, we want to find scalars s and t such that:

4x, x, y = sa + tb

In other words, we want to solve the system of equations:

2s + 3t = 4x
3s + 3t = x
s + 5t = y

We can solve this system by row reducing the augmented matrix:

[2 3 | 4x]
[3 3 | x]
[1 5 | y]

Using elementary row operations, we can obtain the following row echelon form:

[2 3 | 4x]
[0 -3/2 | -5x/2]
[0 0 | z]

where z = y + x + 2x/3.

So, for 4x, x, y to be in V, the system of equations must have a solution, which means that z = 0. Therefore, the condition that must be true is:

y + x + 2x/3 = 0

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The average temperature change per hour should be represented by the value such as = 3.2 °f /hr

The temperature decrease of 20.8f° = 6.5 hrs

The decrease of temperature of xf° = 1 hr

Mathematically;

20.8°f = 6.5 hrs

X °f = 1 HR

Make X the subject of formula;

X = 20.8/6.5

= 3.2 °f /hr

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7) Given the extreme values in a set of data as 5 and 19, the truth about the data set is C. The range is 14.

8) The mean of the data set 5, 9, 15, 18, 22 is B. 14.

9) The mode for the following set of data, 4, 5, 5, 6, 7, 7, 8, 12, is A. there is none.

The range is the difference between the extreme values of a data set.

This difference is computed by subtracting the minimum value from the maximum value.

The mean represents the average value of a data set, computed by dividing the total value by the number of items.

The mode is one value in the data set that occurs most.  There cannot be more than one mode in a data set.

Range between 5 and 19 = 14 (19 - 5)

Mean of 5, 9, 15, 18, 22 = 13.8 (69 ÷ 5) = 14

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The probability of a point chosen uniformly at random in the region [tex]-1\leq x\leq1[/tex] and [tex]-1\leq y\leq1[/tex]  to be classified as '+' is [tex](\frac{4 - A_{minus}}{4})[/tex].

To draw the Voronoi tile for the given data set and find the probability of a point being classified as '+', follow these steps:

1. Plot the data points: Plot the points (-1,-1), (-1,1), (1,-1), (1,1), and (0,0) on a graph. Label (0,0) as '-' and the other points as '+'.

2. Construct the Voronoi diagram:

For each pair of neighboring '+' points, draw a line that is equidistant from both points and bisects the line connecting them. These lines will divide the space into regions called Voronoi tiles, where each tile contains one data point, and any point within that tile is closer to the data point it contains than to any other data point.

3. Identify the tile containing the '-' point:

In this case, the Voronoi tile surrounding (0,0) will be the region that is closer to the '-' point than to any '+' point.

4. Calculate the area of the Voronoi tile containing the '-' point:

Since we are considering the region [tex]-1\leq x\leq1[/tex] and [tex]-1\leq y\leq1[/tex], find the area of the intersection of this region with the Voronoi tile containing the '-' point.

5. Calculate the total area of the considered region: The total area of the considered region is

(-1 to 1)(-1 to 1) = 2(2) = 4 square units.

6. Determine the probability of a point being classified as '+':

The probability of a point chosen uniformly at random in the considered region being classified as '+' is equal to the ratio of the area of the region not covered by the '-' Voronoi tile to the total area of the considered region.

Let's say the area of the '-' Voronoi tile is [tex]A_{minus}[/tex]. Then, the probability of a point being classified as '+' is [tex](\frac{4 - A_{minus}}{4})[/tex].

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[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=96 \end{cases}\implies V=\cfrac{\pi r^2 (96)}{3} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~\hspace{9em}\stackrel{\textit{since we know both Volumes are equal}}{\cfrac{4\pi r^3}{3}~~ = ~~\cfrac{\pi r^2 (96)}{3}}[/tex]

[tex]4\pi r^3=\pi r^2(96)\implies 4\pi r^2\cdot r=\pi r^2(96)\implies r=\cfrac{\pi r^2(96)}{4\pi r^2}\implies \boxed{r=24} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{ \textit{\LARGE cone} }{\cfrac{\pi (24)^2(96)}{3}}\implies \stackrel{ \textit{\LARGE sphere} }{\cfrac{4\pi (24)^3}{3}}\implies18432\pi ~~ \approx ~~ \text{\LARGE 57905.84}~units^3[/tex]

The solution to the given set of equations is Infinite solutions.

We have the equation

R: -3y = -3x -9

S: y = x + 3

Now, solving the equation R and S as

-3y = -3x - 9

3y =  3x + 9

_________

   0 = 0 + 0

   0 = 0

Also, -3/1 = 3/(-1) = 9/(-3)

-3/1 = -3/1 = -3/1

Thus, the equation have Infinite many solutions.

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The graph of a Square ABCD with vertices A(-7,5) B(-4,7) C(-2,4) and D(-5,2) is represent upper square and after 90° counterclockwise rotation the lower square represents ABCD in above figure.

A quadrilateral is a polygon that has number of four sides. This also implies that a quadrilateral has exactly four vertices, and exactly four angles. We have to graph a square with vertices A(-7,5), B(-4,7) ,C(-2,4) and D(-5,2) 90 counterclockwise. Now, steps to draw the square :

In case of rotating a figure of 90 degrees counterclockwise, each point of the figure has to be changed from (x, y) to (-y, x) and graph the rotated figure. So, now the vertices of square be A(-7,5), B(-4,7) ,C(-2,4) and D(-5,2). Now, draw the square for these point, lower square in above figure. Both graphs of square ABCD present in above figure.

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Complete question:

The above figure complete the question.

graph and label 9 and 10 and their given rotation about the origin. Give the coordinates of the images.

Square ABCD with vertices A(-7,5) B(-4,7) C(-2,4) and D(-5,2) 90 counterclockwise.

The possible outcomes of the experiment is {RR, BB, GG, WW, BB, RW, RB, RG, RW, RG, WB, WG}

The sample space for the experiment gave:

{RR, BB, GG, WW, BB, RW, RB, RG, RW, RG, WB, WG}

where each element of the set represents a different pair of socks, and the first letter represents the colour of the sock on the left foot and the second letter represents the colour of the sock on the right foot.

The possible outcomes of the experiment are the elements of the sample space, which are the different pairs of socks that can be selected. For example, selecting the red socks would be represented by the outcome RR, selecting the blue and white socks would be represented by the outcome BW, and so on.

Recall that

Probability = number of outcomes/total number of outcomes

Then, the probability of selecting a blue pair of socks will be:

P(blue) = number of outcomes with blue socks / total number of outcomes

Since there are only two outcomes with blue socks (BB and WB), then:

P(blue) = 2/12 = 1/6

P(green) = number of outcomes with green socks / total number of outcomes

P(green) = 2/12 = 1/6

P(not red) = number of outcomes without red socks / total number of outcomes

P(not red) = 10/12 = 5/6

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The approximate probability that the randomly selected point will lie inside the square is,

≈ 13.3%

Since, Area of square with side of 5 mm is:

A = a² = (5 mm)² = 25 mm²

Now, Find total area of the figure:

A(total) = A(trapezoid) + A(triangle)

A(total) = (b₁ + b₂)h/2 + bh/2

A(total) = (14 + 18)(17 - 12)/2 + 18 x 12/2

           = 80 + 108 = 188

Hence, Find the percent value of the ratio of areas of the square and full figure, which determines the probability we are looking for:

= 25/188  x 100%

= 13.2978723404 %

≈ 13.3%

Thus,  the approximate probability that the randomly selected point will lie inside the square is,

≈ 13.3%

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Answer:

Step-by-step explanation:

You want q and the probability of 3 successes in 8 trials if the probability of success is 0.45.

The value designated q is the complement of p, the probability of success.

  q = 1 -p

  q = 1 -0.45

  q = 0.55

The probability of 3 successes is ...

  P(3 of 8) = 8C3·p^3·q^(8-3) = 56·0.45^3·0.55^5

  P(3 of 8) ≈ 56°0.091125·0.050328 ≈ 0.256826

The probability of exactly 3 successes in 8 trials is about 0.2568.

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The angle is 86.8 degrees and its complement is 3.2 degrees.

let x be the angle and y be the Complementary angle.

If the angles are complementary, then their sum is 90 degrees.

x + y = 90................(1)

and, the angle measures 83.6 degrees more than its complement.

x = y + 83.6

y + 83.6 + y = 90

Solving the equation for y we get

2y + 83.6 = 90

2y = 90 - 83.6

2y = 6.4

y= 3.2

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The probability that a randomly selected adult has an IQ greater than 134.4 is 0.025.

We have,

To solve this problem, we need to standardize the IQ score using the

z-score formula:

z = (x - μ) / σ

where x is the IQ score, μ is the mean, and σ is the standard deviation.

Substituting the values we have:

z = (134.4 - 101.1) / 17

z = 1.96

We can then use a standard normal distribution table or a calculator to find the probability of a z-score being greater than 1.96.

Using a standard normal distribution table, we find that the probability of a z-score being greater than 1.96 is approximately 0.025.

Therefore,

The probability that a randomly selected adult has an IQ greater than 134.4 is 0.025.

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