How many numbers did Jane write?
Jane wrote all the natural numbers from m to k (where k>m).

The slope of  following points  (3, 3) and (6,1)  is -2/3 .

What is Linear equation ?

Linear equation can be defined as the equation in which the highest degree is one.

Given  points are (3, 3) and (6,1)

we know that slope = y2-y1 / x2-x1

y2-y1 = 1-3

The difference of y -intercept is -2

= -2

x2-x1 = 6-(3)

= 6-3

The difference of x-intercept is 3

= 3

So, slope could be y2-y1 / x2-x1  that is -2/3.

Therefore, The slope of  following points  (3, 3) and (6,1) is -2/3 .

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The slope of the least-squares line is:  0.30

The correct answer is an option (A)

We have following information:

correlation between X and Y (r) = 0.9

mean of the values of X ([tex]\bar{X}[/tex]) = 6.5

mean of the values of Y ([tex]\bar{Y}[/tex]) = 6.6

the standard deviation of the values of X ([tex]S_{X}[/tex]) = 3.6

the standard deviation of the values of Y  ([tex]S_{Y}[/tex]) = 1.2

We know that the slope of a least squares regression can be calculated by the formula,

slope m  = r [tex]\frac{S_\bar{Y}}{S_\bar{X}}[/tex],

where, r is the correlation coefficient

Let us assume that for given scenario,  m represents the slope of of the least-squares line.

Using above formula,

⇒ m  = r [tex]\frac{S_\bar{Y}}{S_\bar{X}}[/tex]

⇒ m  = 0.9 ×  [tex]\frac{1.2}{3.6}[/tex]

⇒ m = 0.9 ×  0.333

⇒ m = 0.2997

⇒ m ≈ 0.3

Thus, the slope is 0.30

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The complete question is:

A researcher wishes to study how the average weight Y (in kilograms) of children changes during the first year of life. He plots these averages versus the age X (in months) and decides to fit a least- squares regression line to the data with X as the explanatory variable and Y as the response variable. He computes the following quantities.

r = correlation between X and Y = 0.9

X = mean of the values of X = 6.5

Y = mean of the values of Y = 6.6

S_(X) = standard deviation of the values of X = 3.6

S_(Y) = standard deviation of the values of Y = 1.2

The slope of the least-squares line is: ?

A. 0.30   B. 0.88   C. 1.01   D. 2.70    E. 3.00

Answer:

4.725 ft

Step-by-step explanation:

well if i see this right the shadow for claire is 1.60x the person's hight

The arrow diagram that represents a mapping include the following: arrow diagram C.

The domain of this arrow diagram is equal to {a, c, d}

The range of this arrow diagram is equal to {q, r, s}.

In Mathematics, a domain can be defined as the set of all real numbers for which a particular function is defined.

This ultimately implies that, a domain is the set of all possible input numerical values (x-values) to a function and the domain of a graph comprises all the input numerical values (real numbers) which are located on the x-coordinate.

In this scenario, we can reasonably infer and logically deduce that only the above mentioned relation (arrow diagram C) represents a function because each input variable (domain) has exactly an output variable (range) as illustrated in the diagram.

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Answer: Let x be the weight of the bag over the limit in pounds.

We know that both airlines have the same weight limit and Carrie paid the same baggage fee at each airline, so we can set up the following equation:

25 + 4x = 28 + 3x

Solving for x:

x = 25

So, Carrie's bag was 25 pounds over the weight limit.

To find the fee for each airline, we can substitute x into the equations for the fees:

Fee for the first airline: 25 + 4(25) = $125

Fee for the second airline: 28 + 3(25) = $103

Therefore, Carrie paid $125 in fees for the first airline and $103 in fees for the second airline.

Step-by-step explanation:

Answer: - 15 = N

Step-by-step explanation:

2 negatives make a positive so it would be -35 + 20 = -15 Have a nice day ;)

The output column, S, of the given input-output table was created from the student ID 21444368 using an algorithm where if the student ID digit is an even integer, a 0 is placed in the corresponding row of column S, and if the student ID digit is an odd integer, a 1 is placed in the corresponding row of column S.

Student ID: 81725225

P Q R S

1 1 1 0

1 1 0 1

1 0 1 1

0 1 0 0

0 0 1 1

0 1 1 0

0 0 0 1

Student ID: 81725225

P: The first digit of my student ID is 8 so it is an even integer, therefore I place a 0 in row 1 of column P.

Q: The second digit of my student ID is 1 so it is an odd integer, therefore I place a 1 in row 2 of column Q.

R: The third digit of my student ID is 7 so it is an odd integer, therefore I place a 1 in row 3 of column R.

S: The fourth digit of my student ID is 2 so it is an even integer, therefore I place a 0 in row 4 of column S.

The resulting table is:

P Q R S

1 1 1 0

1 1 0 1

1 0 1 1

0 1 0 0

0 0 1 1

0 1 1 0

0 0 0 1

The output column, S, of the given input-output table was created from the student ID 21444368 using an algorithm where if the student ID digit is an even integer, a 0 is placed in the corresponding row of column S, and if the student ID digit is an odd integer, a 1 is placed in the corresponding row of column S. The same algorithm was applied to my student ID 81725225, resulting in the table shown above.

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Answer: As Mrs Lopez required 40 ounce

Step-by-step explanation:

The equation for finding cost y for x liters of petrol is y = 93x

An equation is a mathematical statement that shows that two mathematical expressions are equal.

Given that, A liter of petrol cost £93,

Since, the unit rate of the petrol is £93 per liter

If we say let x be the number of liters we need, the total cost will have to pay will be = 93x

And, if total cost is y, then,

y = 93x

Hence, the equation for finding cost y for x liters of petrol is y = 93x

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The largest δ value of the given limit [tex]\lim_{x \to 1^+} \frac{1}{x^2-1}=\infty[/tex] and M=1000 is calculated. The required δ value is [tex]\delta=\sqrt{1+10^{-3}}-1[/tex].

When no values are defined, a limit is used to assign values to certain functions. It is possible to assess the limits using algebraic techniques. A collection of guidelines for manipulating limits when using other operators is known as the algebra of limits. Given the expression  [tex]\lim_{x \to 1^+} \frac{1}{x^2-1}=\infty[/tex]  and M=1000. When x>1, we write this expression as,

[tex]\begin{aligned} \frac{1}{x^2-1}& > 1000\\x^2-1& < 1000^{-1}=10^{-3}\\x^2& < 10^{-3}+1\\x& < \sqrt{1+10^{-3}}\end{aligned}[/tex]

Now, choosing, [tex]\delta=\sqrt{1+10^{-3}}-1[/tex] when we have δ > 0. Therefore, [tex]\sqrt{1+10^{-3}} > 1[/tex]. Then, [tex]x\in(1, 1+\delta)[/tex]. This is given by,

[tex]\begin{array}{ccc}1& < x& < 1+\delta\\1& < x& < \sqrt{1+10^{-3}}\end{array}[/tex]

This implies, [tex]\frac{1}{x^2-1} > 1000=M[/tex].

Therefore, the largest δ value will be [tex]\sqrt{1+10^{-3}}-1[/tex].

The complete question is -

For each limit, use graphs and algebra to approximate the largest δ or the smallest magnitude N that corresponds to the given value of ∈ or M, according to the appropriate formal limit definition.

[tex]\lim_{x\to 1^+} \frac{1}{x^2-1}=\infty[/tex], M=1000, find largest δ > 0.

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Answer:

What is the question please respond so I may give a complete response

The solution of the given system of equations is (4, 7).

A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.

The given system of equations are y=3x-5 -------(I) and y=1/2 x+5 -------(II)

Substitute equation (I) in (II), we get

3x-5=1/2 x+5

3x-1/2 x=5+5

2.5x=10

x=4

Substitute x=4 in equation (I), we get

y=3(4)-5

y=7

So, solution is (4, 7)

Therefore, the solution of the given system of equations is (4, 7).

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Answer:

ii) Reduce :

[tex]$\begin{aligned} 5 \ln x-2 \ln (x y) & =\ln x^5-\ln (x y)^2 \\ & =\ln \left(\frac{x^5}{(x y)^2}\right) \\ & =\ln \left(\frac{x^5}{x^2 y^2}\right) \\ & =\ln \left(\frac{x^3}{y^2}\right)\end{aligned}$[/tex]

iii ) Expand :

[tex]\begin{aligned}\log _3\left(\frac{x^5}{y}\right)^4 & =4 \log _3\left(\frac{x^5}{y}\right) \\& =4 \log _3 x^5-4 \log _3 y \\& =20 \log _3 x-4 \log _3 y\end{aligned}[/tex]

Answer:

[tex]\textsf{ii)} \quad \ln \left(\dfrac{x^3}{y^2}\right)[/tex]

[tex]\textsf{iii)} \quad 20 \log_3 x-4 \log_3y[/tex]

Step-by-step explanation:

Given expression:

[tex]5 \ln x-2 \ln (xy)[/tex]

[tex]\textsf{Apply the power law}: \quad n \ln x=\ln x^n[/tex]

[tex]\implies \ln (x^5) - \ln ((xy)^2)[/tex]

[tex]\textsf{Apply the quotient law}: \quad \ln x - \ln y=\ln \left(\dfrac{x}{y}\right)[/tex]

[tex]\implies \ln \left(\dfrac{x^5}{(xy)^2}\right)[/tex]

This is where the error occurred in the original calculation as the log quotient law was applied incorrectly.

[tex]\textsf{Apply exponent rule} \quad (ab)^c=a^{c}b^c\quad \textsf{to the denominator}:[/tex]

[tex]\implies \ln \left(\dfrac{x^5}{x^2y^2}\right)[/tex]

Simplify the fraction:

[tex]\implies \ln \left(\dfrac{x^3}{y^2}\right)[/tex]

Given expression:

[tex]\log_3\left(\dfrac{x^5}{y}\right)^4[/tex]

[tex]\textsf{Apply the log power law}: \quad \log_ax^n=n\log_ax[/tex]

[tex]\implies 4\log_3\left(\dfrac{x^5}{y}\right)[/tex]

[tex]\textsf{Apply the log quotient law}: \quad n\log_a \left(\dfrac{x}{y}\right)=n\log_ax - n\log_ay[/tex]

[tex]\implies 4 \log_3 x^5-4 \log_3y[/tex]

This is where the error occurred in the original calculation as the coefficient 4 was not applied to the both logarithms when applying the log quotient law.

[tex]\textsf{Apply the log power law}: \quad \log_ax^n=n\log_ax[/tex]

[tex]\implies 5 \cdot 4 \log_3 x-4 \log_3y[/tex]

[tex]\implies 20 \log_3 x-4 \log_3y[/tex]

The solution to this equation: 42 +82=x² is x = 12.

A square number, also known as a perfect square, is an integer that is the square of another integer; in other words, it is the result of multiplying another integer by itself. For instance, 9 is a square number because it equals 32 and can be written as 3 3.

Instead of the product n n, the equivalent exponentiation n2, which is typically pronounced "n squared," is used to represent a number's square. The name of the shape is where the term "square number" originated. The area of a unit square (1 × 1) is the unit of area.

The solution to this equation: 42 +82=x²

x = √(42 +82)

x = √(144)

x = 12

Thus, The solution to this equation: 42 +82=x² is x = 12.

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X has a geometric distribution because it is the number of trials (tosses) required until a success (catching the ball) occurs.

b) The probability that it will take exactly two tosses for Sophie to catch a ball is P(x=2) = (0.2)(0.8)^1 = 0.16. This is calculated by multiplying the probability of success (0.2) by the probability of failure (0.8) raised to the power of the number of trials (2-1=1)

c) To find the probability that more than three tosses will be required, we need to find the probability that Sophie does not catch the ball on the first three tosses and then catch it on the fourth toss.

This can be found by summing the probabilities of Sophie not catching the ball on the first three tosses (0.8^3 = 0.512) and Sophie catching the ball on the fourth toss (0.8^3*0.2 = 0.1024). The probability that more than three tosses will be required is P(x>3) = 1 - P(x≤3) = 1 - (0.512 + 0.1024) = 0.3856.

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Answer: 3 11/12

Step-by-step explanation:

Rewrite the equation with the parts seperated:

4 + 2/3 − 3/4

Find the LCD of 2/3 and 3/4

(Which is 12)

Keep going

8/12−9/12= −1/12

Finally, subtract 4 − 1/12 to get 3 11/12

Hope this helps!

The images of the points after the transformation are (6, 1), (3, 6), (5, 2) and (-8, -4)

r(x-axis) o T(2, -3) (4, 2)

This implies that we reflect the point (4, 2) across the x-axis and then translate it by (x + 2, y - 3)

Mathematically, this is represented as

Image = (x + 2, 3 - y)

Substitute the known values in the above equation, so, we have the following representation

Image = (4 + 2, 3 - 2)

Evaluate

Image = (6, 1)

T(2, -5) o R 90 CCW (4, 2)

This implies that we translate the point (4, 2) by (x + 2, y - 5) and then rotate it by 90 degrees counterclockwise

Mathematically, this is represented as

Image = (5 - y, x + 2)

Substitute the known values in the above equation, so, we have the following representation

Image = (5 - 2, 4 + 2)

Evaluate

Image = (3, 6)

T(1, 4) o r(y-axis) (4, 2)

This implies that we translate the point (4, 2) by (x + 1, y + 4) and then reflect it across the y-axis

Mathematically, this is represented as

Image = (x + 1, 4 - y)

Substitute the known values in the above equation, so, we have the following representation

Image = (4 + 1, 4 - 2)

Evaluate

Image = (5, 2)

r(y-axis) o T(1, 4) (4, 2)

This implies that we reflect the point (4, 2) across the y-axis and then translate it by (x + 4, y + 2)

Mathematically, this is represented as

Image = (-x - 4, y + 2)

Substitute the known values in the above equation, so, we have the following representation

Image = (- 4 - 4, 2 + 2)

Evaluate

Image = (-8, 4)

Hence, the image is (-8, -4)

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The measure of the missing angles are 100° and 30°

A polygon with three vertices, angles and sides.

Given is a triangle XYW, whose side WZ is extended to the point z,

We know that,

The sum of two opposite interior angles of a triangle is equal to the measure of exterior angle,

∠ 1 + 50° = 150°

∠ 1 = 100°

We know that, sum of the interior angles of a triangle is 180°

∠ 1  + ∠ 2 + 50° = 180°

150°-180° = ∠ 2

∠ 2 = 30°

Hence, the measure of the missing angles are 100° and 30°

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If the calculated Cp and Cpk values are greater than 1, the process is considered capable. If Pp and Ppk are greater than 1, the process is considered capable.

To calculate Cp, Cpk, Pp, and Ppk using MINITAB, you would need to input the data into the software and run the appropriate statistical analysis.

Cp and Cpk are measures of a process's capability, they are calculated as follows:

Where,

Pp and Ppk are similar to Cp and Cpk, but they are calculated using the range of the data instead of the standard deviation.

Where R is the range of the data.

If the calculated Cp and Cpk values are greater than 1, the process is considered capable. If Pp and Ppk are greater than 1, the process is considered capable.

It is worth noting that Capability analysis can only be performed on stable and in-control processes. If the process is not in control, analyzing process capability is not meaningful.

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The equation of plane is x - 3y - 4z = 27.

a plane produced by the application of vectors perpendicular to a normal. The implication of this is that, given a vector a, b, c, we may deduce that all planes perpendicular to it have the form ax+by+cz=d and that any surface with this form is a plane perpendicular to a, b, c.

The equation for a plane perpendicular to a given line with direction ratios a, b, and c> and passing through a point is a(x-x1) + b(y-y1) + c(z-z1) = 0. (x1, y1, z1).

The equation of the plane traveling through the points x, y, and z on a plane perpendicular to the line whose parametric equation is provided

Equation of a plane through the point (-4,-3,8) is

a(x - 4) + b(y - 3) + c(z + 8) = 0

Since it is perpendicular to the line :

x = -4+t, y=-3-3t, z= -4t,

t = x + 4 = -y - 3 / 3 = -z/4

therefore coefficients a, b, c must be proportional to 1, - 3 & (-4) respectively.

Therefore, putting the values of a, b & c in eq.(1), we get the required equation of the plane as ;

λ(x - 4) -3 λ(y - 3) - 4λ(z + 8) = 0 (λ is a constant)

⇒ x - 4 - 3y + 9 -4z - 32 = 0

⇒ x - 3y - 4z = 27

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The solution for the system of equations 8x + 3y = 4 and 2x + y = 2 by elimination are x = -1, y = 4

we shall write the equations as:

8x + 3y = 4...(1)

2x + y = 2...(2)

multiply equation (2) by 3 to get equation (3)

3(2x + y) = 3 × 2

6x + 3y = 6...(3)

subtract equation (1) from (3) to eliminate y

6x + 3y - 8x - 3y = 6 - 4

-2x = 2

divide through by -2

x = -1

put the value -1 for x in equation (2) to get y

2(-1) + y = 2

-2 + y = 2

y = 2 + 2 {add 2 to both sides}

y = 4

Therefore, the solution for the system of equations 8x + 3y = 4 and 2x + y = 2 by elimination are x = -1, y = 4.

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Step-by-step explanation:

If a company is expanding its team at a rate of 10% per year, and it currently has 200 members of staff, we can calculate the number of staff after 1 year using the formula:

number of staff after 1 year = 200 + (10/100) * 200

number of staff after 1 year = 200 + 20

number of staff after 1 year = 220

So, there will be 220 members of staff after 1 year if there are initially 200 members of staff.

Answer:

Step-by-step explanation:

tha answer is 69

Note that the figure with an angle of 300° on a circle in standard position is: "An angle in standard position with the terminal side at two-thirds of a full revolution counterclockwise." (Option D)

A standard position angle is one measured counterclockwise from the positive x-axis in a coordinate plane.

This signifies that the angle's starting side is on the positive x-axis, while the angle's terminal side is in the first or fourth quadrant. The angle's vertex is placed at the origin.

Angles are commonly measured in degrees or radians, with a full rotation being comparable to 360 degrees or 2 radians. Angles may now be readily graphed and compared to one another in a coordinate plane.

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Answer:

he takes a stretch every 4 miles

Step-by-step explanation: 12/3 = 4

Answer:

16.58

Step-by-step explanation:

as you can see from the graph, there's a monthly fixed cost on Y axis.  so you need to figure out 1: how much per minute. and 2. the monthly fixed fee.  you've got 2 scenarios = 2 equations, for 2 variables. so you can solve both.

assume: x=per min rate; m = monthly fixed cost.

1. 14.06 =52x+m

2. 17.18 =78x+m

use equation 2 minus equation 1 & you get

3.12 = 26 x.  so x = 0.12

plug this into equation 1: 14.06 =52*0.12 +m so m=7.82

now for 73 mins the cost = 73*0.12 +7.82 =16.58

Here we are interested in finding the slope of the line. From the given graph we can see that that the line passes through points (6,0) and (0,-8) . We know that slope can be calculated by ,

[tex]\longrightarrow slope \ (m)=\dfrac{y_2-y_1}{x_2-x_1} =\tan\theta [/tex]

[tex]\longrightarrow m =\dfrac{y_2-y_1}{x_2-x_1}\\[/tex]

[tex]\longrightarrow m =\dfrac{0-(-8)}{6-0}\\[/tex]

[tex]\longrightarrow m =\dfrac{8}{6}\\ [/tex]

[tex]\longrightarrow m =\dfrac{4}{3}\\ [/tex]

[tex]\longrightarrow \underline{\underline{ m = 1.\bar{3}}}[/tex]

And we are done!