Find and copy a group of words that tells you What energy is used to power night mail.​

Answer:

Explanation:

To find the class width, we need to divide the range of the data (which is the difference between the maximum and minimum values) by the number of classes.

The range of the data is 91 - 8 = 83

The class width is 83 / 7 = 11.857142857142857 (rounded to one decimal place, it becomes 11.9)

To find the lower class limits, we start from the minimum value and add the class width repeatedly until we reach the upper limit.

Lower class limits:

Class 1: 8 (minimum value)

Class 2: 8 + 11.9 = 19.9

Class 3: 19.9 + 11.9 = 31.8

Class 4: 31.8 + 11.9 = 43.7

Class 5: 43.7 + 11.9 = 55.6

Class 6: 55.6 + 11.9 = 67.5

Class 7: 67.5 + 11.9 = 79.4

To find the upper class limits, we add the class width to each lower class limit.

Upper class limits:

Class 1: 8 + 11.9 = 19.9

Class 2: 19.9 + 11.9 = 31.8

Class 3: 31.8 + 11.9 = 43.7

Class 4: 43.7 + 11.9 = 55.6

Class 5: 55.6 + 11.9 = 67.5

Class 6: 67.5 + 11.9 = 79.4

Class 7: 79.4 + 11.9 = 91.3 (maximum value)

So, the class width is 11.9, the lower class limits are 8, 19.9, 31.8, 43.7, 55.6, 67.5, 79.4, and the upper class limits are 19.9, 31.8, 43.7, 55.6, 67.5, 79.4, 91.3.

We must divide the data's range—that is, the difference between the maximum and minimum values—by the number of classes in order to determine the class width. The data's range is 91 - 8 = 83.

Thus, Class width is the distance between any class's (category's) upper and bottom bounds. It may also refer to one of the following more precisely, depending on the author.

The difference between the lower limits of two successive classes, or the difference between the upper limits of two successive (neighboring) classes.

An important element of a frequency distribution table is class breadth. A teacher documenting the percentage of pupils that received A's (90+), B's (80-89), C's (70-79), etc. on a test is a nice example of a frequency distribution table.

Thus, We must divide the data's range—that is, the difference between the maximum and minimum values—by the number of classes in order to determine the class width. The data's range is 91 - 8 = 83.

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129 hamburgers were sold on Tuesday.

Let's start by using algebra to represent the information given in the problem.

We are told that the total number of hamburgers and cheeseburgers sold is 387.

If we let X be the number of hamburgers sold, then the number of cheeseburgers sold is 2X (since we are told that the number of cheeseburgers sold was twice the number of hamburgers sold).

So the total number of burgers sold can be expressed as:

X + 2X = 387

Simplifying the left-hand side of the equation, we get:

3X = 387

Dividing both sides by 3, we get:

X = 129

Therefore, 129 hamburgers were sold on Tuesday.

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Franny needs to pay an insurance premium of $250.00 per year

An insurance premium is the amount of money that an individual or business pays to an insurance company in exchange for insurance coverage. It is the price of the insurance policy and is usually paid on a regular basis, such as monthly or annually.

Franny bought an insurance contract for her computer. The contract covers theft of the computer, but does not cover damages caused by faulty wiring. If Franny's computer is stolen, she can file a claim with the insurance company, and the insurance company will pay the rest of the cost to buy another computer, with a deductible of $500.00. Franny needs to pay a total premium of $250.00 per year for the insurance coverage. The contract started on July 1 and is valid for one year.

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Answer:

Explanation:

Answer:

i got u i have your email

Step-by-step explanation:

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:

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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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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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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.

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:

Explanation:

Here are the results for the data:

Mean: The mean, or average, of the data can be calculated by summing up all the values and dividing by the number of values:

(11+16+10+30+24+5+6+12+11+45+9+8+3+4+35+31)/16 = 201/16 = 12.5625

So, the mean number of days spent by COVID-19 patients in the ICU is 12.5625 days.

Range: The range of the data is the difference between the maximum and minimum values:

45 - 3 = 42

So, the range of the data is 42 days.

Interquartile Range: The interquartile range (IQR) is a measure of the dispersion of the data that is less sensitive to outliers than the range. To calculate the IQR, we first need to find the median (Q2), first quartile (Q1), and third quartile (Q3) of the data:

Q1 = (6+8)/2 = 7

Q2 = (11+12)/2 = 11.5

Q3 = (24+30)/2 = 27

The IQR is the difference between Q3 and Q1:

IQR = Q3 - Q1 = 27 - 7 = 20

Variance and Standard Deviation: Variance is a measure of the dispersion of the data that is used to calculate the standard deviation. The formula for variance is:

Variance = sum of squared deviations from the mean / number of values

First, we need to calculate the deviations from the mean:

11 - 12.5625 = -1.5625

16 - 12.5625 = 3.4375

10 - 12.5625 = -2.5625

...

The sum of the squared deviations from the mean is:

Variance = 596.9375/16 = 37.93359375

The standard deviation is the square root of the variance:

Standard deviation = √Variance = √37.93359375 = 6.15

Coefficient of Variation: The coefficient of variation (CV) is a measure of the relative variability of the data, expressed as a percentage of the mean. The formula for the CV is:

CV = (Standard deviation / mean) * 100

CV = (6.15 / 12.5625) * 100 = 49.03%

Comment on Results:

The mean number of days spent by COVID-19 patients in the ICU at the University of Ghana Medical Centre is 12.5625 days. The range of the data is 42 days, while the interquartile range is 20 days. The variance is 37.93 and the standard deviation is 6.15. The coefficient of variation is 49.03%, which indicates a relatively high degree of variability in the data. These results show that the number of days spent by COVID-19 patients in the ICU at the University of Ghana Medical Centre can vary widely, with some patients spending as few as 3 days and others spending as many as 45 days in the ICU.

Answer:

declined in popularity following the late 16th century.

Explanation:

The reason why Miguel de Cervantes never quite rose to the same level of popularity as Lope de Vega during the Golden Age of Spanish drama in the late 16th century may be due to differences in the styles of their plays. While Lope de Vega's plays were dominated by fast action and didn't focus much on characterization, Cervantes' plays were characterized by slow action and a greater focus on characterization. This may have made Cervantes' plays less appealing to audiences of the time who were more interested in fast-paced, action-packed plays. Additionally, the popularity of a playwright can also depend on many other factors such as the cultural context, political climate, and personal reputation of the playwright.

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:

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...

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 X-1 aircraft's mission is based on To put a man on the moon, travel at the speed of light, and launch an aeroplane into space

The latest box office sensation Hidden Figures (2017) sheds light on the previously mostly untold tale of the women who worked as computers on NASA's Project Mercury in the 1960s. The main three actors in the film are Taraji P. Henson, Janelle Monáe, and Octavia Spencer.

When the women's car breaks down on the way to Langley Research Center and a police comes up asking for identification, Katherine hands him her card and murmurs seriously, "NASA, sir," in one of Hidden Figures' greatest scenes.

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