and one 10p. How much more must he save? 10 A train journey from London to Leed takes 2h 35min. At what time do these trains arrive at Leeds if they leave London at a 11:25 b 18:45?​

(i) The smallest angular speeds that cause the toast to hit and then topple to be butter-side down 3.84 rad / s

(ii) The largest angular speeds that cause the toast to hit and then topple to be butter-side down 11.51 rad / 5

The distance from the counter to the floor, d = 82 cm = 0.82 m

Rotation is less than 1 rev.

The toast rotates at a constant angular speed as it falls, and the toast is falling with a constant acceleration (gravitational acceleration).

Using the kinematic equations of motion to calculate the time taken by the toast to hit the floor

[tex]$d & =v_i t+\frac{1}{2} g t^2 \\[/tex]

[tex]$d & =0+\frac{1}{2} g t^2 \\[/tex]

Therefore [tex]$t & =\sqrt{\frac{2 d}{g}}=\sqrt{\frac{2 \times 0. 82}{9.8}[/tex]

= 0.409 s

Where

[tex]$v_i$[/tex] is the initial speed of the toast, [tex]$v_i=0$[/tex] because the toast is falling from rest.

t the time that the toast takes to hit the floor.

g is the gravitational acceleration.

d is the distance between the counter and the floor.

Part (a)The toast is accidentally pushed over the edge of the counter with the butter side up, then the toast rotates as it falls. If the toast hits the ground and then topples to be butter-side down, it has to land on one of its edges. The smallest angle, in this case, is [tex]$\frac{1}{4}$[/tex] revolution and corresponds to the smallest angular speed

[tex]$\omega_{\min } & =\frac{\Delta \theta}{\Delta t} \\[/tex]

[tex]$\omega_{\min } & =\frac{0.25 \mathrm{rev}}{\Delta t}=\frac{0.25 \times 2 \pi}{\Delta t} \\[/tex]

Therefore [tex]$ \omega_{\min } & =\frac{0.5 \pi}{0.409}[/tex]

= 3.84 rad/s

Where

[tex]$\omega_{\min }$[/tex] is the minimum angular speed for the toast to land butter-side down. [tex]$\Delta \theta$[/tex] is the smallest angle for the toast to land butter-side down in radians.

[tex]$\omega_{\min }$[/tex] = 3.84 rad / s

Part b:

The toast is accidentally pushed over the edge of the counter with the butter side up, then the toast rotates as it falls. If the toast hits the ground and then topples to be butter-side down, it has to land on one of its edges. The largest angle, in this case, is [tex]$\frac{3}{4}$[/tex] revolution and corresponds to the largest angular speed

[tex]$\omega_{\max } & =\frac{\Delta \theta}{\Delta t} \\[/tex]

[tex]$\omega_{\max } & =\frac{0.75 \mathrm{rev}}{\Delta t}=\frac{0.75 \times 2 \pi}{\Delta t} \\[/tex]

Therefore [tex]$ \omega_{\max } & =\frac{1.5 \pi}{0.409}[/tex]

= 11.51 rad / s

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When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the floor is 82 cm and for rotation less than 1 rev, what are the (i) smallest and (ii) largest angular speeds that cause the toast to hit and then topple to be butter-side down?

The domain of the rational expression: x-5/3x-6 is all real numbers except 2. Option B is the correct answer.

A polynomial fraction with at least one variable in the denominator is referred to as a "rational expression." The expression is only linear or a polynomial if the variables are only in the numerator. Rational expressions can be used for pretty much anything that can be done with conventional fractions.

To determine a rational function's domain: Take the expression's denominator. Put a zero in that denominator. For the zeroes in the denominator, solve the ensuing equation.

All other x-values make up the domain.

The rational expression is given as:

x - 5 / 3x - 6

The rational expression does not exists when the denominator is equal to 0.

That is,

3x - 6 = 0

3x = 6

x = 2

The domain of a expression is all the input values for which the expression exists.

The given rational expression does not exist at x = 2.

Hence, the domain of the rational expression: x-5/3x-6 is all real numbers except 2. Option B is the correct answer.

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Answer:

2) 486 ft²

4) 38 ft²

5) 27 ft²

6) 149 ft²

Step-by-step explanation:

Just subtract the area of the unshaded part from the shaded part (as if the unshaded part wasn't there)

2) [shaded] (18 • 31.5) – [unshaded] (9 • 9) = 567 – 81 = 486

4) [shaded] (9 • 6) – [unshaded] (4 • 4) = 54 – 16 = 38

5) [shaded] (6 • 6) – [unshaded] (3 • 3) = 36 – 9 = 27

6) [shaded] (14 • 14) – [unshaded] (7 • 7)

= 196 – 49 = 149

The interval of convergence for this power series is (-∞, ∞) and the radius of convergence is infinite.

The power series representation of the function f(x) = 4 - 3x centered at x = 0 is given by the formula:

[tex]f(x) = [infinity]n=0 (4 (-3)^n * x^n)[/tex]

This power series can be used to represent the function for values of x in the interval of convergence. The interval of convergence for this power series is calculated by taking the limit of the absolute value of the coefficient of the highest power of x as n approaches infinity. In this case, the coefficient of the highest power of x is [tex]4(-3)^n[/tex], so the limit of the absolute value of this coefficient is [tex]|4(-3)^∞|[/tex] = 0. Thus, the interval of convergence for this power series is (-∞, ∞).

We can also calculate the radius of convergence for this power series. The radius of convergence is the distance from the center of the series, in this case x = 0, to the point at which the series diverges. To calculate the radius of convergence we can use the ratio test. The ratio test states that if [tex]lim |a(n+1)/a(n)| < 1,[/tex] then the series converges. The ratio of any two consecutive terms in this series is [tex]|4(-3)^(n+1)/4(-3)^n| = |-3|[/tex], which is less than 1. Thus, the radius of convergence for this power series is infinite.

In conclusion, the power series representation of the function f(x) = 4 - 3x centered at x = 0 is given by the formula: [tex]f(x) = [infinity]n=0 (4 (-3)^n * x^n)[/tex]. The interval of convergence for this power series is (-∞, ∞) and the radius of convergence is infinite.

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As per the Poisson distribution, the expected value of V is given by [tex]e^{-\mu} * H(\mu).[/tex]

The Poisson distribution is a discrete probability distribution that models the number of events occurring in a given time interval or space.

The probability of observing k events in this interval is given by the Poisson probability mass function:

[tex]P(k; \mu) = (e^{-\mu} * \mu^k) / k![/tex]

where μ is the mean number of events in the interval. The Poisson distribution has some important properties, including the fact that its mean and variance are both equal to μ.

Now, let us consider a new random variable V, which is defined as V = 1/(1+U). We want to find the expected value of V, which is denoted by E(V).

To do this, we need to use the definition of the expected value. For a discrete random variable X with probability mass function p(x), the expected value is defined as:

E(X) = Σ x * p(x)

where the summation is taken over all possible values of X.

Using this definition, we can find E(V) as follows:

E(V) = Σ v * P(V = v)

where the summation is taken over all possible values of V.

To find P(V = v), we need to use the transformation method. This method involves finding the probability mass function of U and then using it to find the probability mass function of V.

Since U is a Poisson random variable with mean μ, its probability mass function is given by:

[tex]P(U = k) = (e^{-\mu} * \mu^k) / k![/tex]

Now, let us find the probability mass function of V. We have:

V = 1/(1+U) => 1/V = 1+U => U = 1/V - 1

Using this transformation, we can find the probability mass function of V as follows:

P(V = v) = P(U = 1/v - 1) = [tex]e^{-\mu} * \mu^{(1/v - 1)} / (1/v - 1)![/tex]

Now, we can use this probability mass function to find the expected value of V:

E(V) = Σ v * P(V = v)

[tex]= > \sum v * (e^{-\mu} * \mu^{1/v - 1}) / (1/v - 1)![/tex]

To simplify this expression, we can use the fact that the sum of the reciprocals of the first n positive integers is given by the nth harmonic number, Hn:

1/1 + 1/2 + 1/3 + ... + 1/n = Hn

Using this identity, we can rewrite E(V) as:

E(V) = [tex]\sum v * (e^{-\mu} * \mu^{1/v - 1}) / (1/v - 1)![/tex]

[tex]= e^{-\mu} * \sum (\mu/v)^{1/v} * (v-1)![/tex]

=> [tex]= e^{-\mu} * H(\mu)[/tex]

where H(μ) is the μth harmonic number.

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The expression "(7532 + 100y2) + 10(10y2 - 110)" can be written in the form "ay^2 + b" (a and b are constants). The required value of a + b is 6632.

Expanding the given expression, we have:

(7532 + 100y^2) + 10(10y^2 - 110)

= 7532 + 100y^2 + 100y^2 - 1100 (distributing the factor of 10)

= 200y^2 + 6432

So, we have expressed the given expression in the form ay^2 + b, where a = 200 and b = 6432. The sum of a and b is:

a + b = 200 + 6432 = 6632

Therefore, the value of a + b is 6632.

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The answer is (C) the absolute value of the z-score for the midterm exam grade is greater than for the final grade, so the midterm grade is more unusual.

To calculate the z-scores for the midterm and final exam grades, we first need to calculate the mean and standard deviation for each data set.

For the midterm exam grades:

mean = (80 + 78 + 85 + 82 + 79 + 79 + 78 + 86 + 80 + 84) / 10 = 81.1

standard deviation = sqrt([ (80-81.1)^2 + (78-81.1)^2 + ... + (84-81.1)^2 ] / 9) = 2.336

For the final exam grades:

mean = (81 + 88 + 68 + 69 + 69 + 81 + 82 + 86 + 76 + 71) / 10 = 77.1

standard deviation = sqrt([ (81-77.1)^2 + (88-77.1)^2 + ... + (71-77.1)^2 ] / 9) = 7.042

Now, we can calculate the z-scores for Jaymes's grade of 86 on each exam:

z-score for midterm grade = (86 - 81.1) / 2.336 = 2.094

z-score for final exam grade = (86 - 77.1) / 7.042 = 1.259

To determine which grade is more unusual, we need to compare the absolute values of the z-scores. Since the absolute value of the z-score for the midterm grade is greater than for the final exam grade (|2.094| > |1.259|), we can conclude that the midterm grade is more unusual for Jaymes. Therefore, the answer is (C).

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_____The given question is incomplete, the complete question is given below:

The midterm and final exam grades for a statistics course are provided in the data set below. Jaymes, a student in the class, scored 86 on both exams. Treat the given data sets as samples. Jaymes's wants to know which grade is more unusual, the midterm grade or the final exam grade. Use Use a TI-83, TI-83 Plus, or TI-84 calculator to calculate the z-scores corresponding to each grade. Round your answer to three decimal places.

Midterm 80, 78, 85, 82, 79, 79, 78, 86, 80, 84,

final 81, 88, 68, 69, 69, 81, 82, 86, 76, 71

(a) Using a direct proof, we can demonstrate that the assertion "For all integers a and b, if an is even and b is a multiple of 3, then ab is a multiple of 6" is true.

(b) The statement's opposite is "For all numbers a and b, if ab is a multiple of 6, then an is even and b is a multiple of 3.

Assume that both a and b are multiples of three and that an is an even number. Then, we can write a = 2k and b = 3n for some integers k and n, respectively. Adding these two equations together results in:

ab=(2k, 3n), 6kn

Due to the fact that k and n are both numbers, their product 6kn is also an integer. Ab is a multiple of 6, proving the assertion because of this.

(b) The statement's opposite is (b) "For all numbers a and b, if ab is a multiple of 6, then an is even and b is a multiple of 3." The following example refutes the converse assertion, which is untrue:

Let a and b each be three. Therefore, ab is a multiple of 6 and equals 12. The converse assertion is untrue because an is not an even number and b is not a multiple of three.

We have thus demonstrated the validity of the initial claim and its supporting evidence while demonstrating the falsity of the opposite claim.

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The cement and gravel must be used in construction is 60 and 80.

Suppose that we've got two quantities with measurements as 'a' and 'b'

Then, their ratio(ratio of a to b) a:b

or [tex]\dfrac{a}{b}[/tex]

We usually cancel out the common factors from both the numerator and the denominator of the fraction we obtained. Numerator is the upper quantity in the fraction and denominator is the lower quantity in the fraction).

Suppose that we've got a = 6, and b= 4, then:

[tex]a:b = 6:2 = \dfrac{6}{2} = \dfrac{2 \times 3}{2 \times 1} = \dfrac{3}{1} = 3\\or\\a : b = 3 : 1 = 3/1 = 3[/tex]

Remember that the ratio should always be taken of quantities with same unit of measurement. Also, ratio is a unitless(no units) quantity.

Given that;

Ratio of cement:sand:gravel= 1:2.5:3.7

The weight of sand bag 150lb.

Let & be the proportion of cement in a mixture. The ratio of cement to sand is given as 1 : 2.5. With a 150-lb sand, then the proportion of cement is given by

1/2.5 = x/150

150=2.5x

x=150/2.5

x=1500/25

x=60

y= 80

Therefore, by the given ratio answer will be 60 and 80.

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With probability that person has high blood pressure = 0.3 and the probability that person is a runner = 0.4, they are not independent events. So, Correct option is D .

Independence between two events is defined as follows, if two events A and B are independent, then the occurrence of one event (A) has no effect on the probability of occurrence of the other event (B). This can be mathematically represented as:

P(A ∩ B) = P(A) * P(B)

where P(A ∩ B) is the probability of both events A and B happening at the same time.

In this question, we are given the probabilities of two events: H (high blood pressure) and R (being a runner). The probability of H is 0.3 and the probability of R is 0.4. We are also given that the probability of H and R happening at the same time is 0.2.

Now, to check if H and R are independent, we need to see if the equation

P(H ∩ R) = P(H) * P(R)

holds true.

Plugging in the given values, we get:

0.2 = 0.3 * 0.4

This equation is false, so we can conclude that H and R are not independent events. This means that the occurrence of one event affects the probability of occurrence of the other event.

Option d) states that H and R are independent events, which we have just shown to be false.

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the probability that the sample mean cost for the 36 schools is between $25,575 and $26,500 is 54.14%.

Percentage is a way of expressing a number as a fraction of 100. The symbol for percentage is "%". For example, if you have 75%, it means 75 out of 100 or 0.75 as a decimal.

Given by the question.

We can solve these problems by using the central limit theorem and the formula for the standard error of the mean.

The standard error of the mean is:

SE = σ/√n

where σ is the population standard deviation, n is the sample size, and √ is the square root.

Using the given values, we have:

SE = 3204/√36 = 534

To find the probability that the sample mean cost for the 36 schools is less than $25,000, we need to calculate the z-score and then use a z-table to find the probability.

z = (25000 - 26489) / 534 = -2.79

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -2.79 is 0.0026 or 0.26%.

Therefore, the probability that the sample mean cost for the 36 schools is less than $25,000 is 0.26%.

To find the probability that the sample mean cost for the 36 schools is more than $26,000, we need to calculate the z-score and then use a z-table to find the probability.

z = (26000 - 26489) / 534 = -0.91

Using a standard normal distribution table or calculator, we find that the probability of a z-score more than -0.91 is 0.8186 or 81.86%.

Therefore, the probability that the sample mean cost for the 36 schools is more than $26,000 is 81.86%.

To find the probability that the sample mean cost for the 36 schools is between $25,575 and $26,500, we need to calculate the z-scores for both values and then use a z-table to find the probabilities.

z1 = (25575 - 26489) / 534 = -1.73

z2 = (26500 - 26489) / 534 = 0.21

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -1.73 is 0.0418 and the probability of a z-score less than 0.21 is 0.5832.

Therefore, the probability that the sample mean cost for the 36 schools is between $25,575 and $26,500 is the difference between these probabilities:

0.5832 - 0.0418 = 0.5414

Multiplying by 100, we get a probability of 54.14%.

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The point (7, -8) is located in the third quadrant (Quadrant III) of the coordinate plane, as the x-coordinate is positive and the y-coordinate is negative.

In a coordinate plane, a quadrant is one of the four regions that are formed by dividing the plane into four equal parts by the x-axis and y-axis. The x-axis and y-axis intersect at the origin (0, 0), and the four regions are labeled as Quadrant I, Quadrant II, Quadrant III, and Quadrant IV

Therefore, Each quadrant has its own characteristics and is used in different applications, such as plotting points, graphing functions, and solving problems in mathematics and science.

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Third-degree Taylor polynomial for f(x) = e centered at a = 1 using a graphing calculator estimate the value of eˣ near x = 1.

A polynomial is an expression consisting of variables and coefficients, which involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents.

Now, let's focus on the Taylor polynomial. The Taylor polynomial of degree n for a function f(x) centered at a is given by the formula:

[tex]Tn(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^{2/2!} + f'''(a)(x - a)^{3/3!} + ... + f^{n}(a)(x - a)^{n/n!}[/tex]

In this formula, f'(a), f''(a), f'''(a), and f^(n)(a) are the first, second, third, and nth derivatives of f(x), respectively, evaluated at a. The symbol n! denotes the factorial of n, which is the product of all positive integers from 1 to n.

In our case, we're given the function f(x) = e and a = 1. The first derivative of e^x is e^x, the second derivative is e^x, and the third derivative is also eˣ. Evaluating these derivatives at a = 1, we get:

f(1) = e¹ = e

f'(1) = e¹ = e

f''(1) = e¹ = e

f'''(1) = e¹ = e

Substituting these values into the formula for the Taylor polynomial, we get:

[tex]T3(x) = e + e(x - 1) + e(x - 1)^{2/2!} + e(x - 1)^{3/3!}[/tex]

Simplifying this expression, we get:

[tex]T3(x) = e + e(x - 1) + e(x - 1)^{2/2} + e(x - 1)^{3/6}[/tex]

This is the third-degree Taylor polynomial for f(x) = e centered at a = 1. It can be used to approximate the value of eˣ near x = 1.

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Complete Question:

A graphing calculator is recommended. Find the Taylor polynomial T3(x) for the function f centered at the number a. f(x) = e, a = 1.

The value of x in the given figure is 4.

Vertically opposite angles are angles that are opposite one another at a specific vertex and are created by two straight intersecting lines. Vertically, opposite angles are equal to each other. These are sometimes called vertical angles.

Given that are two lines PS and TS intersecting at point Q, making angles PQT and RQS,

We need to find the value of x,

Here, angles PQT and RQS, are vertically opposite angles, and angles which are vertically opposite are equal,

Therefore,

∠ PQT = ∠ RQS

6x-1 = 23

6x = 24

x = 4

Hence, the value of x in the given figure is 4.

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The solid is a cylinder with a typical disk or washer. It is obtained by rotating the region bounded by[tex]y = 6 - x^2[/tex]and y = 2 about the x-axis. The volume is 48π.

The region to be rotated is bounded by the equations [tex]y = 6 - x^2[/tex] and y = 2. The region is bounded horizontally by x = 0 and x = 2 and vertically by y = 2 and [tex]y = 6 - x^2.[/tex] To find the volume of the solid we can use the formula V = ∫2π0∫2y0rdrdθ. The integral inside is the area of a disk with radius r and the integral outside is the circumference of a circle with radius r. We substitute [tex]y = 6 - x^2[/tex] into the integral, giving us [tex]V = ∫2π0∫2(6 - x^2)0rdrdθ[/tex]. By solving the integral we get V = 48π.

y = 2

x² = 6×4 = 24

V = ∫2π0∫2y0rdrdθ

V = 48π

The solid is a cylinder with a typical disk or washer and is obtained by rotating the region bounded by [tex]y = 6 - x^2[/tex] and y = 2 about the x-axis.

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The vectors are dependent. One set of numbers that makes the equation true is c1 = -8/77, c2 = 22/77, and c3 = 0.

To decide if the vectors [77 22 22], [24 12 8], and [75 24 22] are directly reliant, we really want to find scalars c1, c2, and c3 with the end goal that:

c1[77 22 22] + c2[24 12 8] + c3[75 24 22] = [0 0 0]

We can set up an arrangement of straight conditions:

77c1 + 24c2 + 75c3 = 0

22c1 + 12c2 + 24c3 = 0

22c1 + 8c2 + 22c3 = 0

Utilizing column decrease, we can address this arrangement of conditions to track down the upsides of c1, c2, and c3:

[1 24/77 75/77 | 0]

[0 1 - 1 | 0]

[0 0 1 | 0]

The last column of the line decreased expanded network lets us know that c3 = 0. Subbing this back into the initial two conditions, we get:

77c1 + 24c2 = 0

22c1 + 12c2 = 0

Tackling for c1 and c2, we get:

c1 = - 8/77

c2 = 22/77

Consequently, the vectors [77 22 22], [24 12 8], and [75 24 22] are straightly ward, and we can find scalars that make the condition [0 0 0] = c1[77 22 22] + c2[24 12 8] + c3[75 24 22] valid.

Specifically, we have:

-8/77 [77 22 22] + 22/77 [24 12 8] = [-3/77 0 0]

Thus, one bunch of numbers that makes the condition genuine is c1 = - 8/77, c2 = 22/77, and c3 = 0, which gives a non-zero vector [-3/77 0 0] on the left-hand side.

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A nontrivial linear combination of the vectors that equals the zero vector, we can conclude that the set is linearly dependent.

The set of vectors is linearly dependent because V3 can be written as a linear combination of V1, V2, and V4:

V3 = 2V1 - 3V2 + V4

To see this, we can write out each component of the vectors:

V1 = | 2 | , V2 = | 0 | , V3 = | -1 | , V4 = | 1 |

| 1 | | 1 | | 2 | | 3 |

Then we can plug in the values and see that the equation holds:

2V1 - 3V2 + V4 = 2| 2 | - 3| 0 | + | 1 | = | 4 | - | 0 | + | 1 | = | 5 |

| 1 | | 1 | | 3 | | 1 | | 1 | | 2 | | 3 | | 1 |

Since we have found a nontrivial linear combination of the vectors that equals the zero vector, we can conclude that the set is linearly dependent.

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The density of the liquid is 2g/cm³

The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume). The mass of atoms, their size, and how they are arranged determine the density of a substance. Density equals the mass of the substance divided by its volume; D = m/v.

Where D is the density

m is the mass and

v is the volume

mass = 100kg = 100000 g( 1kg = 1000g)

50l = 50 × 1000 = 50000cm³

density = 100000/50000

= 2g/cm³

therefore the density of the liquid is 2 g/cm³

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The value of the car after 1 year will be $22,250 when it had a depreciation of 11%.

Depreciation is a term used in accounting to describe two different aspects of the same idea: first, the actual decline in fair value of an asset, such as the annual decline in value of factory equipment due to use and wear, and second, the allocation in accounting statements of the asset's initial cost to the periods in which it is used (depreciation with the matching principle).

Depreciation is the process of reallocating, or "writing down," the cost of a tangible item (such as equipment) over the course of that asset's useful life.

It also refers to the decline in asset value. Long-term assets are depreciated by businesses for accounting and tax reasons.

So, we know that the price of the car is:

$25000

Depreciation is 11%.

The value of the car after 1 year will be:

25000/100 * 11 = $2,750

25000 - 2,750 = $22,250

Therefore, the value of the car after 1 year will be $22,250 when it had a depreciation of 11%.

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Complete question:

You buy a new car that cost $25000. The car depreciates at a rate of 11% per year. What will be the value f the car after 1 year?

The car that would the customer buy is option (c) Car 2, because it has a lower sample standard​ deviation, hence more predictable gas mileage.

The standard deviation is a measure of how much the data deviates from the mean.

Looking at the data sets, we can see that Car 1 has a wider range of gas mileage, with a minimum of 157 miles and a maximum of 318 miles per 10 gallons of gas. In comparison, Car 2 has a narrower range, with a minimum of 201 miles and a maximum of 279 miles per 10 gallons of gas. The range is a measure of spread, and we can see that Car 1 has a larger spread.

To further analyze the spread, we can calculate the sample standard deviation for each data set.  A higher standard deviation indicates more variability in the data. In this case, Car 1 has a sample standard deviation of 43.6 miles per 10 gallons of gas, while Car 2 has a sample standard deviation of 22.9 miles per 10 gallons of gas. This means that Car 1 has more variability in its gas mileage.

In terms of center, we can calculate the sample mean for each data set. The sample mean is a measure of the average gas mileage. Car 1 has a sample mean of 247.5 miles per 10 gallons of gas, while Car 2 has a sample mean of 249.7 miles per 10 gallons of gas. We can also calculate the median, which is the middle value of the data set. Car 1 has a median of 252.5 miles per 10 gallons of gas, while Car 2 has a median of 252.0 miles per 10 gallons of gas.

Based on these measures, we can see that there is very little difference between the two cars in terms of center. However, Car 2 has a lower sample standard deviation, indicating more predictable gas mileage. Therefore, the customer should choose Car 2 because it has more consistent gas mileage.

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The correct description that best explains the domain of (g o f) (x) is: The elements in the domain of f(x) for which g(f(x)) is defined.

From the question, we have the following parameters that can be used in our computation:

(g o f) (x)

The expression (g o f)(x) is a composite function

The domain of (g o f)(x) is the set of all possible inputs x for which the composition (g o f)(x) is defined.

For the composition (g o f)(x) to be defined, it is necessary that the input x belongs to the domain of f(x), and the output of f(x) belongs to the domain of g(x).

Hence, the domain of (g o f)(x) consists of all the elements in the domain of f(x) for which g(f(x)) is defined

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Answer:

Step-by-step explanation:

Answer:

-3 ≤ x ≤ 3

Step-by-step explanation:
The domain just means all the possible inputs/ x values of a function.

Answer:

Domain is everything under x

Step-by-step explanation:

D: {-3, -2, -1. 0, 1, 3}

Answer:

Step-by-step explanation:

He can walk 125km in 2 and a half hours

Answer:

125km

Step-by-step explanation:

If he walks 50km per hour and he walks 2 hours it's 2 x 50km = 100km

if walks for another half hour the half of 50km is 25km so 100km + 25km = 125km

We would expect John to weigh approximately 199.8 pounds at the end of week 8 if he continues to burn 6100 calories per week and maintains a proper diet.

Finding the best linear relationship between the independent and dependent variables is the aim of linear regression. To put it another way, it seeks out the line (or plane, or hyperplane in higher dimensions) that best fits the data. The dependent variable's future values for new values of the independent variable can then be predicted using this line.

We may build a linear regression model using the provided data, with activity (in calories) serving as the independent variable and weight (in pounds) serving as the dependent variable. Using this model, we can then forecast John's weight at the conclusion of week eight depending on the number of calories he burned in that week.

Using a spreadsheet or statistical software, we can calculate the regression equation:

Weight = -0.0049 x Exercise + 229.85

We can plug in the exercise value for week 8 (6100 calories) and solve for weight:

Weight = -0.0049 x 6100 + 229.85

Weight ≈ 199.8 pounds

Therefore, we would expect John to weigh approximately 199.8 pounds at the end of week 8 if he continues to burn 6100 calories per week and maintains a proper diet.

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The dividends per share for the preferred stock are $2.50, $0.80, and $3.00 for 20Y1, 20Y2, and 20Y3, respectively.

The dividends per share for the common stock are $0.75, $0.24, and $0.90 for 20Y1, 20Y2, and 20Y3, respectively.

To determine the dividends per share for preferred and common stock, we need to divide the total dividends by the number of shares outstanding for each type of stock.

For the preferred stock:

20Y1: $37,500 ÷ 15,000 shares = $2.50 per share

20Y2: $12,000 ÷ 15,000 shares = $0.80 per share

20Y3: $45,000 ÷ 15,000 shares = $3.00 per share

For the common stock:

20Y1: $37,500 ÷ 50,000 shares = $0.75 per share

20Y2: $12,000 ÷ 50,000 shares = $0.24 per share

20Y3: $45,000 ÷ 50,000 shares = $0.90 per share

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Answer:

1. To round to the nearest tenth, we'll need to look at the digit to the right of the tenths place, the hundredths place. Since there's a three in the hundredths place, we will round down to 4.9. The mean absolute deviation of these eight values rounded to the nearest tenth is 4.9.

2. 1

Order the numbers in the data set and find the median.

2

Subtract the median from each number in the data set.

3

Take the absolute value of each difference.

4

Add up all of the positive differences.

5

Divide this sum by the number of data points in the set.

3.The lower quartile, or first quartile (Q1), is the value under which 25% of data points are found when they are arranged in increasing order. The upper quartile, or third quartile (Q3), is the value under which 75% of data points are found when arranged in increasing order.

4.

Step-by-step explanation:

ATMs offer a convenient and fast way for people to perform financial transactions, such as withdrawing cash, depositing checks, and transferring funds.

Convenience: ATMs are available 24/7, making it easy to access your money and perform transactions outside of normal business hours.

Speed: ATMs are typically faster than bank tellers since you don't have to wait in line and can complete transactions quickly.

Privacy: ATMs offer a level of privacy that bank tellers may not, especially if you need to perform sensitive transactions like withdrawing large amounts of cash.

Fees: Some ATMs charge fees for using their services, especially if you are not a customer of that particular bank.

Limited services: While ATMs offer many basic banking services, they may not be able to perform more complex transactions that require the assistance of a bank teller.

Technical issues: ATMs can sometimes experience technical issues, such as running out of cash or malfunctioning, which can be frustrating for users.

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Information provided in the lesson is not in the question so i gave a general view

A bar graph or a column chart would be the best type of graph to show the number of students who earned extra credit each day this week.

The visual display of data (often grouped) in the shape of vertical or horizontal rectangular bars, with the length of the bars corresponding to the measure of the data, is called a bar graph. Bar charts are another name for them.

The bar graph or column chart will allow you to easily compare the number of students who earned extra credit on each day of the week and see any trends or patterns that may exist.

The vertical bars on the graph represent the number of students who earned extra credit, and the horizontal axis shows the days of the week.

You could also consider using a line graph to show the trend in the number of students earning extra credit over the course of the week,

but this would not be as effective in comparing the number of students who earned extra credit each day.

Therefore, the bar graph or column chart is the required graph.

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