Suppose a normal distribution has a mean of 48 and a standard deviation of 2. What is
the probability that a data value is between 46 and 47? Round your answer to the nearest
tenth of a percent.
VO A. 15.0%
О B. 13.0%
O C. 16.0%
D. 14.0%
15.0% is correct

The probability that, out of 18 independent samples received from one river, just two were contaminated is 0.8438.

In the following 18 samples to be evaluated,

Let X = the number of samples that now the pollutant is present in.

Thus, with p = 0.10 and n = 18, X is a binomial random variable.

Using the binomial theorem:

[tex](^{n} _{r} ) p^{x} q^{n-x}[/tex]

p = 0.10

q = 1 - 0.10 = 0.9

n = 18

The likelihood that only two samples out of 18 obtained in different ways from the same river were polluted

P(x = 2) = [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{18-2}[/tex]

=  [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{16}[/tex]

= 153 x 0.01 x 0.1853

= 0.8438

Thus, the probability that, out of 18 separate samples received from one river, just two were contaminated is 0.8438.

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The scale of the map in the question is 1 inch = 50 miles.

A scale in a map is a relation that tells us how many units each unit in the map represents. In this case, we know that the distance between two points A and B on the map is 5 inches, while the actual distance between these two places is 250 miles.

Then we start with the relation:

5 inches = 250 miles.

But to get the scale of the map we need to see how many miles one inch represents in the map, then we can divide both sides of the equation by 5 to geT:

5 in = 250 mi

1 in = 250mi/5

1 in = 50 mi

The scale of the map is 1 inch to 50 miles.

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The difference in price between the two shops for 300 cm of ribbon is £3.60 - £2.40 = £1.20. Therefore, the answer is £1.20.

How to solve and what is Selling?

We can use proportions to find the cost of 300 cm of ribbon at each shop, and then subtract the cost at Shop B from the cost at Shop A to find the difference in price:

For Shop A:

140 cm of ribbon cost £1.68, so 1 cm of ribbon cost £1.68/140 = £0.012.

Therefore, 300 cm of ribbon would cost £0.012 x 300 = £3.60.

For Shop B:

215 cm of ribbon cost £1.72, so 1 cm of ribbon cost £1.72/215 = £0.008.

Therefore, 300 cm of ribbon would cost £0.008 x 300 = £2.40.

The difference in price between the two shops for 300 cm of ribbon is £3.60 - £2.40 = £1.20. Therefore, the answer is £1.20.

Selling is the process of exchanging goods or services for money or other valuable consideration. In business, selling is an essential part of the marketing and sales process, and involves identifying potential customers or clients, communicating with them about the features and benefits of the product or service being sold, and negotiating a price or other terms of the sale.

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Given that (1, 2, 3] System{1, 4, 7,6] for a system known to be LTI, the impulse response of the system: h[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2]

To compute the impulse response h[n] of a linear time-invariant (LTI) system given its input-output relationship, we can use the convolution sum:

y[n] = x[n] * h[n]

y[n] = (1/2)*(x[n] + 2x[n-1] + 3x[n-2])

y[n] = (1/2)*(δ[n] + 2δ[n-1] + 3δ[n-2])

y[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2]

Thus, the impulse response of the system is:

h[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2],where δ[n] is the impulse signal.

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Answer: approximately 68,404 liters.

Step-by-step explanation:

To solve the problem, we first need to find the cross-sectional area of the pipe, which we can calculate using the formula for the area of a circle:

A = πr^2

where r is the radius of the pipe, which is half the diameter. So, in this case, the radius is 2.4 cm / 2 = 1.2 cm.

A = π(1.2 cm)^2

A ≈ 4.5239 cm^2

Next, we can use the formula for volume flow rate to find the volume of water that is discharged per second:

Q = Av

where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the velocity of the water. In this case, we have:

Q = (4.5239 cm^2)(2.8 m/s)

Q ≈ 0.01266 m^3/s

To find the volume of water discharged in one and a half hours (which is 5400 seconds), we can multiply the volume flow rate by the time:

V = Qt

V = (0.01266 m^3/s)(5400 s)

V ≈ 68.404 m^3

Finally, to convert the volume from cubic meters to liters, we can multiply by 1000:

V = 68.404 m^3 × 1000 L/m^3

V ≈ 68,404 L

Therefore, the volume of water discharged in one and a half hours is approximately 68,404 liters.

The expression (2^8) to the power of 3 as a power of 2 is  4096^2

Given the following expression

(2^8) to the power of 3

To write (2^8) to the power of 3 as a power of 2, we can use the rule of exponents that says:

(a^b)^c = (a^c)^b

Applying this rule to (2^8)^3, we get:

(2^8)^3 = (2^3)^8

Simplifying the expression, we get:

(2^8)^3 = (8)^8

So, we have

(2^8)^3 = (8^4)^2

Simplify

(2^8)^3 = 4096^2

Therefore, (2^8)^3 can be written as 4096^2.

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Every real root of a polynomial with rational coefficients is also a root of a polynomial with integer coefficients.

Suppose that c is a root of the polynomial equation:

[tex]q_n[/tex] × [tex]x^n[/tex] + q_{n-1} × [tex]x^{n-1}[/tex] + ... + q1 × x + q0 = 0

where [tex]q_i[/tex] are rational coefficients. Since c is a root of this polynomial equation, we have:

[tex]q_n[/tex] × [tex]c^n[/tex] + [tex]q_{n-1}[/tex] × [tex]c^{n-1}[/tex] + ... + q1 × c + q0 = 0

Multiplying both sides of the equation by the common denominator of the coefficients [tex]q_i[/tex], we can obtain an equation with integer coefficients. Let d be the least common multiple of the denominators of the coefficients [tex]q_i[/tex]. Then we can write:

d × ([tex]q_n[/tex] × [tex]c^n[/tex] + [tex]q_{n-1}[/tex] × [tex]c^{n-1}[/tex] + ... + q1 × c + q0) = 0

Expanding the left-hand side of the equation, we obtain a polynomial with integer coefficients:

d × [tex]q_n[/tex] × [tex]x^n[/tex] + d × [tex]q_{n-1}[/tex] × [tex]x^{n-1}[/tex] + ... + d × q1 × x + d × q0 = 0

Since c is a root of the original polynomial equation, it is also a root of this polynomial with integer coefficients. Therefore, we have shown that if c is a root of a polynomial with rational coefficients, then it is also a root of a polynomial with integer coefficients.

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Answer:

Step-by-step explanation:

Let x be the amount invested in the first account, which pays 14% interest. Then the amount invested in the second account, which pays 11% interest, is 6800 - x.

The interest earned on the first account is 0.14x, and the interest earned on the second account is 0.11(6800 - x). The total interest earned is the sum of these two amounts, so we have:

0.14x + 0.11(6800 - x) = 856

Simplifying and solving for x, we get:

0.14x + 748 - 0.11x = 856

0.03x = 108

x = 3600

Therefore, Asaad invested $3600 in the first account and $3200 (6800 - 3600) in the second account.

Answer:

The numbers 8 and lower are possible values of x.

Step-by-step explanation:

The inequality [tex]x\leq 8[/tex] means x is less than or equal to 8. Therefore, 8 is a possible value of x.

Answer:

Step-by-step explanation:

True. 1 is an odd positive integer less than 11 and is an element of set B.

Answer:

4

Step-by-step explanation:

Answer:

Step-by-step explanation:

To solve this problem, we need to use the equations of motion for constant acceleration, constant velocity, and constant deceleration. We'll break the problem into several parts and use these equations to find the distance traveled in each part. Then, we'll add up the distances to get the total distance traveled.

First, we need to convert the units of speed from km/h to m/s, since the equations of motion use meters per second. We have:

Initial speed (u) = 0 km/h = 0 m/s

Final speed (v) = 40 km/h = 11.11 m/s

Constant speed = 40 km/h = 11.11 m/s (for 4 minutes)

Final speed before hill = 30 km/h = 8.33 m/s

Speed at top of hill = 0 m/s

Acceleration (a) = (v-u)/t = (11.11-0)/(3*60) = 0.0611 m/s^2

PART 1: ACCELERATION PHASE

Time taken (t) = 3 minutes = 180 seconds

Distance traveled (s) = ut + (1/2)at^2

s = 0 + (1/2)0.0611(180^2) = 331.83 meters

PART 2: CONSTANT SPEED PHASE

Time taken (t) = 4 minutes = 240 seconds

Distance traveled (s) = vt

s = 11.11*240 = 2666.4 meters

PART 3: DECELERATION PHASE

Time taken (t) = 1 minute = 60 seconds

Deceleration (a) = (v-u)/t = (8.33-11.11)/60 = -0.0461 m/s^2 (negative since it's deceleration)

Distance traveled (s) = vt + (1/2)at^2

s = 8.3360 + (1/2)(-0.0461)*(60^2) = 494.7 meters

PART 4: CONSTANT SPEED PHASE

Time taken (t) = 10 minutes = 600 seconds

Distance traveled (s) = vt

s = 8.33*600 = 4998 meters

PART 5: DECELERATION PHASE TO STOP

Time taken (t) = 2 minutes = 120 seconds

Initial speed (u) = 8.33 m/s

Final speed (v) = 0 m/s

Deceleration (a) = (v-u)/t = (0-8.33)/120 = -0.0694 m/s^2

Distance traveled (s) = vt + (1/2)at^2

s = 8.33120 + (1/2)(-0.0694)*(120^2) = 733.3 meters

TOTAL DISTANCE TRAVELED:

Adding up the distances from each part, we get:

Total distance = 331.83 + 2666.4 + 494.7 + 4998 + 733.3 = 9184.23 meters = 9.18 km (rounded to two decimal places)

Therefore, the cyclist has traveled approximately 9.18 km.

One pair of functions that satisfies the given condition is:

Exponential function: [tex]f(x) = 1.46^x,[/tex] Quadratic function: [tex]g(x) = x^2[/tex]

In mathematics, an expression is a combination of numbers, variables, and mathematical operations such as addition, subtraction, multiplication, and division. Expressions can also include functions, brackets, and other symbols.

Let's consider the two functions:

Exponential function: [tex]f(x) = a^x, where a > 1[/tex]

Quadratic function: [tex]g(x) = x^2[/tex]

We want to find the pair of functions for which the exponential is consistently growing at a faster rate than the quadratic over the interval 0 ≤ x ≤ 5.

To determine this, we can compare the growth rates of the two functions by looking at their derivatives.

The derivative of the exponential function is:[tex]f'(x) = a^x * ln(a)[/tex]

The derivative of the quadratic function is: [tex]g'(x) = 2x[/tex]

To compare the growth rates of the two functions, we need to compare their derivatives. We want to find the value of x for which the exponential function is growing faster than the quadratic function, i.e., where f'(x) > [tex]g'(x).\\f'(x) > g'(x)\\a^x * ln(a) > 2x[/tex]

Now, we can solve for x:

[tex]a^x * ln(a) > 2xln(a)/2 * a^x > x[/tex]

Since we want to find the pair of functions for which the exponential is consistently growing at a faster rate than the quadratic over the interval 0 ≤ x ≤ 5, we need to find a value of a such that the inequality ln(a)/2 * [tex]a^5 > 5[/tex] is true for all values of a > 1.

We can use a graphing calculator or a numerical solver to find the value of a that satisfies this inequality. One possible solution is a ≈ 1.46.

Therefore, one pair of functions that satisfies the given condition is:

Exponential function: [tex]f(x) = 1.46^x,[/tex] Quadratic function: [tex]g(x) = x^2[/tex]

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For the given polynomials in factored form;

(a) the leading term is x³, the zeroes are -2,3 and 5 , the graph will be a cubic function passing through x-axis at -2, 3 and 5.
(b) the leading term is 2x², the zeros are -1, 4 the graph will be a quadratic function passing through x-axis at -1 and 4.

Part(a) : The polynomial in factored form is : f(x) = (x + 2)(x - 3)(x - 5)

The Leading Term is : x³; The Zeros are : -2, 3, 5.

The General Shape: The graph of the polynomial will be a cubic function that passes through the x-axis at -2, 3, and 5.

The function will approach negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity.

Part(b) : The Polynomial in factored form is : f(x) = 2(x + 1)(x - 4)

The Leading Term is : 2x²;  The Zeros are : -1, 4.

The General Shape: The graph of the polynomial will be a quadratic function that passes through the x-axis at -1 and 4. The function will open upwards since the leading coefficient is positive.

The function will approach negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity.

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The given question is incomplete, the complete question is

For each polynomial in factored form show the leading term, the zeros on the x-axis, and the general shape of the polynomial.

(a) f(x) = (x + 2)(x - 3)(x - 5)

(b) f(x) = 2(x + 1)(x - 4).

Answer: x = 5 and x = 1.

Step-by-step explanation:

To solve fg(x) = gf(x), we need to find the expressions for fg(x) and gf(x) and then set them equal to each other.

fg(x) = f(g(x)) = f(x² + 2) = 2(x² + 2) - 3 = 2x² + 1

gf(x) = g(f(x)) = g(2x - 3) = (2x - 3)² + 2 = 4x² - 12x + 11

Now we set fg(x) equal to gf(x) and solve for x:

2x² + 1 = 4x² - 12x + 11

2x² - 12x + 10 = 0

Dividing both sides by 2 gives:

x² - 6x + 5 = 0

This quadratic equation factors as:

(x - 5)(x - 1) = 0

So the solutions are x = 5 and x = 1.

Therefore, the solutions to fg(x) = gf(x) are x = 5 and x = 1.

Answering the presented question, we may conclude that This greater expressions scale makes it easier to see the variations between the ratings of the three college students in a extra correct and informative way.

In mathematics, an expression is a collection of integers, variables, and complex mathematical (such as arithmetic, subtraction, multiplication, division, multiplications, and so on) that describes a quantity or value. Phrases can be simple, such as "3 + 4," or complicated, such as They may also contain functions like "sin(x)" or "log(y)". Expressions can be evaluated by swapping the variables with their values and performing the arithmetic operations in the order specified. If x = 2, for example, the formula "3x + 5" equals 3(2) + 5 = 11. Expressions are commonly used in mathematics to describe real-world situations, construct equations, and simplify complicated mathematical topics.

The difference in look between the two charts is due to the choice of the scales on the x and y-axes. In the left chart, the y-axis starts offevolved at 800 and has a range of only 200 points, whilst the x-axis starts offevolved at 1300 and has a vary of 200 points. This compressed scale makes the variations between the ratings of the three students appear larger than they absolutely are.

On the other hand, the proper chart has a y-axis that starts at zero and has a vary of 800 points, whilst the x-axis begins at 1200 and has a range of 800 points. This greater expanded scale makes it easier to see the variations between the ratings of the three college students in a extra correct and informative way.

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The answer of the given question based on solving  13 to the power of 16 is 13 to the power of 16 is 3,947,868,257,259,789. and the simplified expression is 7 + 2ab + br.

A expression is a combination of symbols or values that represents a particular concept or computation.

In mathematics, an expression is a combination of numbers, variables, operators, and/or functions that can be evaluated to produce a numerical result.

an expression is a way to represent an idea, computation, or meaning using a set of symbols or words.

To solve 13 to the power of 16, we can start by multiplying 13 by itself 16 times:

13 × 13 = 169

169 × 13 = 2197

2197 × 13 = 28,561

28,561 × 13 = 371,293

371,293 × 13 = 4,826,389

4,826,389 × 13 = 62,748,857

62,748,857 × 13 = 815,730,721

815,730,721 × 13 = 10,604,807,473

10,604,807,473 × 13 = 137,858,491,849

137,858,491,849 × 13 = 1,792,160,390,737

1,792,160,390,737 × 13 = 23,303,986,079,681

23,303,986,079,681 × 13 = 303,305,489,096,753

303,305,489,096,753 × 13 = 3,947,868,257,259,789

Therefore, 13 to the power of 16 is 3,947,868,257,259,789.

As for the second expression, we can simplify it using the distributive property of multiplication:

2(2 + ab) + b(r + 3) = 4 + 2ab + br + 3b

Simplifying further, we can combine the constant terms:

2(2 + ab) + b(r + 3) = 7 + 2ab + br

So the simplified expression is 7 + 2ab + br.

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If a is a zero of the polynomial P(x), then (x-a) must be a factor of P(x) and [tex](x-a)^m[/tex] be a factor of P(x) when we factor P completely.

The values of x that fulfil the formula f(x) = 0 are the zeros of a polynomial. The polynomial's zeros are the x values for which the function's value, f(x), equals zero in this case. The degree of the equation f(x) = 0 determines how many zeros a polynomial has.

The locations when a polynomial equals 0 overall are known as its zeros. In layman's terms, we may state that a polynomial's zeros are variable values at which the polynomial equals 0. The zeros of a polynomial are often referred to as the equation's roots and are frequently written as,, and. A few techniques for locating polynomial zeros include grouping, factoring, and employing algebraic expressions.

(a) if we have zero at x=a of polynomial P(x)

then,  (x-a)  must be factor of P(x).

(b) if we have zero at x=a of polynomial P(x)

with multiplicity=m

then,  [tex](x-a)^m[/tex]  must be factor of P(x).

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Therefore , the solution of the given problem of unitary method comes out to be $500.

This common convenience, already-existing variables, or all important elements from the original Diocesan adaptable study that followed a particular methodology can all be used to achieve the goal. Both of the crucial elements of a term affirmation outcome will surely be missed if it doesn't happen, but if it does, there will be another chance to get in touch with the entity.

Here,

Reginald's total moving costs must first be determined in order to determine the moving expense deduction he may claim on his 2022 tax return.

The full cost of relocating is:

=> $450 (gasoline) + $250 (truck rental) + $100 (tow trailer) + $90 (food) + $35 (Starbucks) + $300 (motel lodging) + $175 (storage)

=> $1,400

Reginald's moving expenditures were covered by the government for $900, so the following are his actual moving costs:

=> $1,400 - $900 = $500

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Therefore , the solution of the given problem of unitary method comes out to be  the attraction sold 943 child tickets and 736 adult tickets on that particular day.

It is possible to accomplish the objective by using previously recognized variables, this common convenience, or all essential components from a prior malleable study that adhered to a specific methodology. If the expression assertion result occurs, it will be able to get in touch with the entity again; if it does not, both crucial systems will undoubtedly miss the statement.

Here,

Assume the attraction sold x tickets for adults and y tickets for kids.

Based on the supplied data, we can construct the following two equations:

=>  x + y = 1679 (equation 1, representing the total number of tickets sold)

=>  35x + 20y = 44620 (equation 2, representing the total revenue generated)

Using the elimination technique, we can find the values of x and y.

When we divide equation 1 by 20, we obtain:

=>  20x + 20y = 33580 (equation 3)

Equation 3 is obtained by subtracting equation 2 to yield:

=>  15x = 11040

=>  x = 736

When we enter x = 736 into equation 1, we obtain:

=>  736 + y = 1679

=> y = 943

As a result, the attraction sold 943 child tickets and 736 adult tickets on that particular day.

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Answer: Questionnaire refers to a research instrument, in which a series of question, is typed or printed along with the choice of answers, expected to be marked by the respondents, used for survey or statistical study. It consists of aformalisedd set of questions, in a definite order on a form, which are mailed to the respondents or manually delivered to them for answers. The respondents are supposed to read, comprehend and give their responses, in the space provided.

A ‘Pilot Study’ is advised to be conducted to test the questionnaire before using this method. A pilot survey is nothing but a preliminary study or say rehearsal to know the time, cost, efforts, reliability and so forth involved in it.

The interview is a data collection method wherein a direct, in-depth conversation between interviewer and respondent takes place. It is carried out with a purpose like a survey, research, and the like, where both the two parties participate in the one to one interaction. Under this method, oral-verbal stimuli are presented and replied by way of oral-verbal responses.

It is considered as one of the best methods for collecting data because it allows two way exchange of information, the interviewer gets to know about the respondent, and the respondent learns about the interviewer. There are two types of interview:

Personal Interview: A type of interview, wherein there is a face to face question-answer session between the interviewer and interviewee, is conducted.

Telephonic Interview: This method involves contacting the interviewee and asking questions to them on the telephone itself.

Probability of winning at home is 0.67, Probability of winning away is 0.58.

To calculate the probability of home and away wins, we need to divide the number of wins by the total number of games played at home and away respectively.

Probability of winning at home = (Number of wins at home) / (Total number of games played at home)

= 8 / 12

= 0.67

Probability of winning away = (Number of wins away) / (Total number of games played away)

= 7 / 12

= 0.58

Based on these results, Sasha's team has a higher probability of winning at home (0.67) compared to away games (0.58).

Therefore, her belief that her team plays better away is not supported by the data.

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Answer:

3x(x - 4y)

Step-by-step explanation:

3x² - 12xy ← factor out 3x from each term

= 3x(x- 4y)

The required answers are 1) [tex]$$A = x^2 + 28x + 192$$[/tex] 2) 300000 3) [tex]$$x^2 + 28x + 192 = (x + 14 - 2\sqrt{19})(x + 14 + 2\sqrt{19})$$[/tex].

area of the land purchased is given as 480m², and the dimensions of the land are (x+12)mx(x+16)m. Therefore, the area of the land can be expressed as:

[tex]$$A = (x+12)(x+16)$$[/tex]

Expanding this expression, we get:

[tex]$$A = x^2 + 28x + 192$$[/tex]

Hence, the area of the land purchased is given by the polynomial expression [tex]$x^2 + 28x + 192$[/tex].

The total budget for the expenses is Rs. 5,00,000. If Mr. Chand's daughter is ready to share 3/5 of the expenses, then the fraction of the expenses she will pay is:

[tex]$\frac{3}{5}=\frac{x}{500000}$$[/tex]

Simplifying this expression, we get:

[tex]$x = \frac{3}{5}\times 500000 = 300000$$[/tex]

Therefore, Mr. Chand's daughter will pay Rs. 3,00,000 towards the expenses.

We can solve the polynomial [tex]$x^2 + 28x + 192$[/tex] into different factors by using the quadratic formula:

[tex]$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$[/tex]

Here, the coefficients of the polynomial are:

[tex]$$a = 1, \quad b = 28, \quad c = 192$$[/tex]

Substituting these values in the quadratic formula, we get:

[tex]$x = \frac{-28 \pm \sqrt{28^2 - 4\times 1 \times 192}}{2\times 1}$$[/tex]

Simplifying this expression, we get:

[tex]$$x = -14 \pm 2\sqrt{19}$$[/tex]

Therefore, the polynomial [tex]$x^2 + 28x + 192$[/tex] can be factored as:

[tex]$$x^2 + 28x + 192 = (x - (-14 + 2\sqrt{19}))(x - (-14 - 2\sqrt{19}))$$[/tex]

or

[tex]$$x^2 + 28x + 192 = (x + 14 - 2\sqrt{19})(x + 14 + 2\sqrt{19})$$[/tex]

So, we have factored the polynomial into two factors.

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The value of [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}[/tex] is -1500.

What is function?

A mathematical phrase, rule, or law that establishes the link between an independent variable and a dependent variable (the dependent variable). In mathematics, functions exist everywhere, and they are crucial for constructing physical links in the sciences.

Here the given function is,

[tex]f(x)=(x^2+6x+9)^{50}-4x+3[/tex]

For any r , [tex](x^2+6x+9)^{50}-4x+3[/tex] is satisfy, Then take f(x)=0 then

=> [tex](x^2+6x+9)^{50}-4x+3=0[/tex]

Take x=r then

=> [tex](r^2+6r+9)^{50}-4r+3=0[/tex]

=> [tex](r^2+6r+9)^{50}=4r-3[/tex]

=> [tex]((r+3)^2)^{50}=4r-3[/tex]

=> [tex](r+3)^{100}=4r-3[/tex]

Then,

=> [tex]\sum_{i=1}^{100} (r_i+3)^{100}=\sum_{i=1}^{100} (4r_i-3)[/tex] = 4 × sum of roots - 300

Expanding [tex](x+3)^{100}-4x+3[/tex] using the binomial theorem, we get

=> f(x) = [tex]x^{100}+300x^{99}+....[/tex]

So sum of roots = -300 then

=> [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100} =[/tex] 4*(-300)-300=-1200-300

=> -1500.

Hence , The value of [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}[/tex] is -1500.

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When the data is skewed, it is untrue that the mean and standard deviation are better indicators of the centre and spread.

The median is a better tool to use to locate the centre when it is skewed right or left with high or low outliers. The IQR is the most accurate indicator of spread when the median is the centre. When the mean is the centre, the standard deviation should be utilised because it gauges how far a data point is from the mean.

The standard deviation will be greatly overstated in cases when the distribution of the data is highly skewed, making it a poor choice as a measure of variability.

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Answer:

[tex]5x+11[/tex]

Step-by-step explanation:

Step 1: Distribute

[tex]3x+6+2x+5[/tex]

Step 2: Add like terms

[tex]5x+11[/tex] < your answer

Consider the line segment AB shown the locations for point C makes ABC a right triangle with hypotenuse AB

In a coordinate pair the points are as represented as (x, y).

The point that forms the right triangle is located by tracing the point on the x axis of of the point A and the point on the y axis of the point B.  This is done below

therefore we can say that the point C that forms the right triangle is

C (2, 3)

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Answer: 44 miles

Step-by-step explanation:

Let's call the number of miles Kevin can travel "m". We can set up an equation using the given information:

4.25 + 0.47m ≤ 25

Solving for "m", we can begin by subtracting 4.25 from both sides:

0.47m ≤ 20.75

Then, divide both sides by 0.47:

m ≤ 44.15

Therefore, Kevin can travel at most 44 miles without exceeding his spending limit.

All answers are mentioned below.

A balance sheet is a financial statement that provides a snapshot of a company's financial position at a specific point in time. It presents the company's assets, liabilities, and equity, and shows how these are financed. The balance sheet follows the accounting equation, which states that the total assets must equal the sum of liabilities and equity.

(a) Trading, Profit & Loss and Profit & Loss Appropriation Account for the year ended December 31, 1996

Trading Account

$

Sales (51,550 - 5,550 + 5,385) 51,385

Less: Cost of Goods Sold

Opening Stock 5,550

Add: Purchases 29,525

35,075

Less: Closing Stock 5,385

Cost of Goods Sold 29,690

Gross Profit 21,695

Profit & Loss Account

$

Gross Profit 21,695

Less: Expenses

Carriage Inward 100

Discount Allowed 1,000

Wages & Salaries 17,650

Motor Vehicle Repairs 5,850

Telephones 350

Rent & Rates 1,800

Advertising 2,750

Carriage Outward 1,000

Bank Charges 550

Depreciation

Land & Building (30,000/50 years) 600

Motor Vehicle (20% x 20,000) 4,000

34,550

Net Loss (12,855)

Profit & Loss Appropriation Account

$

Net Loss (12,855)

Add: Interest on Capital (5% x $90,000) 4,500

(8,355)

To be shared equally between Knox and Cox

Knox (4,178)

Cox (4,178)

Additional Salary for Cox (5,000)

Carried Forward (13,356)

(b) Partners' Current Account

Knox Cox

$ $

To Opening Balance 8,185 4,400

By Share of Net Loss (4,178) (4,178)

By Interest on Capital 2,500 2,000

By Drawings (1,200) (500)

By Salary 0 (5,000)

To Closing Balance 5,307 (3,278)

(c) Balance Sheet as at December 31, 1996

Fixed Assets

Land & Building (100,000 - 30,000) 70,000

Motor Vehicles (20,000 - 4,000) 16,000

86,000

Current Assets

Stock 5,385

Debtors 7,250

Bank 3,200

Cash 510

16,345

Current Liabilities

Creditors (7,650)

Net Current Assets 8,695

94,695

Less: Partners' Capital and Current Accounts

Knox Capital (50,000 + 5,307) 55,307

Cox Capital (40,000 - 3,278) 36,722

92,029

2,666

Notes:

Depreciation of Land and Building: 30,000/50 years = 600 per year

Depreciation of Motor Vehicles: 20% x 20,000 = 4,000

Interest on Capital: Knox - 5% x 50,000 = 2,500, Cox - 5% x 40,000 = 2,000

Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50

Additional Salary for Cox is treated as an expense and charged to the Profit & Loss Account.

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(a) Trading, Profit & Loss and Profit for the year ended December 31, 1996 is mentioned below. (b) Knox Cox, Opening Balance 8,185 4,400, respectively.(c) Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50

A balance sheet is a financial statement that provides a snapshot of a company's financial position at a specific point in time. It presents the company's assets, liabilities, and equity, and shows how these are financed. The balance sheet follows the accounting equation, which states that the total assets must equal the sum of liabilities and equity.

(a) Trading, Profit & Loss and Profit & Loss Appropriation Account for the year ended December 31, 1996

Trading Account

$

Sales (51,550 - 5,550 + 5,385) 51,385

Less: Cost of Goods Sold

Opening Stock 5,550

Add: Purchases 29,525

35,075

Less: Closing Stock 5,385

Cost of Goods Sold 29,690

Gross Profit 21,695

Profit & Loss Account

$

Gross Profit 21,695

Less: Expenses

Carriage Inward 100

Discount Allowed 1,000

Wages & Salaries 17,650

Motor Vehicle Repairs 5,850

Telephones 350

Rent & Rates 1,800

Advertising 2,750

Carriage Outward 1,000

Bank Charges 550

Depreciation

Land & Building (30,000/50 years) 600

Motor Vehicle (20% x 20,000) 4,000

34,550

Net Loss (12,855)

Profit & Loss Appropriation Account

Net Loss (12,855)

Add: Interest on Capital (5% x $90,000) 4,500

(8,355)

To be shared equally between Knox and Cox

Knox (4,178)

Cox (4,178)

Additional Salary for Cox (5,000)

Carried Forward (13,356)

(b) Partners' Current Account

                                    Knox Cox

                                          $ $

To Opening Balance 8,185 4,400

By Share of Net Loss (4,178) (4,178)

By Interest on Capital 2,500 2,000

By Drawings               (1,200) (500)

By Salary                         0 (5,000)

To Closing Balance     5,307 (3,278)

(c) Balance Sheet as at December 31, 1996

Fixed Assets

Land & Building (100,000 - 30,000) 70,000

Motor Vehicles (20,000 - 4,000) 16,000

86,000

Current Assets

Stock 5,385

Debtors 7,250

Bank 3,200

Cash 510

16,345

Current Liabilities

Creditors (7,650)

Net Current Assets 8,695

94,695

Less: Partners' Capital and Current Accounts

Knox Capital (50,000 + 5,307) 55,307

Cox Capital (40,000 - 3,278) 36,722

92,029

2,666

Notes:

Depreciation of Land and Building: 30,000/50 years = 600 per year

Depreciation of Motor Vehicles: 20% x 20,000 = 4,000

Interest on Capital: Knox - 5% x 50,000 = 2,500, Cox - 5% x 40,000 = 2,000

Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50

Additional Salary for Cox is treated as an expense and charged to the Profit & Loss Account.

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