Let q be the group of rational numbers under addition and let q* be the group of nonzero rational numbers under multiplication. In q, list the elements in k 1 2l. In q*, list the elements in k 1 2l.

Answer:

The median is 3.5!

Step-by-step explanation:

You either got it right, or you guessed! I arranged the data set from lowest to highest value, and then chose the middle one. For this one, there were two middle values : 3 and 4. So, the median would be in the middle, middle, which is 3.5 :)

The probability that X is between 134.4 and 140.1 is approximately 0.4076, which is closest to the option 0.4069.

To calculate this probability, we need to standardize the values using the standard normal distribution. First, we find the z-scores for the lower and upper bounds:

z1 = (134.4 - 137.0) / 5.3 ≈ -0.4906

z2 = (140.1 - 137.0) / 5.3 ≈ 0.5849

Next, we use a standard normal distribution table or a calculator to find the probabilities associated with these z-scores. The probability corresponding to z1 is P(Z < -0.4906) ≈ 0.3131, and the probability corresponding to z2 is P(Z < 0.5849) ≈ 0.7207.

To find the probability that X is between 134.4 and 140.1, we subtract the probability associated with the lower bound from the probability associated with the upper bound:

P(134.4 < X < 140.1) = P(Z < 0.5849) - P(Z < -0.4906) ≈ 0.7207 - 0.3131 = 0.4076.

As a result, the chance that X lies between 134.4 and 140.1 is roughly 0.4076, which is closest to the value of 0.4069 for the option.

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Exactly what was your answer for a

Answer:

654000 Rupees

Step-by-step explanation:

If we assume that the opera house was full every day then it would be 872 x 50 for the price of 1 day

872 x 50 = 43600

We multiple that number by the amount of days open so is would be 43600 x 15

43600 x 15 = 654000

Therefore it will make 654000 Rupees

The example provided is an example of empirical probability.

Empirical probability, also known as experimental probability, is based on actual observations or data gathered from experiments, surveys, or real-world events. In this case, the probability that the elderly adults prefer a poodle is determined through a survey, which involves collecting data from the group of elderly adults about their favorite breeds of dogs. The conclusion that the probability is about 30% is based on the data obtained from the survey, making it an empirical probability.

Therefore, the example given is an example of empirical probability because it is based on data collected from a survey of elderly adults' favorite breeds of dogs

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Answer: Input values are 8,-2,12,9, and 18, so select 9 and -2.

Answer:

The answer is 10.16

Step-by-step explanation:

I do the asigment

Answer:

See answers below

Step-by-step explanation:

A. 3/4 * 3 = 2.25 <= 4 = YES

B. 11/12 * 4 = 3.67 <= 4 = YES

C. 7/8 * 5 = 4.38 >= 4 = NO

D. 5/6 * 6 = 5 >= 4 = NO

E. 1/2 * 8 = 4 <= 4 = Yes

Mark brainliest if this helped

Mark brainliest if this helped

Answer:

$0.40

Step-by-step explanation:

Price of 3 lemons = $1.20

Price of 1 lemon = $1.20 \(\div\) 3 = $0.40

Answer:

x to the power of 67000 to the 5th power.

Step-by-step explanation:

Answer:

7 and 14  

Step-by-step explanation:

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Answer:

7 an 43

Step-by-step explanation:

Answer:

\(y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}\)

Step-by-step explanation:

Given the second-order differential equation. Solve by using variation of parameters.

\(y''+2y'+y=e^{-t}\ln(t)\)

(1) - Solve the DE as if it were homogeneous to find the homogeneous solution

\(y''+2y'+y=e^{-t}\ln(t) \Longrightarrow y''+2y'+y=0\\\\\text{The characteristic equation} \rightarrow m^2+2m+1=0, \ \text{solve for m}\\\\m^2+2m+1=0\\\\\Longrightarrow (m+1)(m+1)=0\\\\\therefore \boxed{m=-1,-1}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}\)

Notice we have repeated/duplicate roots, form the homogeneous solution.

\(\boxed{\boxed{y_h=c_1e^{-t}+c_2te^{-t}}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now using the method of variation of parameters, please follow along very carefully.

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(1 of 2):}}\\ \text{Given a DE in the form} \rightarrow ay''+by"+cy=g(t) \\ \text{1. Obtain the homogenous solution.} \\ \Rightarrow y_h=c_1y_1+c_2y_2+...+c_ny_n \\ \\ \text{2. Find the Wronskain Determinant.} \\ |W|=$\left|\begin{array}{cccc}y_1 & y_2 & \dots & y_n \\y_1' & y_2' & \dots & y_n' \\\vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)}\end{array}\right|$ \\ \\ \end{array}\right}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(2 of 2):}}\\ \text{3. Find} \ W_1, \ W_2, \dots, \ W_n.\\ \\ \text{4. Find} \ u_1, \ u_2, \dots, \ u_n. \\ \Rightarrow u_n= \int\frac{W_n}{|W|} \\ \\ \text{5. Form the particular solution.} \\ \Rightarrow y_p=u_1y_1+u_2y_2+ \dots+ u_ny_n \\ \\ \text{6. Form the general solution.}\\ y_{gen.}=y_h+y_p\end{array}\right}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(2) - Finding the Wronksian determinant

\(|W|= \left|\begin{array}{ccc}e^{-t}&te^{-t}\\-e^{-t}&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t}-te^{-t})-(te^{-t})(-e^{-t})\\\\\Longrightarrow (e^{-2t}-te^{-2t})-(-te^{-2t})\\\\\therefore \boxed{|W|=e^{-2t}}\)

(3) - Finding W_1 and W_2

\(W_1=\left|\begin{array}{ccc}0&y_2\\g(t)&y_2'\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}0&te^{-t}\\e^{-t} \ln(t)&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow 0-(te^{-t})(e^{-t} \ln(t))\\\\\therefore \boxed{W_1=-t\ln(t)e^{-2t}}\)

\(W_2=\left|\begin{array}{ccc}y_1&0\\y_1'&g(t)\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}e^{-t}&0\\-e^(-t)&e^{-t} \ln(t)\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t} \ln(t))-0\\\\\therefore \boxed{W_2=\ln(t)e^{-2t}}\)

(4) - Finding u_1 and u_2

\(u_1=\int \frac{W_1}{|W|}; \text{Recall:} \ W_1=-t\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{-t\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow -\int t\ln(t)dt \ \text{(Apply integration by parts)}\\\\\\\boxed{\left\begin{array}{ccc}\text{\underline{Integration by Parts:}}\\\\uv-\int vdu\end{array}\right }\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=tdt \rightarrow v=\frac{1}{2}t^2 \\\\\)

\(\Longrightarrow -\Big[(\ln(t))(\frac{1}{2}t^2)-\int [(\frac{1}{2}t^2)(\frac{1}{t}dt)]\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\int (t)dt\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\cdot\frac{1}{2}t^2 \Big]\\\\\therefore \boxed{u_1=\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t)}\)

\(u_2=\int \frac{W_2}{|W|}; \text{Recall:} \ W_2=\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow \int \ln(t)dt \ \text{(Once again, apply integration by parts)}\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=1dt \rightarrow v=t \\\\\Longrightarrow (\ln(t))(t)-\int[(t)(\frac{1}{t}dt )] \\\\\Longrightarrow t\ln(t)-\int 1dt\\\\\therefore \boxed{u_2=t \ln(t)-t}\)

(5) - Form the particular solution

\(y_p=u_1y_1+u_2y_2\\\\\Longrightarrow (\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t))(e^{-t})+(t \ln(t)-t)(te^{-t})\\\\\Longrightarrow\frac{1}{4}t^2e^{-t}-\frac{1}{2}t^2\ln(t)e^{-t}+ t^2\ln(t)e^{-t}-t^2e^{-t}\\\\\therefore \boxed{ y_p=\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}\)

(6) - Form the solution

\(y_{gen.}=y_h+y_p\\\\\therefore\boxed{\boxed{y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}}\)

Thus, the given DE is solved.

Answer:

240 students

Step-by-step explanation:

     To find the minimum number of students, we have to find the LCM of 10, 12, 16

10 = 2 * 5

12 = 2 * 2 * 3

16 = 2 * 2 * 2 * 2

LCM = 2 * 2 *  2 * 2 * 3 * 5

       = 240

Answer:

2756/53 = 52

52 = 520 to the nearest tenth

52 = 52 to the nearest hundredth

52 = 52 to the nearest thousandth

= 0 to the nearest tenth

= 0 to the nearest hundredth

= 0 to the nearest thousandth

Step-by-step explanation:

I hope this helps you in some way.

Answer:

47 zeros

Step-by-step explanation:

Answer:

  0

Step-by-step explanation:

There are an infinite number of equivalent expressions. Perhaps the simplest is the value of the expression:

  4 - 4 = 0

The length of the rectangle is 19 cm.

The formula for the area of a rectangle,

Area = Length x Width

Given that the area is 336 cm squared.

So, we can set up an equation,

⇒ 336 = (w + 5)w

where w represents the width of the rectangle.

Expanding this equation,

⇒ 336 = w² + 5w

Moving all terms to one side:

⇒ w² + 5w - 336 = 0

This is a quadratic equation that we can solve using the quadratic formula,

⇒ w = (-5 ± √(5² - 4(1)(-336))) / (2(1))

⇒ w = (-5 ± 23) / 2

We'll take the positive value,

⇒ w = 14

So, the width of the rectangle is 14 cm.

We also know that the length is 5 cm more than the width,

Therefore,

⇒ l = w + 5

⇒ l = 14 + 5

⇒ l = 19

Therefore, the length of the rectangle is 19 cm.

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Answer:x^2+3x-4

Step-by-step explanation:

False. Cronbach's alpha is a measure of internal consistency reliability, not a measure of the correlation between scale items.

It assesses the extent to which the items in a scale or questionnaire are measuring the same underlying construct.

Cronbach's alpha is calculated based on the inter-item correlations among the items in a scale. It provides a measure of the average correlation between all possible pairs of items in the scale. The range of Cronbach's alpha is between 0 and 1, with higher values indicating greater internal consistency or reliability.

Essentially, Cronbach's alpha quantifies the extent to which the items in a scale are consistently measuring the same construct or concept. It assesses how well the items "hang together" as a reliable measurement tool.

While correlation between scale items is related to internal consistency, Cronbach's alpha specifically measures the degree to which the items are interrelated and provides a single coefficient that reflects the overall reliability of the scale. It does not directly indicate the extent of item-item correlations or the strength of individual item contributions to the scale.

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To determine the probability that neither card shows an even number with replacement, we need to calculate the probability of drawing an odd number on each card and multiply the probabilities together.

Let's assume we have a standard deck of 52 playing cards, where half of them are even numbers (2, 4, 6, 8, 10) and the other half are odd numbers (1, 3, 5, 7, 9).

Since the cards are replaced after each draw, the probability of drawing an odd number on each card is 1/2. Therefore, the probability that neither card shows an even number is:

P(neither card shows an even number) = P(odd on card 1) * P(odd on card 2) = 1/2 * 1/2 = 1/4

So, the probability that neither card shows an even number, with replacement, is 1/4.

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The type directions that lie in the (110) plane are Crystal planes are equivalent planes that represent a group of crystal planes with a common set of atomic indexes.

Crystallographers use Miller indices to identify crystallographic planes. A crystal is a three-dimensional structure with a repeating pattern of atoms or ions.In a crystal, planes of atoms, ions, or molecules are stacked in a consistent, repeating pattern. Miller indices are a mathematical way of representing these crystal planes.

Miller indices are the inverses of the fractional intercepts of a crystal plane on the three axes of a Cartesian coordinate system.Let us now determine which of the type directions lie in the (110) plane.[101] is not in the (110) plane because it has an x-intercept of 1, a y-intercept of 0, and a z-intercept of 1. So, this direction does not lie in the (110) plane.[110] is in the (110) plane since it has an x-intercept of 1, a y-intercept of 1, and a z-intercept of 0.

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Answer:

50

Step-by-step explanation:

180-130=50

Answer:

105

Step-by-step explanation:

Answer:

\(14 \times 0.5 = 70 \\ 70\)

An estimator is said to be unbiased if the expected value of the estimator is equal to the true value of the parameter being estimated.

Unbiased estimators are useful because they provide an estimate that, on average, is equal to the true value of the parameter, making them desirable for making accurate inferences about a population.

In statistics, an estimator is a statistic used to estimate the value of an unknown parameter in a population. An estimator is said to be unbiased if, on average, it produces an estimate that is equal to the true value of the parameter. An unbiased estimator is desirable because it provides an estimate that is, on average, accurate, which is important for making inferences about a population.

Biased estimators, on the other hand, tend to consistently overestimate or underestimate the true value of the parameter, which can lead to incorrect conclusions. Therefore, unbiased estimators are useful because they provide more accurate estimates, which can lead to more reliable inferences about a population.

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Answer:

7 boxes of pencils, 5 boxes of erasers

Step-by-step explanation:

LCM of 10 and 14 is 70.

10 = 2 × 5, 14 = 2 × 7

70 ÷ 10 = 7 boxes of pencils

70 ÷ 14 = 5 boxes of erasers

Answer:

x is -9

CB is 10

Step-by-step explanation:

In order to solve this question we need to figure out what x is first

First we do

2(x+19)=x+29

Then we get 2x+28=x+29

subtract and divide to get

x=-9

Plug in the x to get CB

-9+19=10

Taring a kitchen scale before weighing an ingredient is essential to obtain accurate measurements by removing the weight of the container, streamlining the process, and ensuring precise and reliable results in culinary preparations.

Taring a kitchen scale before weighing an ingredient is necessary to accurately measure the weight of the ingredient without including the weight of the container or vessel in which it is placed. Taring essentially resets the scale to zero, accounting for the weight of the container so that only the weight of the ingredient being added is measured.

By taring the scale, you eliminate the need to manually subtract the weight of the container from the final measurement. This allows for more precise and efficient measurements in recipes or other culinary applications.

Taring is particularly important when working with small or precise quantities of ingredients, where even a slight variation in weight can significantly impact the final outcome of a dish. It ensures that the weight of the container does not contribute to the measurement, providing accurate and reliable results.

Additionally, taring simplifies the weighing process by eliminating the need to calculate or estimate the weight of the container separately. It saves time and reduces the chances of errors in measurements, promoting consistency and precision in cooking and baking.

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Answer:

C

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Answer:

-24

Step-by-step explanation:

First find f(3), which is the value of the blue line when x =3

f(3) = -2

Then find g(-1) which is the value of the red line when x= -1

g(-1) = 6

-6 f(3) - 6*g(-1)

-6*-2 - 6(6)

12 - 36

-24