Add rational expressions with unlike denominators

x-5/4x^2-25x-21 + 4x^2+15x+9

The player with the most attempted goals is Alex Morgan.

The number listed in the "GAS" column is the number of goals attempted by each player in the 2021 season.

To determine which player attempted to make the most goals, we would need to divide the number of goals attempted by the number of games played to find the average number of goals attempted per game.

For example, let's consider Alex Morgan:

GAS: 5

Games Played: 2

Average Goals Attempted per Game = GAS ÷ Games Played = 5 ÷ 2 = 2.5

let's perform the calculation for each player:

Based on these calculations, Alex Morgan attempted to make the most goals, with an average of 2.5 goals attempted per game.

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Answer:

Explanation:

Answer:

i got u i have your email

Step-by-step explanation:

Answer:

Explanation:

left parenthesis, 7. 5, comma, 2, right parenthesis

(-3, -4)(−3,−4)left parenthesis, minus, 3, comma, minus, 4, right parenthesis

(0, -2)(0,−2)left parenthesis, 0, comma, minus, 2, right parenthesis

Quadrant I, Quadrant II, Quadrant III, Quadrant IV, xxx-axis, yyy-axis

Answer:

1:true, and its a sign of Jesus saving us sinners.

2:true

3:false, you shouldn't correct sinners, its like saying; you just sinned, let me tell you what you did wrong.

4:true very true

5:true, you gave them a home they needed when they were desperate.

6:false

7:true

Explanation:

If I'm wrong I'm sorry. I hope I helped...

The regression prediction error for height at 6 from height at 18 is 1.5.

One of the methods most frequently used to assess the accuracy of forecasts is root mean square error, also known as root mean square deviation. It illustrates the Euclidean distance between measured true values and forecasts. The Root Mean Square Error (RMSE) statistic calculates the amount of error between two data sets. In other words, it contrasts a value that was anticipated with one that was observed or known.

A measurement of the variation between known locations and locations that have been digitalized or interpolated. RMS error is calculated by taking the square root of the difference between the known and unknown points, which is then multiplied by the number of test points.

√((1-r^2 )

The r.m.s. error for this regression's height at 18 versus height at 6 prediction is as follows:

√(1-〖0.8〗^2 ).2.5=√(1-0.64 ).2.5=√0.36 .2.5=0.6 ⋅2.5=1.5

The regression prediction error for height at 6 from height at 18 is:

√(1-〖0.8〗^2 ).1.7=√(1-0.64 ).1.7=√0.36 .1.7=0.6 ⋅1.7=1.02

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Answer:

9

Explanation:

To find the positive solution to the equation x^2 + 10 = 91, we'll start by isolating x^2 on one side of the equation. To do this, we'll subtract 10 from both sides:

x^2 + 10 - 10 = 91 - 10

x^2 = 81

Next, we'll take the square root of both sides of the equation to solve for x:

√(x^2) = √(81)

x = ±9

Since we are looking for the positive solution, x = 9.

Answer:

Explanation:

The sentence "I place this clipping into the basket next to me and crouch under" describes an action being taken by the speaker. The phrase "next to me" is a prepositional phrase that modifies the noun "basket".

In English, when a prepositional phrase modifies a noun, it is common to separate the noun and the prepositional phrase with a comma. This helps to clarify the relationship between the noun and the prepositional phrase, making the sentence easier to understand for the reader.

Therefore, option C: "basket, next to me" is the best choice because it correctly separates the noun "basket" and the prepositional phrase "next to me" with a comma. This makes the sentence clearer and easier to understand.

0.0294 is the proportion of college students sleep for more than 8.6 hours.

An equivalence of two ratios is a proportion. We use proportions to construct equivalent ratios and find solutions to problems involving unknown numbers.If a change in one variable can be expected from a shift in the other, two factors are considered to be in proportion. For secondary school chemists, there are two different kinds of proportional relationships that are important: direct proportion and inverse proportion.

Accordingly, you can answer the task as follows:

population mean(u) =8.6 hours

& standard deviation( s)= 35/60 =0.583 hours

P(X>9.7) =  P(Z> ((9.7-8.6) / 0.583)

= P(Z>1.89)  

= 1 - P(Z<1.89)

= 1 - 0.9706    

= 0.0294

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The estimate is considered accurate if it is within 1.3 cm, or 8.72 cm, as indicated by the results about the accuracy of estimates of s discovered using the range rule of thumb.

While a high standard deviation indicates that the data are much more spread, a small standard deviation suggests that the data are clustered around the mean. The standard deviation is a measure of how variable your data set is on average. It displays the standard deviation of the each score with respect to the mean.

Standard deviation is important because it makes measurements easier to understand when the data is dispersed. The more equally distributed the data is, the bigger overall standard deviation of a data will be.

Determine the mean first. Step 2: Determine the square of the variation from the mean of each data point. In Step 3, add the values from Step 2. By the overall amount of information collected in step 4, divide.

The range rule of thumb is a rough estimation of a data set's standard deviation based on the data's range. The estimate comes from:

s ≈ R / 4

where R denotes the data range, which is the difference between the data set's maximum and minimum values. The guys' heights in this instance range from 191.9 to 157.0 cm, or 34.9 cm.

We estimate the standard deviation to be as follows using the range rule of thumb:

s ≈ 34.9 / 4 = 8.72 cm

We can observe that the estimate is within 1.3 cm of the real standard deviation by comparing it to the estimated standard deviation, which is 7.99 cm. This shows that the standard deviation estimate derived from the range rule of thumb is rather accurate.

It is important to remember that this rule of thumb is only a rough estimate, and that there are more exact ways to get the standard deviation, such as the calculation based on the mean and variance, which should be used in most situations.

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Answer:

The value of x and y in the equation will be 3 2/3 and 12 2/3 respectively.

An equation simply has to do with the statement that illustrates the variables given. In this case, it is vital to note that two or more components are considered in order to be able to describe the scenario. It is important to note that an equation is the mathematical statement which can be made up of two expressions which are connected by an equal sign.

-x + y = 8

2x - y = 3

From equation i, y = 8 + x

2x - y = 3

2x - (8 + x) = 3

2x - 8 + x = 3

3x = 11

x = 11/3

y = 8 + 11/3

y = 11 2/3

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Answer:

Explanation:

To solve this problem, we need to consider the effect of temperature on the measured length, as well as the effect of tape elongation due to the applied tension. The corrected total length can be calculated using the following steps:

Step 1: Find the thermal expansion of the tape at the two temperatures.

The thermal expansion of the tape can be calculated using the formula:

ΔL = αLΔT

where ΔL is the change in length, α is the coefficient of thermal expansion, L is the original length, and ΔT is the change in temperature. We can assume that the change in temperature is the same for the entire tape, so we only need to calculate ΔL for one half of the line. Using the given values, we get:

ΔL = αLΔT

= (12*10^-6 per oc) * (1000 m) * (5/2) * (20-15)

= 0.15 m

So the tape will expand by 0.15 m at a temperature of 20°C.

Similarly, at a temperature of 26°C, we have:

ΔL = αLΔT

= (12*10^-6 per oc) * (1000 m) * (5/2) * (26-15)

= 0.27 m

So the tape will expand by 0.27 m at a temperature of 26°C.

Step 2: Find the elongation of the tape due to the applied tension.

The elongation of the tape due to the applied tension can be calculated using the formula:

ΔL = (F * L) / (E * A)

where ΔL is the change in length, F is the applied force, L is the original length of the tape, E is the Young's modulus of the tape material, and A is the cross-sectional area of the tape. We can assume that the force is the same for the entire tape, so we only need to calculate ΔL for one half of the line. Using the given values, we get:

A = (3 mm) * (1.25 mm)

= 3.75 mm^2

= 0.375 cm^2

ΔL = (F * L) / (E * A)

= (22 kg) * (50 m) / (2.11*10^6 kg/cm^2 * 0.375 cm^2)

= 0.111 m

So the tape will elongate by 0.111 m due to the applied tension.

Step 3: Find the corrected total length.

To find the corrected total length, we need to subtract the thermal expansion of the tape at each temperature and add the elongation due to the applied tension. Since we are measuring only half of the line at each temperature, we need to double the result to get the total length. Using the values calculated in steps 1 and 2, we get:

Corrected length at 20°C = (2000 m / 2) - 0.15 m + 0.111 m

= 995.96 m

Corrected length at 26°C = (2000 m / 2) - 0.27 m + 0.111 m

= 994.84 m

Corrected total length = 2 * (995.96 m + 994.84 m)

= 3981.6 m

Therefore, the corrected total length of the line is approximately 3981.6 meters.

The corrected total length of the line is approximately 2000.07 meters.

A material's response to a change in temperature is measured by a coefficient of thermal expansion, which is frequently denoted by the symbol.

To calculate the corrected total length, we need to consider the effects of tape expansion and temperature on the tape and the measured line. Here are the steps to follow:

Calculate the expansion of the tape due to temperature difference:

The temperature difference between 15°C and 20°C is 5°C, and the temperature difference between 15°C and 26°C is 11°C.

The coefficient of expansion is [tex]12X10^-^6[/tex] per °C.

The length of the tape is 50m, and its width and thickness are 3mm and 1.25mm, respectively.

The expansion of the tape due to the temperature difference is:

ΔL = L₀ * α * ΔT

where L₀ is the initial length of the tape (50m), α is the coefficient of expansion ([tex]12X10^-^6[/tex] per °C), and ΔT is the temperature difference (5°C for the first half and 11°C for the second half).

ΔL1 = 50m x [tex]12X10^-^6[/tex]/°C * 5°C = [tex]310^-^3[/tex] m

ΔL2 = 50m x [tex]12X10^-^6[/tex]/°C * 11°C = [tex]6.610^-^3[/tex] m

Calculate the tension in the tape due to the applied pull:

The weight of the tape is 7.75 g/[tex]cm^3[/tex], and its dimensions are 3mm (width) and 1.25mm (thickness).

The mass of the tape is:

m = ρ x V = 7.75 g/[tex]cm^3[/tex] x 3mm * 1.25mm x 50m = 0.73125 kg

The tension in the tape due to the applied pull of 22kg is:

T = F/A = 22kg x g / (3mm * 50m) = 14.67 N

The stress in the tape is:

σ = T/A = T / (3mm x 1.25mm) = 3.72 x [tex]10^7 N/m^2[/tex]

Calculate the strain in the tape due to the stress:

The Young's modulus of the tape is E = [tex]2.1110^6 kg/cm^2[/tex] = [tex]2.11X10^1^0 N/m^2.[/tex]

The strain in the tape is:

ε = σ / E = [tex]3.72X10^7[/tex]N/[tex]m^2[/tex] / [tex]2.11X10^1^0 N/m^2[/tex] = [tex]1.76X10^-^3[/tex]

The elongation of the tape due to the strain is:

δL = L₀ * ε = 50m x 1.76 x [tex]10^-^3[/tex] = 0.088 m

Calculate the corrected length of the line:

The length of the line measured with the tape is 2 km = 2000 m.

Now, let's calculate the corrected total length of the line. The original measured length of the line is 2 km, or 2000 m. Therefore, the corrected total length of the line is:

Lc = 2000 m + ΔL1 + ΔL2 + ε1L1 + ε2L2

Lc = 2000 m + 0.009 m + 0.0198 m + 0.00340 * 1025 m + 0.00340 * 1025 m

Lc = 2000.0686 m

Thus, the corrected total length of the line is approximately 2000.07 meters.

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The value of y from the given equation is 26.

We can substitute the second equation into the first equation to solve for y. Doing so, we get:

Now that we have x = 9, we can substitute it into either of the equations to find y. Let's use the second equation:

So the value of y is 26 if (x, y) satisfies the given system of equations.

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If you're encountering the message while trying to sign and submit your FAFSA, there are a few things you can try to resolve the issue:

Clear your browser's cache and cookies, and then try to submit the FAFSA again.

Try using a different web browser, or try accessing the FAFSA website from a different device (e.g., a different computer, tablet, or smartphone).

Double-check that you have completed all required fields on the FAFSA form and that you have correctly entered all requested information.

Wait a few hours and try again, as the issue may be temporary and resolved on its own.

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Building, Print-on-Demand, Equipment Upkeep, Dropshipping, and Owning an Online Store.

Furthermore, according to Schumpeter, the sources of chances for this type of entrepreneurship are found in changes in socio-demographics, politics, and technology.

The launch of a new company venture is the best illustration of entrepreneurship. The market is opened up to new ideas by entrepreneurs, who are frequently credited as innovators or sources of fresh concepts.

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Answer:

Its 17% per person

Explanation:

To estimate the standard error, we can use a bootstrap resampling method. We randomly sample with replacement from the original sample and compute the proportion of agreement in each new sample.

Enter the numbers "Agree" as 382, "Disagree" as 578, and "Undecided" as 40 in the corresponding fields after clicking "Enter data."

Select "Bootstrap" under "Random samples" after clicking.

Set the number of resamples to 1000.

Look for the standard error in the "Bootstrap standard error" column after clicking "Compute." About 0.016 is the estimated standard error.

Sample percentage +/- z*, confidence interval (standard error)

where z is the desired level of confidence's associated z-score. The z-score for a 95% confidence interval is 1.96.

The sample proportion of agreement is 382/1000 = 0.382.

Thus, the 95% confidence interval for the proportion of the population to agree with the statement is:

0.382 +/- 1.96*(0.016) = (0.351, 0.413)

Therefore, the 95% confidence interval is (0.351, 0.413).

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Answer:

Energy is used to power the "night mail" is not specified in the works of J.R.R. Tolkien, who was the author of "The Lord of the Rings" and "The Hobbit." The "night mail" is a reference to a postal service that delivers letters and packages at night, as described in the poem "Night Mail" by W. H. Auden. However, there is no information available about what energy source was used to power this service.