the sum of two consecutive numbers divided by their positive difference difference is equal to nine find the larger number​

1. the z* values based on a standard normal distribution (a) z* = 1.28, (b) z* = 1.39, and (c) z* = 1.75. 2. the z* values based on a standard normal distribution (a) z* = 1.44, (b) z* = 1.64, (c) z* = 2.05

1. (a) For an 80% confidence interval for a proportion, we need to find the z* value that cuts off 10% in each tail. Using a standard normal table or calculator, we find that z* = 1.28.
  (b) For an 82% confidence interval for a slope, we need to find the z* value that cuts off 9% in each tail. Using a standard normal table or calculator, we find that z* = 1.39.
  (c) For a 92% confidence interval for a standard deviation, we need to find the z* value that cuts off 4% in each tail. Using a standard normal table or calculator, we find that z* = 1.75.
2. (a) For an 86% confidence interval for a correlation, we need to find the z* value that cuts off 7% in each tail. Using a standard normal table or calculator, we find that z* = 1.44.
   (b) For a 90% confidence interval for a difference in proportions, we need to find the z* value that cuts off 5% in each tail. Using a standard normal table or calculator, we find that z* = 1.64.
   (c) For a 96% confidence interval for a proportion, we need to find the z* value that cuts off 2% in each tail. Using a standard normal table or calculator, we find that z* = 2.05.

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The 95% confidence interval for the mean time for all players is given as follows:

(8.1, 10.9).

The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the equation presented as follows:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

The variables of the equation are listed as follows:

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 10 - 1 = 9 df, is t = 2.2622.

The parameters are given as follows:

[tex]\overline{x} = 9.5, n = 10, s = 1.98[/tex]

The lower bound of the interval is given as follows:

[tex]9.5 - 2.2622 \times \frac{1.98}{\sqrt{10}} = 8.1[/tex]

The upper bound is given as follows:

[tex]9.5 + 2.2622 \times \frac{1.98}{\sqrt{10}} = 10.9[/tex]

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The value of x is 13

To determine the value of the variable, we need to know the properties of a triangle;

These properties are;

From the information given, we have that;

The angles given are;

Angle 59

Angle 79

Angle 2x + 16

Now, equate the angles, we have;

59 + 79 + 2x + 16 = 180

collect the like terms, we have;

2x = 180 - 154

subtract the values

2x = 26

x = 13

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The family of curves with polar equations r = 1 c cos() can be rewritten as x = c cos() and y = c sin(), using the conversion formulas r cos() = x and r sin() = y. This means that each curve in the family is a circle centered at the origin with radius c, rotated by an angle of 90 degrees.

As c changes, the radius of each circle changes, and therefore the size of each circle changes. If c is positive, the circle will be in the first and third quadrants of the Cartesian plane, and if c is negative, the circle will be in the second and fourth quadrants.

When c = 0, we get a circle of radius 0, which is just the single point at the origin. This makes sense, since cos(0) = 1 and all other values of cos() are between -1 and 1, so the equation r = 1 c cos() can only be satisfied when c = 0 if cos() = 0, which occurs only at = 0 and = pi.

In summary, as c changes, the family of curves with polar equations r = 1 c cos() changes in size and position, but remains circular in shape. When c = 0, the curve is just a single point at the origin.

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Answer: false

Step-by-step explanation:

The value of the algebraic expression is -4m³ -22m² +28m²n +3mn -4

In mathematics, an algebraic expression is an expression built up from constant algebraic numbers, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, 3x² − 2xy + c is an algebraic expression.

The given expression is

−12m2 +8m3 +3mn−12m3n2 −2+14m2n+6m3n2 −11m2 −2+14m2n+6m3n2 −11m2

Rearrange this by collecting the like terms to have

8m³ - 12m² -11m² -11m² -12m³n² +6m³n² +6m³n² +14m²n + 14m²n + 3mn -2 -2

Simplify further to have

-4m³ -22m² +28m²n +3mn -4

In conclusion the expression gives -4m³ -22m² +28m²n +3mn -4

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The value of the range of function y = sin x is,

⇒ Range = -1 ≤ y ≤ 1

Since, A relation between sets of inputs which having exactly one output each is called a function.

And, an expression, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

Here, The function is,

y = sin x

Now, We know that;

The range of y = sin x is,

⇒ Range = -1 ≤ x ≤ 1

Hence, Option A is true.

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A. Matrix form as follows: Y = Xβ + ε

B. This is because one column of X can be expressed as a linear combination of the other column, resulting in a rank of 1.

C. β1 =[tex](X1^2 + X2^2)^{-1 }(X1Y) / (X1^2 + X2^2)^{-1} (X1^2)[/tex]

β2 = [tex](X1^2 + X2^2)^{-1} (X2Y) / (X1^2 + X2^2)^{-1} (X2^2)[/tex]

A. The multiple regression model with two independent variables and no intercept can be represented in matrix form as follows:

Y = Xβ + ε

where,

Y is an n x 1 vector of dependent variable observations

X is an n x 2 matrix of independent variable observations

β is a 2 x 1 vector of regression coefficients for the independent variables

ε is an n x 1 vector of errors

The matrix X contains the independent variable observations for each of the n observations, with each row representing an observation, and each column representing a different independent variable.

B. The rank of X is determined by the number of linearly independent rows in X. Since there is no intercept in the model, the columns of X are centered around zero, and therefore, the rank of X cannot exceed 1. This is because one column of X can be expressed as a linear combination of the other column, resulting in a rank of 1.

C. The least square estimators of β can be calculated as:

β = [tex](X^T X)^-1 X^T Y[/tex]

where, X^T is the transpose of X and[tex](X^T X)^-1[/tex] is the inverse of the matrix product of [tex]X^T[/tex] and X.

Expanding this formula, we get:

β =[tex][(X1^2 + X2^2) X1 X2]^-1 [X1 X2] [Y][/tex]

where,

X1 and X2 are n x 1 vectors representing the two independent variables in the model

[tex]X1^2[/tex] and[tex]X2^2[/tex]are n x n diagonal matrices with the squares of the elements of X1 and X2, respectively

[X1 X2] is an n x 2 matrix containing the two independent variables

[Y] is an n x 1 vector of dependent variable observations

Simplifying this expression, we get:

β1 =[tex](X1^2 + X2^2)^{-1 }(X1Y) / (X1^2 + X2^2)^{-1} (X1^2)[/tex]

β2 = [tex](X1^2 + X2^2)^{-1} (X2Y) / (X1^2 + X2^2)^{-1} (X2^2)[/tex]

where,

β1 is the least square estimator for the regression coefficient of X1

β2 is the least square estimator for the regression coefficient of X2

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The solution to the given initial-value problem is:

[tex]x(t) = $\frac{1}{2}$e$^{-2t}$ $\begin{bmatrix}2\ -1\end{bmatrix}$ + $\frac{3}{2}$e$^{5t}$ $\begin{bmatrix}1\ 3\end{bmatrix}$ + $\begin{bmatrix}2\ -1\end{bmatrix}$[/tex].

To solve the given initial-value problem:

[tex]x' = $\begin{bmatrix}-1 & -2\ 3 & 4\end{bmatrix}$x + $\begin{bmatrix}5\ 5\end{bmatrix}$, x(0) = $\begin{bmatrix}-3\ 7\end{bmatrix}$[/tex]

First, we find the solution to the homogeneous system:

[tex]x' = $\begin{bmatrix}-1 & -2\ 3 & 4\end{bmatrix}$x[/tex]

The characteristic equation is:

[tex]|$\begin{bmatrix}-1-\lambda & -2\ 3 & 4-\lambda\end{bmatrix}$| = $\lambda^2-3\lambda-10 = 0$[/tex]

Solving the above quadratic equation, we get:

[tex]\lambda_1 = -2$ and $\lambda_2 = 5$[/tex]

The corresponding eigenvectors are:

[tex]v_1 = $\begin{bmatrix}2\ -1\end{bmatrix}$ and v_2 = $\begin{bmatrix}1\ 3\end{bmatrix}$[/tex]

Therefore, the general solution to the homogeneous system is:

[tex]xh(t) = c1e$^{-2t}$ $\begin{bmatrix}2\ -1\end{bmatrix}$ + c2e$^{5t}$ $\begin{bmatrix}1\ 3\end{bmatrix}$[/tex]

Next, we find the particular solution to the non-homogeneous system. We assume the solution to be of the form:

xp(t) = A

Substituting this in the given equation, we get:

[tex]A = $\begin{bmatrix}-1 & -2\ 3 & 4\end{bmatrix}$A + $\begin{bmatrix}5\ 5\end{bmatrix}$[/tex]

Solving for A, we get:

[tex]A = $\begin{bmatrix}2\ -1\end{bmatrix}$[/tex]

Therefore, the particular solution is:

[tex]xp(t) = $\begin{bmatrix}2\ -1\end{bmatrix}$[/tex]

The general solution to the non-homogeneous system is given by:

[tex]x(t) = xh(t) + xp(t) = c1e$^{-2t}$ $\begin{bmatrix}2\ -1\end{bmatrix}$ + c2e$^{5t}$ $\begin{bmatrix}1\ 3\end{bmatrix}$ + $\begin{bmatrix}2\ -1\end{bmatrix}$[/tex]

Using the initial condition [tex]x(0) = $\begin{bmatrix}-3\ 7\end{bmatrix}$,[/tex]we get:

[tex]c_1$\begin{bmatrix}2\ -1\end{bmatrix}$ + c_2$\begin{bmatrix}1\ 3\end{bmatrix}$ + $\begin{bmatrix}2\ -1\end{bmatrix}$ = $\begin{bmatrix}-3\ 7\end{bmatrix}$[/tex]

Solving for c₁ and c₂, we get:

[tex]c_1 = $\frac{1}{2}$ and c_2 = $\frac{3}{2}$[/tex]

Therefore, the solution to the given initial-value problem is:

[tex]x(t) = $\frac{1}{2}$e$^{-2t}$ $\begin{bmatrix}2\ -1\end{bmatrix}$ + $\frac{3}{2}$e$^{5t}$ $\begin{bmatrix}1\ 3\end{bmatrix}$ + $\begin{bmatrix}2\ -1\end{bmatrix}$.[/tex]

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If the probability of recovering from a stomach disease is 0.7, then the probability of not recovering is 0.3.

Out of 20 people who contracted the disease, the probability that any one person will recover is 0.7.

To calculate the probability that all 20 people will recover, we need to multiply 0.7 by itself 20 times (0.7^20), which equals 0.00079792266.

This means that there is less than 1% chance that all 20 people will recover from the disease.

On the other hand, the probability that at least one person will not recover is the same as the probability of not all 20 people recovering, which is 1-0.00079792266, or approximately 0.999.

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Answer: Point-slope form equation:

Using the point-slope form equation, which is y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope, we can substitute the given values to find the equation.

Point = (-9, 3)

Slope = -2/3

Using the point-slope form equation:

y - 3 = (-2/3)(x - (-9))

Simplifying:

y - 3 = (-2/3)(x + 9)

Expanding:

y - 3 = (-2/3)x - 6

Rearranging:

y = (-2/3)x - 3

Therefore, the equation of the line is y = (-2/3)x - 3.

Parallel to y = 5x - 2:

The parallel line will have the same slope (5) as the given line because parallel lines have the same slope. The y-intercept is given as -3.

Using the slope-intercept form equation, which is y = mx + b, where m is the slope and b is the y-intercept, we can substitute the given values to find the equation.

Slope = 5

Y-intercept = -3

Therefore, the equation of the line is y = 5x - 3.

Perpendicular to y = (1/2)x + 1:

To find the perpendicular line, we need to take the negative reciprocal of the slope (1/2). The negative reciprocal of a number is obtained by flipping the fraction and changing the sign.

The given line has a slope of 1/2, so the perpendicular line will have a slope of -2 (negative reciprocal of 1/2). The y-intercept is given as 9.

Using the slope-intercept form equation, which is y = mx + b, where m is the slope and b is the y-intercept, we can substitute the given values to find the equation.

Slope = -2

Y-intercept = 9

Therefore, the equation of the line is y = -2x + 9.

The producer surplus at the equilibrium quantity is 5488/3 or approximately 1829.33.

The equilibrium quantity is found by setting the demand equal to the supply:

862.4 - 0.6x² = 0.5x²

Simplifying and solving for x, we get:

1.1x² = 862.4

x² = 784

x = 28

So the equilibrium quantity is 28.

The producer surplus at the equilibrium quantity, we first need to find the equilibrium price.

The demand or supply function to do this and since the supply function is simpler, we'll use that:

s(28) = 0.5(28)²

= 196

So the equilibrium price is 196.

The producer surplus at the equilibrium quantity is the area above the supply curve and below the equilibrium price, up to the quantity of 28. The supply curve is a quadratic function can find this area using integration:

∫[0,28] (196 - 0.5x²) dx

= [196x - (0.5/3)x³] from 0 to 28

= (5488/3)

= 1829.33.

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Answer: To find the area of one loop of the rose r = 3 cos(3θ), we can use the formula:

A = 1/2 ∫θ2 θ1 (f(θ))^2 dθ

where f(θ) is the function that defines the curve, and θ1 and θ2 are the angles that define one loop of the curve.

In this case, the curve completes one loop when θ goes from 0 to π/6 (or from π/6 to π, since the curve is symmetric about the y-axis). Therefore, we can compute the area as:

A = 1/2 ∫0^(π/6) (3cos(3θ))^2 dθ

A = 9/2 ∫0^(π/6) cos^2(3θ) dθ

Using the identity cos^2(θ) = (1 + cos(2θ))/2, we can simplify this to:

A = 9/4 ∫0^(π/6) (1 + cos(6θ)) dθ

A = 9/4 (θ + sin(6θ)/6) ∣∣0^(π/6)

A = 9/4 (π/6 + sin(π)/6)

A = 3π/8 - 3√3/8

Therefore, the area of one loop of the rose r = 3 cos(3θ) is 3π/8 - 3√3/8.

The odds that are in favour of guessing the right answer would be = 1:5. That is option A.

The given multiple choice questions has only 5 possible answers.

This means that when both the correct and wrong answers are added together, the total should be = 5.

That is;

4:1 = 4+1 = 5

1:4 = 1+4 = 5

3:2 = 3+2 = 5

Therefore, 1:5 = 1+5 = 6 which can't be a possible answer as it's more than the total of the multiple choice questions.

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The line integral ∫e^t dx along the line Segment C is equal to 0.

(a) To evaluate the line integral ∫ds along the line segment C from (0,2) to (0,4), we can use the formula for the arc length of a curve in two dimensions.

The formula for the arc length of a curve defined by a vector-valued function r(t) = (x(t), y(t)) on an interval [a, b] is given by:

L = ∫ √(dx/dt)^2 + (dy/dt)^2 dt

In this case, since the line segment C is a straight line parallel to the y-axis, the x-coordinate remains constant at x = 0. Therefore, dx/dt = 0 for all t.

The y-coordinate varies from y = 2 to y = 4 along C, so dy/dt = 2. Integrating √(dx/dt)^2 + (dy/dt)^2 over the interval [a, b] where a and b are the parameter values corresponding to the endpoints of C, we get:

∫ds = ∫ √(dx/dt)^2 + (dy/dt)^2 dt

= ∫ √0 + 2^2 dt

= ∫ 2 dt

= 2t + C

Evaluating this integral over the interval [a, b] = [0, 1], we get:

∫ds = 2t ∣[0,1]

= 2(1) - 2(0)

= 2

Therefore, the line integral ∫ds along the line segment C is equal to 2.

(b) To evaluate the line integral ∫e^t dx along the line segment C, we can use the fact that dx = 0 since the x-coordinate remains constant at x = 0 Therefore, ∫e^t dx = ∫e^t * 0 dt = 0.

Hence, the line integral ∫e^t dx along the line segment C is equal to 0.

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The number of weeks in which Sarah and Asher have saved the same amount of money is 8 weeks.

The number of weeks in which Sarah and Asher have saved the same amount of money is calculated by setting the two equations equal to each other as follows;

12x + 6 = 9x + 30

Simplify the equation by collecting similar terms as follows;

12x - 9x  = 30 - 6

3x = 24

x = 24/3

x = 8

Thus, the number of weeks in which Sarah and Asher have saved the same amount of money is 8 weeks.

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The curl of the vector field F is (cos x - e^x sin z, cos y - e^y sin x, cos z - e^z sin y). The divergence of the vector field F is 0.

To find the curl of the vector field F(x, y, z) = (e^x sin y, e^y sin z, e^z sin x), we use the formula for curl:

curl(F) = (∂Fz/∂y - ∂Fy/∂z, ∂Fx/∂z - ∂Fz/∂x, ∂Fy/∂x - ∂Fx/∂y).

Calculating the partial derivatives:

∂Fz/∂y = e^z cos x, ∂Fy/∂z = e^y cos z,

∂Fx/∂z = e^x cos z, ∂Fz/∂x = e^z cos y,

∂Fy/∂x = e^y cos x, ∂Fx/∂y = e^x cos y.

Substituting these values into the curl formula, we get:

curl(F) = (e^z cos x - e^y cos z, e^x cos z - e^z cos y, e^y cos x - e^x cos y).

Simplifying further, we have:

curl(F) = (cos x - e^x sin z, cos y - e^y sin x, cos z - e^z sin y).

To find the divergence of the vector field F, we use the formula for divergence:

div(F) = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z.

Calculating the partial derivatives:

∂Fx/∂x = e^x sin y, ∂Fy/∂y = e^y sin z, ∂Fz/∂z = e^z sin x.

Adding these values together, we get:

div(F) = e^x sin y + e^y sin z + e^z sin x.

Simplifying further, we have:

div(F) = 0.

Therefore, the divergence of the vector field F is 0.

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The wavelength λ1 for gamma rays of frequency f1 = 5.50×1021 Hz is 5.45 × 10-14 m.

To calculate the wavelength λ1 for gamma rays of frequency f1 = 5.50×1021 Hz, we can use the formula:
λ1 = c/f1

where c is the speed of light in a vacuum, which is approximately 3.00 × 108 m/s.
Substituting the values given, we get:

λ1 = 3.00 × 108 m/s / 5.50 × 1021 Hz
λ1 = 5.45 × 10-14 m

Therefore, the wavelength λ1 for gamma rays of frequency f1 = 5.50×1021 Hz is 5.45 × 10-14 m.

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'Among the given options, the parametric tests are: t test and F test. Option D

Parametric tests assume that the data follows a specific distribution, usually a normal distribution, and make assumptions about the population parameters such as mean and variance. The t test is used to compare means between two groups, and the F test is used for comparing variances or testing the overall significance of a regression model.

The t test and F test are examples of parametric tests. They are used to analyze data that meets certain assumptions, such as normality and homogeneity of variance.

These tests are appropriate when the data follows a specific distribution, such as the normal distribution. They are commonly used to compare means or variances between groups.

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In a 2 x 3 between subjects ANOVA, there are a total of 6 groups. The first factor, with 2 levels, divides the participants into two distinct groups. The second factor, with 3 levels, further divides each of the two groups into three subgroups. This results in a total of 6 groups.

In this design, each group consists of a unique combination of the two factors, ensuring that each participant is assigned to only one group.

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Therefore, the kinetic energy of the bird is 4.05 J.

The kinetic energy of a 0.10 kg bird that is flying at a constant speed of 9.0 m/s is 4.05 J.

Kinetic energy (KE) is the energy possessed by a moving object.

It is the energy required to bring an object of mass m moving at a velocity v to rest.

Kinetic energy (KE) is represented by the formula

KE = 1/2mv².KE = 1/2 x m x v²

Given: mass (m) = 0.10 kg

velocity (v) = 9.0 m/s

KE = 1/2 x m x v²

KE = 1/2 x 0.10 kg x (9.0 m/s)²

KE = 1/2 x 0.10 kg x 81 m²/s²KE = 4.05 J  

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The given parametric equations model the position of an object in circular motion with radius 4 and center at the origin.

The given parametric equations are:

x = 4 cos(t)

y = -4 sin(t)

To understand the motion of the object, we can plot its position in the xy-plane as t varies.

The equation x = 4 cos(t) represents the horizontal position of the object, which varies between -4 and 4 as t varies between 0 and 2π. The equation y = -4 sin(t) represents the vertical position of the object, which varies between -4 and 4 as t varies between 0 and 2π.

Thus, the object moves in a circle of radius 4 centered at the origin, in a counterclockwise direction, completing one revolution in 2π seconds.

We can also find the equation of the circle in Cartesian form by eliminating t from the given equations. Squaring both equations and adding, we get:

x^2 + y^2 = 16

which is the equation of a circle with radius 4 centered at the origin.

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The equivalent ratio to 4 to 6 is 2 to 3.

Student 1: 8 to 12, Student 2: 2 to 3,  Student 3: 10 to 15, Student 4: 40 to 60. The ratio of blue pens to black pens on a teacher's desk is 4 to 6. If we add 4 and 6, we get 10. This means that for every 10 pens, 4 of them are blue and 6 of them are black. We can write this ratio as 4:6 or as a fraction 4/10, which can be simplified to 2/5.To write an equivalent ratio, we need to multiply the numerator and the denominator of the original ratio by the same number. We can multiply both by 2, to get the equivalent ratio of 8:12 or simplify it to 2:3, which is Student 2's answer. Therefore, the equivalent ratio to 4 to 6 is 2 to 3.

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The work done by F over the curve in the direction of increasing t is 3π.

To find the work done by a force vector F over a curve r(t) in the direction of increasing t, we need to evaluate the line integral:

W = ∫ F · dr

where the dot denotes the dot product and the integral is taken over the curve.

In this case, we have:

F = 2y i + 3x j + (x + y) k

r(t) = cos t i + sin t j + tk, 0 ≤ t ≤ 2π

To find dr, we take the derivative of r with respect to t:

dr/dt = -sin t i + cos t j + k

We can now evaluate the dot product F · dr:

F · dr = (2y)(-sin t) + (3x)(cos t) + (x + y)

Substituting the expressions for x and y in terms of t:

x = cos t

y = sin t

We obtain:

F · dr = 3cos^2 t + 2sin t cos t + sin t + cos t

The line integral is then:

W = ∫ F · dr = ∫[0,2π] (3cos^2 t + 2sin t cos t + sin t + cos t) dt

To evaluate this integral, we use the trigonometric identity:

cos^2 t = (1 + cos 2t)/2

Substituting this expression, we obtain:

W = ∫[0,2π] (3/2 + 3/2cos 2t + sin t + 2cos t sin t + cos t) dt

Using trigonometric identities and integrating term by term, we obtain:

W = [3t/2 + (3/4)sin 2t - cos t - cos^2 t] [0,2π]

Simplifying and evaluating the limits of integration, we obtain:

W = 3π

Therefore, the work done by F over the curve in the direction of increasing t is 3π.

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the closing price of the toy company stock on that day was approximately $4.50 per share, and the closing price of the fast food restaurant stock was approximately $32.50 per share.

To solve a first-degree equation, we must find the value of the unknown (which we will call x) and, for this to be possible, just isolate the value of x in equality, that is, x must be alone in one of the members of the equation.

Organize the information:

Organize the information into equations:

Substitute the value of y from Equation 2 into Equation 1:

[tex]32(x + 28) + 78x = 1391\\32x + 896 + 78x = 1391\\110x + 896 = 1391\\110x = 495\\x=4.5[/tex]

Now find the value of Y:

[tex]y = x + 28\\y = 4.50 + 28\\y=32.5[/tex]

Therefore, the closing price of the toy company stock on that day was approximately $4.50 per share, and the closing price of the fast food restaurant stock was approximately $32.50 per share.

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In the triangles, the value of a and b are,

⇒ a = 9.4

⇒ b = 12.1

Since, The Pythagoras theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

WE have to given that;

There are two triangles are shown.

Now, In first triangle,

Base = 5 cm

Perpendicular = 8 cm

Hence, By using Pythagoras theorem we get;

⇒ a² = 8² + 5²

⇒ a² = 64 + 25

⇒ a² = 89

⇒ a  = √89

⇒ a = 9.4

In second triangle,

Hypotenuse = 17 cm

Base = 12 cm

Hence, By using Pythagoras theorem we get;

⇒ 17² = 12² + b²

⇒ 289 = 144 + b²

⇒ b² = 289 - 144

⇒ b  = √145

⇒ b = 12.1

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The cost of the remaining bikes is $51000. The profit made per bike is $300. Therefore, the number of bikes is 65.

If Sarah made a profit of $300 per bike, then let's find the cost of a single bike. The bikes' cost can be found using the given information:

$60000 - cost of 3 bikes = cost of remaining bikes.

So, the cost of the remaining bikes:

= $60000 - $9000

= $51000

Now, if the bikes sold for $70500, then the total profit can be calculated by subtracting the cost of the bikes from the selling price:

Profit = Selling Price - Cost Price

Profit = $70500 - $51000

= $19500

Using the profit made per bike of $300, we can find the number of bikes as follows:

Profit per bike = $300.

So,

Number of bikes = Profit/Profit per bike

A number of bikes = $19500/$300

The number of bikes = 65 bikes

Therefore, 65 bikes were ordered.

Sarah ordered $60000 of bikes from a manufacturer for her bike store. She gave three of the bikes to her kids and decided to sell the remaining bikes for $70500. If Sarah made a profit of $300 per bike, then we need to find how many bikes were ordered. The cost of the remaining bikes is $51000. The profit made per bike is $300. Therefore, the number of bikes is 65.

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Viscosity and osmolarity will both increase if the amount of solutes in the blood increases. The blood is a complex fluid that is constantly circulating throughout the body.

The blood's composition is carefully regulated to ensure that all the body's cells receive the nutrients they need and that waste products are efficiently removed from the body. Viscosity and osmolarity are two critical properties of blood that are affected by the presence of solutes in the blood. Viscosity is a measure of the thickness or resistance to flow of a fluid. Osmolarity, on the other hand, is a measure of the concentration of solutes in a solution.Increased solute concentrations, such as those found in dehydration or in disorders such as polycythemia, can increase blood viscosity and osmolarity. Increased blood viscosity and osmolarity can cause a variety of problems. In the case of blood viscosity, it can cause the blood to flow more slowly, which can lead to problems such as blood clots or even stroke. In the case of osmolarity, it can cause water to be drawn out of cells and into the bloodstream, leading to cell dehydration and other problems.

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To find the point on the curve where the tangent is horizontal, we need to find the value(s) of t for which the derivative of y with respect to x (i.e., dy/dx) is equal to zero.

First, we can find the derivative of y with respect to x using the chain rule:

dy/dx = (dy/dt) / (dx/dt)

We have

dx/dt = 3t^2 - 192

dy/dt = 6t

Therefore:

dy/dx = (dy/dt) / (dx/dt) = (6t) / (3t^2 - 192)

To find the values of t where dy/dx = 0, we need to solve the equation:

6t / (3t^2 - 192) = 0

This equation is satisfied when the numerator is equal to zero, which occurs when t = 0.

To confirm that the tangent is horizontal at t = 0, we can check the second derivative:

d^2y/dx^2 = d/dx (dy/dt) / (dx/dt)

         = [d/dt ((6t) / (3t^2 - 192)) / (dx/dt)] / (dx/dt)

         = (6(3t^2 - 192) - 12t^2) / (3t^2 - 192)^2

         = -36 / 36864

         = -1/1024

Since the second derivative is negative, the curve is concave down at t = 0. Therefore, the point on the curve where the tangent is horizontal is (x,y) = (0, -576).

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