Given m∠ADC= 118°, find m∠ADB.

The given statement "the function g(x)=x3 is a linear transformation from r into r." is false. . A linear transformation must satisfy additivity and homogeneity properties, but the function g(x)=x^3 does not satisfy homogeneity, making it nonlinear.

A linear transformation from R into R is a function that satisfies two properties

Additivity T(u + v) = T(u) + T(v) for all u, v in R

Homogeneity T(cu) = cT(u) for all u in R and all scalars c.

However, the function g(x) = x^3 is not linear because it violates the second property, homogeneity. Specifically, g(cx) = (cx)^3 = c^3x^3, which is not equal to cg(x) = c*x^3, unless c = 1. Therefore, g(x) = x^3 is a nonlinear transformation from R into R.

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--The given question is incomplete, the complete question is given

" the function g(x)=x3 is a linear transformation from r into r. True or false"--

The conditions for inference for a one-proportion z test are met.

Yes, the conditions for inference for a one-proportion z test are met.

The standard women's clothing sizes are designed to fit women between 64 and 68 inches in height.

A dress designer and manufacturer wants to produce clothing so that at least 60% of women clients are covered in this range.

A random sample of 50 of their regular clients had 34 of them with heights between 64 and 68 inches.

A proportion is used to describe the number of times an event occurs in a specified number of trials.

A proportion test is used to test if two proportions are equal or if a single proportion is equal to a specified value.

The test statistic for a one-proportion z test is given by the formula

\(z = \frac{{\hat p - p}}{{\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }}\\\)

where

\(\hat p = \frac{x}{n}\)

is the sample proportion, x is the number of successes, n is the sample size, and p is the hypothesized proportion.

The conditions for inference for a one-proportion z test are:

1. Independence: Sample observations should be independent.

2. Sample size: The sample size should be sufficiently large (n ≥ 10).

3. Success-failure condition: Both np and n(1 - p) should be greater than or equal to 10.

Provided that the sample observations are independent and that the sample size is sufficiently large, the success-failure condition is satisfied by

\($$np = 50 \cdot 0.6 = 30$$\)

\($$n\left( {1 - p} \right) = 50 \cdot 0.4 = 20$$\)

Since both np and n(1 - p) are greater than or equal to 10,

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Explanation:

At the top of the proof, your teacher states "Given" followed by the relevant useful info. We simply repeat that info word for word as statement 1. This is how all proofs start out. We start with what we're given and try to build toward what we want to prove.

So this is why reason 1 is "given". Reason 2 is "given" also. When your teacher gives multiple statements like this, it's handy to break them up into the smallest pieces possible.

--------------------

For reason 3, we use the SAS Similarity rule here because we have congruent angles LPN and LNM (both 90 degrees each) and the sides are proportional due to the statement \(\frac{PN}{NM} = \frac{LP}{LN}\)

Notice how statement 3 builds entirely off from what statements 1 and 2 set up.

(a) Find the dual of the LP.Primal problem isminimize \($b_1y_1+C_1x_1+...+C_nx_n$\) subject to \($a_{11}x_1+a_{12}x_2+...+a_{1n}x_n \leq\) \(b_1$...$a_{m1}x_1+a_{m2}x_2+...+a_{mn}x_n \leq b_m$ and $x_1, x_2,\)..., x_n\(\geq 0$\)

Let us find the dual of the above primal problem.

Dual problem ismaximize \($b_1y_1+...+b_my_m$\)subject to \($a_{11}y_1+a_{21}y_2+...+a_{m1}y_m \leq\)\(C_1$...$a_{1n}y_1+a_{2n}y_2+...+a_{mn}y_m \leq C_n$\)

and\($y_1, y_2, ..., y_m \geq 0$\)

(b) Find the standard form of the LP and dual.Standard form of the primal problem isminimize \($b_1y_1+C_1x_1+...+C_nx_n$\)subject to \($a_{11}x_1+a_{12}x_2+...+a_{1n}x_n +s_1 = b_1$...$a_{m1}x_1+a_{m2}x_2+...+a_{mn}x_n +s_m = b_m$\) and\($x_1, x_2, ..., x_n, s_1, s_2, ..., s_m \geq 0$\)

Standard form of the dual problem ismaximize \($b_1y_1+...+b_my_m$\)subject to \($a_{11}y_1+a_{21}y_2+...+a_{m1}y_m \leq 0$...$a_{1n}y\)

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Answer:

2067in

Step-by-step explanation:

19x26= 494

13x17=221

26x26x2=1352

494+221+1352=2067

a. The polynomials are given as follows:

1. 2x^4 + 18x³ - 4.

2. -8x³ + 4x + 2.

3. 2x² - x - 10

4. 3x³ + 10x² + 9x + 2.

b. The resulting degrees are given as follows:

1. 4.

2. 3.

3. 2.

4. 3.

c. The end behavior of each polynomial is given as follows:

1. As x -> -∞, f(x) -> ∞, as x -> ∞, f(x) -> ∞.

2. As x -> -∞, f(x) -> ∞, as x -> ∞, f(x) -> -∞.

3. As x -> -∞, f(x) -> ∞, as x -> ∞, f(x) -> ∞.

4. As x -> -∞, f(x) -> -∞, as x -> ∞, f(x) -> ∞.

The polynomials are obtained doing the operations and then combining the like terms.

Hence:

(6x³ + 3x² - 4) + (2x^4 - 3x² + 12x³)

2x^4 + (6 + 12)x³ + (3 - 3)x² - 4.

2x^4 + 18x³ - 4.

(10 + 4x - 3x³) - (5x³ + 8)

10 + 4x - 3x³ - 5x³ - 8

-8x³ + 4x + 2.

(x + 2)(2x - 5)

2x² - 5x + 4x - 10

2x² - x - 10

(x + 2)(3x² + 4x + 1)

3x³ + 4x² + x + 6x² + 8x + 2

3x³ + 10x² + 9x + 2.

The degree of a polynomial is given by the highest exponent in it's equation.

The end behavior of a polynomial is calculated as the result of elevating negative infinity and then positive infinity to the highest exponent of the polynomial, considering the leading coefficient.

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the probability of transitioning from state 1 to state 2 after the first transition is:

P(X(t) enters state 2 after the first transition | X(0) = 1) = 1 / 3

To write the system of ordinary differential equations (ODEs) for the transition probabilities Pᵢⱼ(t) of the given continuous-time Markov chain, we need to consider the rate at which the system transitions between different states.

Let Pᵢⱼ(t) represent the probability that the Markov chain is in state j at time t, given that it started in state i at time 0.

The ODEs for the transition probabilities can be written as follows:

dP₁₂(t)/dt = q₁₂ * P₁(t) - q₂₁ * P₂(t)

dP₁₃(t)/dt = q₁₃ * P₁(t) - q₃₁ * P₃(t)

dP₂₁(t)/dt = q₂₁ * P₂(t) - q₁₂ * P₁(t)

dP₂₃(t)/dt = q₂₃ * P₂(t) - q₃₂ * P₃(t)

dP₃₁(t)/dt = q₃₁ * P₃(t) - q₁₃ * P₁(t)

dP₃₂(t)/dt = q₃₂ * P₃(t) - q₂₃ * P₂(t)

where P₁(t), P₂(t), and P₃(t) represent the probabilities of being in states 1, 2, and 3 at time t, respectively.

Now, let's consider the second part of the question: Suppose that the initial state is 1. We want to find the probability that after the first transition, the process X(t) enters state 2.

To calculate this probability, we need to find the transition rate from state 1 to state 2 (q₁₂) and normalize it by the total rate of leaving state 1.

The total rate of leaving state 1 can be calculated as the sum of the rates to transition from state 1 to other states:

total_rate = q₁₂ + q₁₃

Therefore, the probability of transitioning from state 1 to state 2 after the first transition can be calculated as:

P(X(t) enters state 2 after the first transition | X(0) = 1) = q₁₂ / total_rate

In this case, the transition rate q₁₂ is 1, and the total rate q₁₂ + q₁₃ is 1 + 2 = 3.

Therefore, the probability of transitioning from state 1 to state 2 after the first transition is:

P(X(t) enters state 2 after the first transition | X(0) = 1) = 1 / 3

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Answer:

c is the correct answer i took the test

Step-by-step explanation:

The gender that shows greater variability in their playing time is C. Girls show greater variability because the interquartile range for girls is 12 for girls but only 8 for boys.

For the males, the interquartile range will be:

= Third quartile - First quartile

= 17 - 9

= 8

For the females, the interquartile range will be:

= Third quartile - First quartile

= 15 - 3

= 12

Therefore, girls show greater variability because the interquartile range for girls is 12 for girls but only 8 for boys.

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Answer:

26 hours + 6 minutes can also be said as 1560 min + 6 minutes

1560+6/6 = 1566/6 = 261

That is = 4 hours and 21 minutes

By evaluating a cost equation, we will see that Sam needs to pay $4.17

We know that one pound of chocolate costs $6.95, then we have the cost equation:

y = $6.95*x

Where y is the cost and x is the number of pounds of chocolate that you want to buy.

So, if you want to buy 0.6 of a pound, then we have x = 0.6

Then we can evaluate the equation to get:

y = $6.95*0.6

y = $4.17

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The required shape is translated 9 square(s) to the right/left and h square(s) up/down, depending on the signs of 9 and h.

The translation by the vector (9) h moves a shape g squares to the right and h squares up. This means that the shape is translated g squares to the right if 9 is positive, or g squares to the left if 9 is negative. Similarly, the shape is translated h squares up if h is positive, or h squares down if h is negative. So, the complete description of the translation is:

The shape is translated 9 square(s) to the right/left and h square(s) up/down, depending on the signs of 9 and h.

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Answer:

m<C = 110

Step-by-step explanation:

<a = <b

<b = 35 so,

<a = 35

<c = x

<a + <b + <c = 180

35 + 35 + x = 180

70 + x = 180

x = 180 - 70

x = 110

The length of the base is 5 feet

A quadratic equation is a second-degree polynomial equation that can be written in the form "ax² + bx + c = 0", where "x" is the variable, and "a", "b", and "c" are constants. The coefficient "a" cannot be zero, as this would result in a linear equation.

The area of a triangle is gotten from;

A = 1/2bh

2A = bh

A = 35 square feet

70= b (2b +4)

70 = 2b^2 + 4b

2b^2 + 4b - 70 = 0

b = 5 or -7

Since length can not be negative;

b = 5 feet

length = 2(5) + 4 = 14 feet

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The coordinates of the polygon are given, and then dilated by a factor of 1/2.

Hence the new coordinates would be derived as follows;

The new coordinates therefore are derived as;

A'(1, 2)

B'(-2, -4)

C'(0, 2)

D'(6, 1)

This is a reduction

The linear optimization problem is to maximize the objective function 250x + 150y, subject to the constraints x + y ≤ 60, 3x + y ≤ 90, and 2x + y > 30, where x and y are both greater than or equal to 20.

The feasible region can be determined by graphing the constraints and finding the overlapping region that satisfies all the conditions. In this case, the feasible region is the area where the lines x + y = 60, 3x + y = 90, and 2x + y = 30 intersect. This region can be visually represented on a graph.

To find the corner points of the feasible region, we need to find the points of intersection of the lines that form the constraints. By solving the systems of equations, we can find that the corner points are (20, 40), (20, 60), and (30, 30).

The optimal solution and the optimal objective function value can be determined by evaluating the objective function at each corner point and selecting the point that yields the maximum value. By substituting the coordinates of the corner points into the objective function, we find that the maximum value is achieved at (20, 60) with an objective function value of 10,500.

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The answer of the given question based on the histogram of Investment Account Values the correct option is c) 63%.

Investment refers to act of allocating resources, like  money or time, with  expectation of generating income or profit in  future. In other words, it involves sacrificing something of a value in  present, in hope of obtaining something of the greater value in  future

Investments can be taken  in many forms, including stocks, bonds, mutual funds, real estate, and commodities, among others. The choice of investment depends on variety of factors, including investor's risk tolerance, financial goals, and the investment time horizon.

Based on the histogram of Investment Account Values provided, it appears that approximately 63% of investments have a value of $150 or less. This can be determined by adding up the percentage of investments in each bar up to and including the $150 bar.

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The main condition for a setting to be considered binomial is that the probability of success remains the same for each trial, and the other conditions include having 3 possible outcomes for each observation, no variation in outcomes based on the first success, and independence of trials from one another.

A condition for a setting to be considered binomial is that the probability of success is the same for each trial.

In order for a setting to be considered binomial, there are certain conditions that need to be met. The first condition is that the probability of success remains constant for each trial or observation. This means that the likelihood of achieving the desired outcome remains unchanged throughout the entire process.

The second condition states that each observation or trial must have exactly 3 possible outcomes. This implies that there are only three options or choices for each trial, typically categorized as success, failure, or a neutral outcome.

The third condition is that the number of outcomes should not vary based on the occurrence of the first success. This means that the probability of success is not affected or altered by the outcome of previous trials.

Lastly, the fourth condition is that the trials or observations must be independent of one another. This implies that the outcome of one trial should not impact the outcome of subsequent trials.

Therefore, the main condition for a setting to be considered binomial is that the probability of success remains the same for each trial, and the other conditions include having 3 possible outcomes for each observation, no variation in outcomes based on the first success, and independence of trials from one another.

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Answer:

S = 4π(7^2) = 196π square inches

= about 615.75 square inches

Answer:

19.998? or is there something else to it

1. In 2008, there were approximately 28.2 million live Christmas trees sold in the U.S

2. The linear regression equation that models the set of data above is C = 0.28t + 27.28.

3. From 2004 to 2011, the number of Christmas trees sold in the U.S. increased by approximately 0.28 million trees each year.

4. In 2008, there were approximately 28.4 million live Christmas trees sold in the U.S.

5. Why is this the case: A. The data are not perfectly linear. The regression equation gives only an approximation.

By using the data values in the table, the number of live Christmas trees that were sold in the year 2008 can be calculated as follows;

t = 0 + (2008 - 2004) years.

t = 4 years.

At t = 4 years on the table, approximately 28.2 million live Christmas trees sold in the U.S.

Part 2.

Based on the table, we can logically deduce that the y-intercept or initial value is (0, 27.1).

y = mx + b ≡ C = mt + b

b = (547.6 - 538.6)/(105 - 72.2)

b = 0.28

Therefore, the required linear regression equation is given by;

C = 0.28t + 27.28

Part 3.

Based on the slope of the above linear regression equation, the number of Christmas trees sold in the U.S. from 2004 to 2011 increased by approximately 0.28 million trees each year.

Part 4.

For the number of live Christmas trees sold in year 2008, we have:

C(4) = 0.28(4) + 27.28

C(4) = 28.4 million.

Part 5.

The answers to parts 1 and 4 are different because the data set do not have a perfectly linear relationship and the linear regression equation gives only an approximated value, not an exact value.

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The distribution would exhibit a negative skew.

A type of distribution in which more values are concentrated on the right side(tail) of the distribution graph while the left tail of the distribution graph is longer is called negatively skewed distribution.  Or a distribution is said to be negatively skewed or skewed to the left if the scores of the distribution falls mainly to the right of the graph and less to the left i.e. the lower scores are very few as compared to the high scores.

Here we are given that the three students who scored below 140 were dropped that means the some of the low scores are now dropped from the graph. Therefore, if 3 of the students who scored below 140 decide to drop the class, the shape of the distribution will be negative skewed or the distribution would exhibit a negative skew.

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Approximately 17 students who took the optional class passed their driver's test.

Let's solve this problem step by step.

We are given the following information:

28 out of 75 students who took the optional class passed their driver's test.

8 out of 15 students overall passed their driver's test.

3 out of 5 students took the optional class.

To find the number of students who took the optional class and passed, we need to calculate the intersection of these two groups.

First, let's calculate the total number of students who took the optional class:

Total students who took the optional class = (3/5) \(\times\) Total number of students

Total students who took the optional class = (3/5) \(\times\) 75 = 45

Now, let's calculate the total number of students who passed their driver's test:

Total students who passed their driver's test = (8/15) \(\times\) Total number of students

Total students who passed their driver's test = (8/15) \(\times\) 75 = 40

Next, let's find the number of students who both took the optional class and passed their driver's test.

This can be found by taking the intersection of the two groups:

Number of students who took the optional class and passed = (28/75) \(\times\)Total students who took the optional class

Number of students who took the optional class and passed = (28/75) \(\times\)45 = 16.8 (approximated to the nearest whole number).

Therefore, approximately 17 students who took the optional class passed their driver's test.

It's important to note that since we're dealing with whole numbers, the approximated answer is 17, as we cannot have a fraction of a student.

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Answer:

11a^2 + 122a + 11

Step-by-step explanation:

First, multiply a by 11a, which gives you 11a^2. Then you multiply 11 to 11a, which gives you 122a, then last just multiply 11 to 1 and you get 11. Hope this helps.

Answer:

Step-by-step explanation:

1 Determine the smallest number that is divisible by 6 and 8 e×actly. Multiples of 6: 6, 12, 18, 24, 30, 36, 42…. Multiples of 8: 8, 16, 24, 32, 40, 48, 56…. Hence the LCM of 6 and 8 is 24.

The LCM(Lowest Common Multiple)of 6 and 8 is 24.

What is LCM?

In Mathematics, the LCM of any two numbers is the value that is evenly divisible by the two given numbers. The full form of LCM is Least Common Multiple. It is also called the Least Common Divisor (LCD). For example, LCM (4, 5) = 20. Here, the LCM 20 is divisible by both 4 and 5 such that 4 and 5 are called the divisors of 20.

LCM is also used to add or subtract any two fractions when the denominators of the fractions are different. While performing any arithmetic operations such as addition, subtraction with fractions, LCM is used to make the denominators common. This process makes the simplification process easier.

Now,

To find LCM find the Multiples of 6 and 8

i.e. 6,12,18,24,30,36.......... and 8,16,24,32,40....

As Lowest common multiple of both is 24.

hence,

          The LCM of 6 and 8 is 24.

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Answer:

2/5

Step-by-step explanation:

it is the opposite reciprocal of the slope

Pythagorean Theorm Unit? I got you!

It would be C, about 63.07

Answer:

Step-by-step explanation:

When the center of dilation is at the origin, we can multiply the coordinates of all the vertices by the scale factor.

So D(2, 4) turns into D'(7, 14)

E (6, 4) turns into E'(21, 14)

And F(2, 6) turns into F'(7, 21)