DIVIDING ALGEBRAIC FRACTIONS

a^2-7a-18/4a^3 divided by a^2-4a-45/2a^3-4a^2 multiplied by a^2+3a-10/a^2-4+4

Your answer is 19 ok

-3/2 = (-4/7)u - (4/3)

-3/2 + 4/3 = (-4/7)u

-9/6 + 8/6 = (-4/7)u

(-9 + 8)/6 = (-4/7)u

-1/6 = (-4/7)u

(-1/6) * (-7/4) = (-7/4)*(-4/7)u

7/24 = u

The answer: u = 7/24

The current definition of the standard meter of length is based on the distance traveled by light in a vacuum. The speed of light in a vacuum is 186,282 miles per second and in kilometers it is 299,792 kilometers per second. According to the SI unit (which is the international standard unit system for the measurements), the unit of length is meter.

1. If tomorrow is Monday then, today is Sunday.
2. If today is not Sunday then tomorrow is not Monday.
3. If  tomorrow is not Monday then today is not Sunday

If tomorrow is not Monday then today is not Sunday.

The given statement is "If today is Sunday, then tomorrow is Monday".

What is contrapositive statement?

A contrapositive statement occurs when you switch the hypothesis and the conclusion in a statement, and negate both statements. In this example, when we switch the hypothesis and the conclusion, and negate both, the result is: If it is not a polygon, then it is not a triangle.

Here,

1. If tomorrow is Monday then, today is Sunday.

2. If today is not Sunday then tomorrow is not Monday.

3. If tomorrow is not Monday then today is not Sunday.

Therefore, if tomorrow is not Monday then today is not Sunday.

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We need to print 14 copies of a 80 pages report. Thus the total number of pages that needs to be printed is:

80*14=80(10+4)=800+320=1,120 pages.

Paper is bought on reams of 500 pages, so we can buy :

1 ream, that is 500 pages, 

2 reams, that is 500*2=1,000 pages

3 reams, that is 500*3=1,500 pages.


It is clear that to need to buy 3 reams, so that we have enough paper.


3 reams * $3.66 per ream = $10.98 



Answer: $10.98 

To print 14 copies of an 80-page report, you would need 3 reams of paper, costing a total of $10.98.

To calculate the cost of the paper for printing reports, we will need to determine the total number of pages being printed, how many reams that equates to, and then calculate the cost based on the price per ream. First, we calculate the total pages: 80 pages per report  imes 14 copies = 1120 total pages. Next, we determine how many reams of paper are needed: 1120 total pages ÷ 500 pages per ream = 2.24 reams. Since paper is sold by whole reams, we will need to purchase 3 reams to have enough paper.

The cost of the paper will be 3 reams x $3.66 per ream = $10.98. Therefore, the cost for the paper needed to print the 14 reports will be $10.98.

If you devide 52 by 4 you get 13....

5+2= 7 so 7x7=49 so Shelia will spend $49 dollars on 7 rides

7x2= 14+5=19 Sheila will spend 19

Both are used to understand chance and to collect and analyze numerical data.

Statistical data is used to see if conclusions are true and to make future predictions.

Answer:

The reason is that none of these latter statements accurately reflects the data. Scientific data rarely lead to absolute conclusions.

Step-by-step explanation:

The final answer is that there's a 38.29% probability that [tex]\( y \)[/tex]  will fall between 195 and 205.

Let's break down the calculation step by step.

1. First, we need to determine the mean (average) and standard deviation of the data set. Let's assume we have a normal distribution with a mean (μ) of 200 and a standard deviation (σ) of 10.

2. Next, we'll use the Z-score formula to standardize the values of 195 and 205.

[tex]\( Z = \frac{{X - \mu}}{\sigma} \)[/tex]

[tex]For \( X = 195 \):[/tex]

[tex]\( Z_{195} = \frac{{195 - 200}}{10} = -0.5 \)[/tex]

[tex]For \( X = 205 \):[/tex]

[tex]\( Z_{205} = \frac{{205 - 200}}{10} = 0.5 \)[/tex]

3. After standardizing, we can look up the corresponding probabilities in the standard normal distribution table (Z-table) or use a calculator/tool that provides this information.

4. The Z-table shows that the probability of Z being between -0.5 and 0.5 (inclusive) is approximately 0.3829 (or 38.29%).

So, the exact probability that [tex]\( y \)[/tex] will fall between 195 and 205 is 38.29%.

Explanation:

1. We started by identifying the mean and standard deviation of the data set, which are essential parameters for working with normal distributions.

2. Using the Z-score formula, we standardized the values of 195 and 205 to Z-scores, allowing us to compare these values on a standard normal distribution.

3. By referencing the Z-table or using a tool, we found that the probability of Z falling between -0.5 and 0.5 is 0.3829 (38.29%).

4. Therefore, the final answer is that there's a 38.29% probability that [tex]\( y \)[/tex]  will fall between 195 and 205.

Complete Question:
Can you state the exact probability that y will fall between 195 and 205?

I had to do several calculations to get the answer.

First I calculated how many points each incorrect answer decreased:

6 questions * x = 30 points => x = 30 points / 6 questions = 5 points / question.

Then I converted it in an equation of the kind:

( ) ( ) = 5

And concluded it was 1/6 * 30 = 5

So, the numbers must be placed on either of these two forms:

( 1/6 ) ( 30) = 5, or

( 30 ) ( 1/ 6) = 5.

Of course, both of them are correct answers.

well, from 2008 to 2010 is just 2 years, so let's say the initial amount is the 19300 from 2008, how many will it be in 2010?

[tex]\bf \qquad \textit{Amount for Exponential Growth}\\\\ A=I(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &19300\\ r=rate\to 20\%\to \frac{20}{100}\to &0.20\\ t=\textit{elapsed time}\to &2\\ \end{cases} \\\\\\ A=19300(1+0.2)^2[/tex]

Answer: The amount of cars they can expect to wash is 27, 792 in 2010.

Step-by-step explanation:

This problem asks us to calculate the potential growth of the business through a period. One of the ways to solve this problem is to use the formula for exponential growth to determine the number of cars they expect to wash in 2010.

[tex]f(x) = a(1 + r) ^x\\f(x) = 19, 300 (1+0.20)^2\\f(x) = 19, 300 (1.20)^2\\f(x) = 27, 792[/tex]

a = initial growth - 19, 300

r = growth rate - 0.20

x = time (years) - 2 years

We get 27, 792 as our final answer.

$16.32 is the correct answer.

Thanks for the question!

To do the inverse of the distributive property, we need to find the GCF. In this case, it's 6. then, we take the 6 common and divide each term by 6:

6(x + 18)

Hope this helps!

the Washington monument is 555 feet

555/3.28 = 169.21 meters

Answer:

169.164

Step-by-step explanation:

555 / 3.28 = 169.164

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

[tex]\begin{tabular} {|c|c|c|c|} Month&Price per Chip&Month&Price per Chip\\[1ex] January&\$1.90&July&\$1.80\\ February&\$1.61&August&\$1.83\\ March&\$1.60&September&\$1.60\\ April&\$1.85&October&\$1.57\\ May&\$1.90&November&\$1.62\\ June&\$1.95&December&\$1.75 \end{tabular}[/tex]

The forcast for a period [tex]F_{t+1}[/tex] is given by the formular

[tex]F_{t+1}=\alpha A_t+(1-\alpha)F_t[/tex]

where [tex]A_t[/tex] is the actual value for the preceding period and [tex]F_t[/tex] is the forcast for the preceding period.

Part 1A:
Given α ​= 0.1 and the initial forecast for october of ​$1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

[tex]F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804[/tex]

Therefore, the foreast for period 11 is $1.80


Part 1B:

Given α ​= 0.1 and the forecast for november of ​$1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

[tex]F_{12}=\alpha A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ =0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782[/tex]

Therefore, the foreast for period 12 is $1.78



Part 2A:

Given α ​= 0.3 and the initial forecast for october of ​$1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

[tex]F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ =0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703[/tex]

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α ​= 0.3 and the forecast for November of ​$1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

[tex]F_{12}=\alpha A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ =0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676[/tex]

Therefore, the foreast for period 12 is $1.68




Part 3A:

Given α ​= 0.5 and the initial forecast for october of ​$1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

[tex]F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ =0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645[/tex]

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α ​= 0.5 and the forecast for November of ​$1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

[tex]F_{12}=\alpha A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ =0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635[/tex]

Therefore, the forecast for period 12 is $1.64



Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

[tex] \frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16[/tex]

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.1 of October, November and December is given by: 0.157



Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

[tex] \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = \frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = \frac{0.32}{3} \approx0.107[/tex]

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.3 of October, November and December is given by: 0.107




Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

[tex] \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = \frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = \frac{29}{3} \approx0.097[/tex]

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.5 of October, November and December is given by: 0.097

Write this sum as:

[tex]i^{600}+i^{599}+i^{598}+\ldots+i+1=1+i+i^2+i^3+\ldots+i^{599}+i^{600}[/tex]

This is sum of a geometric series with [tex]n=601 [/tex] terms, first term [tex]a=1[/tex] and common ratio [tex]r=i[/tex]. So the sum:

[tex]S=a\dfrac{1-r^n}{1-r}=1\cdot\dfrac{1-i^{601}}{1-i}=\dfrac{1-i\cdot i^{600}}{1-i}=\dfrac{1-i\cdot(i^2)^{300}}{1-i}=\\\\\\=\dfrac{1-i\cdot(-1)^{300}}{1-i}= \dfrac{1-i\cdot1}{1-i}=\dfrac{1-i}{1-i}=\boxed{1}[/tex]

In expanded form it would be:

(7 x 1) + (0/10) + (4/100) =7.04

Thanks for your question!

8 + 3
11

11-7
4 degrees

Hope this helps!

0.6 because, you move your decimal 1 time to the left.

To solve this, we simply need to break down the problem. Doing this will yield us the correct equation.

"8.46 less than"
This shows we are going to have something subtracted by 8.46.
-8.46

"the product of"
This shows we will have a multiplication problem.
*-8.46

"42 and x"
This shows the values that we will be multiplying
42*x-8.46

Now we can simplify
42x-8.46

Using the logic above, we can see that the expression to represent the phrase is 42x-8.46.

Final answer:

The algebraic expression for the word phrase '8.46 less than the product of 42 and x' is '42x - 8.46'. The phrase 'less than' means to subtract the number from the result of the multiplication.

Explanation:

The process of writing an algebraic expression involves converting the given word phrase into mathematical symbols and operations. In this case, the word phrase is '8.46 less than the product of 42 and x'. 'Product' in mathematics means multiplication so 'the product of 42 and x' could be written as '42x'. 'Less than' means to subtract, but the order matters. '8.46 less than' means we subtract 8.46 from our product, not the other way around. Therefore, the algebraic expression for this word phrase is '42x - 8.46'.

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Sent a picture of the solution to the problem (s).

The probability that at least 2 people have the same birthday is 29.37%

Further explanation

Probability is the likelihood of an event occurring. Probability is the number of ways of achieving success. Probability is also the total number of possible outcomes.

Suppose there are 30 people at a party. Do you think any two share the same birthday?

Let's use the random-number table to simulate the birthdays of the 30 people at the party, ignoring leap year.

Let's assume that the year has 365 days. number the days, with 1 representing January 1, 2 representing January 2, and so forth, with 365 representing December 31.

Draw a random sample of 30 days (with replacement). These days represent the birthdays of the people at the party. Would you expect any two of the birthdays to be the same?

[tex]1^{st} people = \frac{365}{365} \\ 2^{nd} people = \frac{364}{365} \\ 3^{rd} people = \frac{363}{365}[/tex]

For 30 people

365! = 365*364*363*...336

So

[tex]= \frac{365*364*363*...336}{(365^{30})}  =  \frac{365!}{(365^{30})}[/tex]

[tex]\frac{365!}{(365^{30})}[/tex] [tex]= \frac{365!/(365-30)!}{365^{30}}[/tex]

[tex]= \frac{365!/335!}{365^{30}} \\ = 0.2937 = 29.37%[/tex]

The probability that at least 2 people have the same birthday is 29.37%

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Answer details

Grade:  9

Subject:  mathematics

Chapter:  probability

Keywords: the same birthday, probability, random sample, party, simulate

Greetings!

To find equivalent ratios, you must divide/multiply both terms of the ratio by the same number:

Examples:
1) 14*2=28
2*2=4
28:4

2) 14/2=7
2/2=1
7:1

3) 14*100=1400
2*100=200
1400:200

Hope this helps.
-Benjamin

24,000/8= 3,000 balls

The machine produce 3,000 balls per hour.

Hope that helps!

Answer: 3,000 balls per hour

Step-by-step explanation:

Given: The number of plastic balls manufactured by a machine = 24,000

Time taken to manufacture 24,000 plastic balls = 8 hours

To calculate the unit rate , we divide the number of plastic balls by the amount of time taken to produce it, we get

[tex]\text{Unit rate}=\dfrac{\text{Number of plastic balls}}{\text{Time}}\\\\\Rightarrow\ \text{Unit rate}=\dfrac{24,000}{8}\\\\\Rightarrow\ \text{Unit rate}=3,000\text{balls per hour}[/tex]

[tex]\bf 81h^8k^6\quad \begin{cases} 81=9^{1\cdot 2}\\ h^8=h^{2\cdot 4}\\ k^6=k^{3\cdot 2} \end{cases}\implies 9^{1\cdot \stackrel{\downarrow }{2}}h^{\stackrel{\downarrow }{2}\cdot 4}k^{3\cdot \stackrel{\downarrow }{2}}\implies (9^1h^4k^3)^{\stackrel{\downarrow }{2}} \\\\\\ (9h^4k^3)^2[/tex]

the blank would be 4 as 4*7 is the same as 7*4

7 x 4=4 x 7

This is known as the commutative property, it basically means when using addition and multiplication the order of the numbers do not matter, because the answer will always turn out the same.

4.

12 - 8 = 4

Hope this helps!

The work done by the weightlifter to lift a barbell and a chain with a total weight of 2,200 N through a vertical distance of 1 meter is 2,200 joules.

The question involves calculating the work done when a weightlifter lifts a barbell from one height to a higher height. To find the work done, we can use the formula W = F  imes d  imes  ext{cos}(\theta), where W is the work, F is the force, d is the distance through which the force acts, and \theta is the angle between the force and the direction of motion. Since the weightlifter is lifting vertically, the angle \theta is 0 degrees, and cos(0) is 1, simplifying the formula to W = F  imes d.

The total weight of the barbell and the chain is the sum of both weights, which is 1,700 N + 500 N = 2,200 N. The barbell is initially 1 meter above the ground and needs to be lifted to a height of 2 meters, so the distance d is 2 m - 1 m = 1 m. Therefore, the work done to lift the barbell and chain is W = 2,200 N  imes 1 m = 2,200 J.

Final answer:

The wet surface area of the boat is approximately 69.7ft^2.

Explanation:

The drag force on a boat varies jointly with the wet surface area and the square of the speed of the boat. This can be represented by the equation:

F = k · A · s^2

where F is the drag force, A is the wet surface area, s is the speed, and k is a constant of variation.

To find the wet surface area of a boat traveling at 7.5mph and experiencing a drag force of 135N, we can set up the following proportion:

(98N) ÷ (50ft^2) = (135N) ÷ (x ft^2)

Cross-multiplying and solving for x, we find that the wet surface area of the boat is approximately 69.7ft^2.

Using the line of best fit, it is found that a fair price for the house is of $176,280.

What does the line of best fit states?

It states that the fair price for a house of x square feet, in thousands of dollars, is given by:

y = 0.074x + 50.48.

In this problem, we have a house of 1700 ft^2, hence the fair price in thousands of dollars is given by:

y = 0.074(1700) + 50.48 = 176.28.

Which is a fair price of $176,280.

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Answer:199,000

Step-by-step explanation: