true or false

1. Christ’s cross stands as a symbol of the universal redeeming love.

2. Human Rights defined as all those rights which are essential for the protection of dignity.

3. Correcting sinners is one of the corporal works of mercy.

4.Jesus came to establish God’s kingdom of peace, justice, and love.

5. To shelter the homeless is one of the spiritual works of mercy.

6.Max Scheler discovered the meaning of life during his time of imprisonment.

7.The Good Samaritan approached Jesus and asked what good must do to gain eternal life.

8.

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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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.

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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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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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:

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:

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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Answer:

Explanation:

Answer:

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Step-by-step explanation: