a team of researchers gets a list of all 500 members of an organization they want to include in the study. they only have funding and time to survey 50 members. they take the list of members, randomly select a starting point, and select every 10th name from the list for the sample. what is the sampling interval?

Answer:

x= 1/50 or 0.02, or x=50^-1

Answer:

[tex]x=\frac{1}{50}[/tex]

Step-by-step explanation:

If you would like to see the steps here they are.

To draw asymptotes, first determine the equation of the line of the asymptote. Then use the equation to plot points and draw the line. For vertical asymptotes, use the x-intercept and for horizontal asymptotes, use the y-intercept.

To draw asymptotes, first you need to determine the equation of the line of the asymptote. The equation of the asymptote is usually in the form of y = ax + b or x = ay + b. The a and b coefficients allow you to determine the slope and intercept of the line. Once you have the equation, use it to plot points and draw the line. For vertical asymptotes, use the x-intercept and for horizontal asymptotes, use the y-intercept. Then, connect the points with a smooth line. Make sure the line does not intercept the graph at any point as this would not be an asymptote. Once you have drawn the line, check to make sure it is indeed an asymptote by comparing it to the given equation. If it matches, you have successfully created an asymptote!

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To graph four lines, you will need to plot at least four points for each line and then draw a line through these points.

A general procedure you can use is as follows:

It can be helpful to use a different color or style (such as a solid line or a dotted line) for each line to make it easier to tell them apart.

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The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin.

In mathematics, a graph is a visual representation or diagram that shows facts or values in an ordered way.

The relationships between two or more items are frequently represented by the points on a graph.

The y = 5 graph is a line that runs parallel to the x-axis at a distance of 5 units from the origin.

Assuming Y-axis as vertical line, X-axis as horizontal line

(see it in attached image)

Graph of y = 5 is the line where y is not changing or y always be 5, only x will change.

Graph of y is the straight line passing through 5 and parallel to X axis

Therefore, a line that is 5 units from the origin and parallel to the x-axis forms the graph of y=5.

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Answer:

Step-by-step explanation:

You want to identify the true statements regarding angles at a transversal crossing parallel lines, and perpendicular bisectors.

The Corresponding Angles theorem tells you that corresponding angles are congruent where a transversal crosses parallel lines. Since vertical angles are congruent, and angles congruent to the same angle are congruent to each other, this also means that alternate interior angles are congruent.

A bisector of a segment passes through its midpoint, a point that is equidistant from the end points. When the bisector is perpendicular to the segment, all points on the perpendicular bisector are equidistant from the segment's endpoints.

Correct Question:-

Solve the following differential equation using Laplace transformation.

[tex]\rm{x''(t)+x'(t)=2,\quad x(0)=3,\quad x'(0)=1}[/tex]

where [tex]\rm{x'(t)=\dfrac{dx}{dt},\quad x''(t)=\dfrac{d^2x}{dt^2}.}[/tex]

[tex]\quad[/tex]

Solution:-

Let,

[tex]\cal{L}\{\rm{x(t)}\}=X(s)[/tex]

[tex]\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=x(t)[/tex]

Given,

[tex]\longrightarrow\rm{x''(t)+x'(t)=2}[/tex]

We take Laplace transformation of both sides of the equation.

[tex]\longrightarrow\cal{L}\{\rm{x''(t)+x'(t)}\}=\cal{L}\{\rm{2}\}[/tex]

[tex]\longrightarrow\cal{L}\{\rm{x''(t)}\}+\cal{L}\{\rm{x'(t)}\}=\cal{L}\{\rm{2}\}[/tex]

We have,

[tex]\cal{L}\{\rm{x''(t)}\}=\rm{s^2\,X(s)-s\,x(0)-x'(0)}[/tex]

[tex]\cal{L}\{\rm{x'(t)}\}=\rm{s\,X(s)-\,x(0)}[/tex]

[tex]\cal{L}\{\rm{2}\}=\rm{\dfrac{2}{s}}[/tex]

Then,

[tex]\small\text{$\longrightarrow\rm{\big[s^2\,X(s)-s\,x(0)-x'(0)\big]+\big[s\,X(s)-x(0)\big]=\dfrac{2}{s}}$}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)-(s+1)\,x(0)-x'(0)=\dfrac{2}{s}}[/tex]

Given that [tex]\rm{x(0)=3}[/tex] and [tex]\rm{x'(0)=1.}[/tex] Then,

[tex]\longrightarrow\rm{s(s+1)\,X(s)-3(s+1)-1=\dfrac{2}{s}}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)-3s-4=\dfrac{2}{s}}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)=\dfrac{2}{s}+3s+4}[/tex]

[tex]\longrightarrow\rm{X(s)=\dfrac{2}{s^2(s+1)}+\dfrac{3s+4}{s(s+1)}}[/tex]

[tex]\longrightarrow\rm{X(s)=\dfrac{3s^2+4s+2}{s^2(s+1)}\quad\dots(1)}[/tex]

We will factorise this expression by partial fractions.

Assume,

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{As+B}{s^2}+\dfrac{C}{s+1}}[/tex]

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{(As+B)(s+1)+Cs^2}{s^2(s+1)}}[/tex]

[tex]\longrightarrow\rm{3s^2+4s+2=(As+B)(s+1)+Cs^2}[/tex]

[tex]\longrightarrow\rm{3s^2+4s+2=(A+C)s^2+(A+B)s+B}[/tex]

Equating corresponding coefficients,

[tex]\rm{A+C=3}[/tex]

[tex]\rm{A+B=4}[/tex]

[tex]\rm{B=2}[/tex]

Solving each equation we get,

[tex]\rm{A=2}[/tex]

[tex]\rm{B=2}[/tex]

[tex]\rm{C=1}[/tex]

Therefore,

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{2s+2}{s^2}+\dfrac{1}{s+1}}[/tex]

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}[/tex]

Then (1) becomes,

[tex]\longrightarrow\rm{X(s)=\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}[/tex]

Now we will inverse Laplace transformation to obtain the solution.

[tex]\longrightarrow\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}\right\}[/tex]

[tex]\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s^2}}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s+1}}\right\}$}[/tex]

[tex]\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1!}{s^{1+1}}}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s-(-1)}}\right\}$}[/tex]

We have,

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}=\rm{1}[/tex]

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1!}{s^{1+1}}}\right\}=\rm{t}[/tex]

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s-(-1)}}\right\}=\rm{e^{-t}=\dfrac{1}{e^t}}[/tex]

Hence,

[tex]\longrightarrow\rm{\underline{\underline{x(t)=2+2t+\dfrac{1}{e^t}}}}[/tex]

This is the solution to our differential equation.

[tex]\quad[/tex]

Some results of Laplace Transformation:-

[tex]\boxed{\begin{array}{c|c}&\\\rm{x(t)}=\cal{L}^{\rm{-1}}\{\rm{X(s)}\}&\rm{X(s)}=\cal{L}\{\rm{x(t)}\}\\\\=============&==================\\\\\rm{\dfrac{d^nx}{dt^n}}&\rm{s^n\,X(s)}-\displaystyle\sum_{\rm{r=0}}^{\rm{n-1}}\rm{s^{n-r-1}\,\dfrac{d^rx}{dt^r}(t=0)}\\\\----------&--------------\\\\\rm{x'(t)}&\rm{s\,X(s)-x(0)}\\\\----------&--------------\\\\\rm{x''(t)}&\rm{s^2\,X(s)-s\,x(0)-x'(0)}\\\\----------&--------------\\\\\rm{a}&\rm{\dfrac{a}{s}}\\\\----------&--------------\\\\\rm{t^n,\ n}\in\mathbb{N}&\rm{\dfrac{n!}{s^{n+1}}}\\\\----------&--------------\\\\\rm{e^{at}}&\rm{\dfrac{1}{s-a}}\\&\end{array}}[/tex]

Answer:

Step-by-step explanation:

i belive it would be 500 divided by 20m

Answer:

5/3

Step-by-step explanation:

The scale factor is the ratio of the side length of the image to the corresponding side in the preimage.

So, the scale factor is [tex]\frac{50}{30}=\frac{5}{3}[/tex].

Answer: -0.7

Step-by-step explanation:

-1.1 - 1.4m + 0.4 = ? - 1.4m

-1.1 + 0.4 = -0.7

-0.7 - 1.4m = -0.7 - 1.4m

f(x) = sinx is an trigonometric function.

A trigonometric function is a type of mathematical function that is defined in terms of the angles of a right triangle. The most common trigonometric functions are sine, cosine, and tangent. The function f(x) = sinx is a sine function, which is defined as the ratio of the length of the side opposite an angle in a right triangle to the length of the hypotenuse. The sine function can be used to calculate the angles of a triangle when the lengths of two sides are known. It is also useful for modeling many real-world phenomena such as the motion of waves and the oscillations of pendulums.

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The function's standard form equation is equivalent to [tex]\frac{-x^2}{4}+1=0[/tex].

A mathematical expression (equation) that can be used to define and represent the relationship between two or more variables on a graph is known as a quadratic function.

The standard form of a quadratic equation in mathematics is given by;

[tex]ax^2+bx+c=0\\y=f(x)=ax^2+bx+c[/tex]

Next, using the information in the previous table, we would create the following system of equations:

[tex]-3=a(-4)^2+b(-4)+c\\-3=16a-4b+c[/tex].......equation (i)

[tex]0=a(-2)^2+b(-2)+c\\0= 4a-2b+c[/tex]........equation(ii)

[tex]1=a(0)^2+b(0)+c\\1=c[/tex].........equation (iii)

Inputting the value of c into equations 1 and 2 results in:

[tex]-4=16a-4b[/tex]......ewuation(iv)

[tex]-1=4a-2b[/tex]......equation(v)

Equation 5 is multiplied by two and solved concurrently to yield:

[tex]-2=8a\\a=\frac{-2}{8} \\a=\frac{-1}{4}[/tex]

What we have for b's value is:

[tex]-1=4a-2b\\-1=4(\frac{-1}{4})-2b\\ -1=-1-2b\\2b=0\\b=0[/tex]

Inputting the values of a, b, and c into a quadratic equation in standard form:

[tex]ax^2+bx+c=0\\\frac{-1}{4}x^2+(0)x+1=0\\ \frac{-x^2}{4} +1=0[/tex]

When x equals 2, we get:

[tex]\frac{-x^2}{4} +1=0\\\frac{-2^2}{4}+1=0\\ \frac{-4}{4}+1=0\\ -1+1=0[/tex]

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An inverse relationship in math is a relationship between two quantities where one quantity increases as the other decreases and vice versa.

An example of an inverse relationship is the relationship between the temperature in degrees Celsius and the temperature in degrees Fahrenheit.

The relationship between temperature in degrees Celsius (C) and temperature in degrees Fahrenheit (F) is defined by the following equation:

F = (9/5)C + 32

This equation shows that as the temperature in degrees Celsius increases, the temperature in degrees Fahrenheit decreases. For example, if the temperature in degrees Celsius is 20, the temperature in degrees Fahrenheit is 68. If the temperature in degrees Celsius is 40, the temperature in degrees Fahrenheit is 104.

Therefore, the relationship between temperature in degrees Celsius and temperature in degrees Fahrenheit is an inverse relationship.

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The correct equation that represents the inequality is f(x)  > √1 - x.

What is the inequality equation?

To graph an inequality, treat the <, ≤, >, or ≥ sign as an = sign, and graph the equation. If the inequality is < or >, graph the equation as a dotted line. If the inequality is ≤ or ≥, graph the equation as a solid line.

An inequality is a mathematical relationship between two expressions and is represented using one of the following: ≤: "less than or equal to" <: "less than" ≠: "not equal to"

We have given the graph that represents the inequality equation,

and the inequality equations:

f(x)  > √1 - x

f(x)  ≤ √1 - x

f(x)  ≥ √1 - x

f(x)  < √1 - x

Hence, the correct equation that represents the inequality is f(x)  > √1 - x.

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The equation that defines the fines with respect to the number of days overdue is written as: y = 0.1x + 1.

The equation that defines the fines with respect to the number of days overdue can be written as a linear equation that is expressed in slope-intercept form as y = mx + b, where the slope or unit rate is m, while the y-intercept or initial value is b.

Using two points from the chart, (2, 1.20) and (5, 1.50), find the slope or unit rate:

Slope / unit rate (m) = change in y / change in x = (1.50 - 1.20) / (5 - 2)

m = 0.3 / 3

m = 0.1

Find the y-intercept by substituting m = 0.1 and (x, y) = (2, 1.20) into y = mx + b:

1.20 = 0.1(2) + b

1.20 = 0.2 + b

1.20 - 0.2 = b

1 = b

To write the equation, substitute m = 0.1 and b = 1 into y = mx + b:

y = 0.1x + 1

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Answer:

x =  - 28/101

Step-by-step explanation:

Answer:

x=-42

Step-by-step explanation:

[tex]-3x-28=98\\(-3x-28)+28=(98)+28\\-3x=126\\\frac{-3x}{-3}=\frac{126}{-3}\\x=-42[/tex]

If a₁ = 4 and an = -4an-1 +5 then the value of a5 is 72, in the given sequence.

A group of enumerated objects where repetitions are allowed and order is crucial is referred to as a sequence in mathematics. Similar to a set, it has members. The sequence's length is expressed in terms of its number of elements.

Given that, sequence  

[tex]a_1=4, a_n=4 a_{n-1}+5[/tex]

Now

[tex]a_2=4 a_1+5[/tex]

[tex]a_2=4 \times 4+5[/tex]

[tex]a_2=21[/tex]

[tex]a_2=4 \times 4+5[/tex]

[tex]a_2=21[/tex]

And the common difference d = 21 - 4 = 17

Therefore [tex]a_5=a_1+4 d[/tex]

[tex]a_5=4+4 \times 17[/tex]

[tex]a_5=4+68[/tex]

[tex]a_5=72[/tex]

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Answer:

1. 7

2. 5/3

3. 2.6

Step-by-step explanation:

The "+" or "-" sign of a number is disregarded in this operation since distance in mathematics is never negative. You can tell what a number's absolute value is by putting it between two vertical bars known as absolute value brackets.

Answer:

1.  7

2.  5/3

3.  2.6

Step-by-step explanation:

The absolute value is the distance between a number and zero

1. We see the distance is 7, so the answer is 7

2. We see the distance as 5/3, so the answer is 5/3

3. We see the distance is 2.6, so the answer is 2.6

Answer:?=93/14

Step-by-step explanation:

the ? is the variable in this equation so we have to solve for it

1. distributive property ->

5(?+3) = 5?+15

?(6+3) = 9?

so 5?+15+9?=108 -> 14?+15=108

2. then solve,

14? -15 = 108 -15

14?=93

?=93/14

Linear graph equations are one-degree equations. It is a straight-line equation. A linear equation with parameters a and b equal to zero has the conventional form ax + by + c = 0.

A linear equation has the following standard form: ax + by + c = 0.

Here, variables x and y and constants a, b, and c are used.

Also, a ≠ 0, and b ≠ 0.

For linear equations, the slope-intercept formula is: y=mx+b

Where m stands for the line's slope and b represents the y-intercept.

The three types of linear equations are point-slope, slope-intercept, and standard form.

When an algebraic equation is graphed, it always produces a straight line since each term has an exponent of 1 or 0. This type of equation is known as a linear equation.

Steps of drawing linear graphs are described below:

Step 1: Using the provided linear equation, determine the value of y with respect to x.

Step 2: Arrange these data in a table format.

Step 3: Create a graph using the points from the database.

Step 4: Connect the points to form a straight line.

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The formula that can be used to calculate the half-life is: [tex]N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1 / 2}}}[/tex]

The half-life is the amount of time required for a quantity (of substance) to reduce to half of its initial value.

The expression is commonly used in nuclear physics to describe how quickly unstable atoms decay radioactively or how long stable atoms remain.

Additionally, a broader definition of the term is used to characterize any exponential decay (or, very infrequently, nonexponential decay).

For instance, the medical sciences may refer to the biological half-life of drugs and other substances in the human body.

Time doubles in proportion to life's exponential expansion.

The formula to calculate half-life: [tex]N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1 / 2}}}[/tex]

Where N(t) is the quantity of the substance remaining, N₀ is the initial quantity of the substance, t is the time elapsed and t₁₎₂ is the half-life of the substance.

Therefore, the formula that can be used to calculate the half-life is: [tex]N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1 / 2}}}[/tex]

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The simulation to estimate the standard deviation of the sampling distribution of the sample median for random samples of size 50 is given as follows:

Sampling is a technique for when a group of elements from a population is taken to represent the entire population.

Hence, for this problem, multiple samples of 50 students are taken, and for each sample, the median is computed and then stored in a list, according to the procedure described at the beginning of the answer.

Then, after the entire procedure runs, we will have a sample of 50 values, which represent the median for each sample. Then the standard deviation is calculated as the square root of the sum of the difference squared between each of these observations and the mean of all these observations, divided by 50, which is the cardinality of the data-set.

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Using (D) the binomial formula, which is used to determine probability when independent events are known, we will get the probability in the scenario.

In an experiment with two possible outcomes, the likelihood of exactly x successes on n repeated trials is known as the binomial probability (commonly called a binomial experiment).

The binomial probability is nCx⋅px⋅(1−p)n−x if the likelihood of success on a single trial is p.

When a process is repeated a certain number of times (for example, in a set of patients), the result for each patient can either be a success or a failure, the binomial distribution model enables us to calculate the probability of witnessing a defined number of "successes."

So, in the given situation we will find the probability with help of the binomial formula which is used to calculate the probability when independent events are given.

Therefore, using (D) the binomial formula, which is used to determine probability when independent events are known, we will get the probability in the scenario.

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Complete question:

We want to determine the probability of obtaining at most 4 successful operations in 10 independent surgical operations where the probability of success is the same for each operation. The appropriate formula to be used is

(A) the mean of the binomial distribution.

(B) the hypergeometric formula.

(C) the mean of the hypergeometric distribution.

(D) the binomial formula.

The sum of all natural numbers from 1 to 100 is 5050. The total number of natural numbers in this range is 100. So, by applying this value in the formula: S = n/2[2a + (n − 1) × d], we get S=5050.

In Java, finding the sum of two or more numbers is very easy. First, declare and initialize two variables to be added. Another variable to store the sum of numbers. Apply mathematical operator (+) between the declared variable and store the result. So you simply make this: sum=sum+num; for the cycle. For example sum is 0, then you add 5 and it becomes sum=0+5 , then you add 6 and it becomes sum = 5 + 6 and so on.

Thus, the sum of all natural numbers 1 to 10 can be calculated using the formula, S= n/2[2a + (n − 1) × d], where, a is the first term, d is the difference between the two consecutive terms, and n is the total number of natural numbers from 1 to 10. Therefore, the sum of the first ten natural numbers is 55.

public class T35{

  public static void main(String[] args) {

       int nmb;

               for(nmb= 1; nmb<= 100; nmb++){

                     System.out.println(nmb);  

       }      

  }

}

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This is not a two-variable linear equation.

If an equation is written in the form axe + by + c=0, where a, b, and c are real integers and the coefficients of x and y, i.e., When both a and b, in turn, do not equal zero, an equation with two variables is said to be linear. Such an equation has a pair of numbers as its solution, one for x and one for y, which further equalizes the two sides of the equation.

Given equation,

[tex]3x^{2} -7y=13[/tex]

In this case, the variable x's degree is 2.

Hence, this is not a two-variable linear equation.

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Answer:

2 feet

Step-by-step explanation:

(60 / 157.5) * 63 inches = 24 inches = 2 feet

Answer:

D) 8 kilograms

Step-by-step explanation:

If one science book has a mass of 2 kilograms, then 4 science books would have a mass of 2 x 4 = 8 kilograms.

The resulting value of the equation after performing integration is , f(x) = -log(x) + log(1) - log(4).

Integrating on both the sides with respect to x.

⇒  ∫ f''(x)dx= ∫x-2dx

⇒ ∫ f''(x) dx = (x-2 + 1)/(-2 + 1) + C1

⇒ f'(x) = x-1/(-1) + C1

Then again performing integration ,

∫ f'(x) dx = -log(x) + C1 + x + C2

f(x) = -log(x) + C1 + x + C2

Now, since we know that f(1) =0

when x = 1 then  f(1) =0

0 = -log(1) + C1 + C2

it is also given that ,f(4) = 0

0 = -log(4) + 4C1+ C2

We will subtract these two equations in order to get C1 and C2.

0 = -log(1) + log(4) - 3C1

3c1 = -log(1) + log(4)

= log(4/1) {since log(a) - log(b)

= log(a/b)}

C1 = log(4)/3

C1 = log(4)/3 and (2)

0 = -log(1) +(4/3)log(4) + C2

log(1) - (4/3)log(4) = C2

Therefore, the equation becomes

f(x) = -log(x) + [log(4)]/3 + log(1) - (4/3)log(4)

f(x) = -log(x) + log(1) - log(4)

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