Mathematics: I need help with number 18 and 20

I need help with number 18 and 20, №14097806, 31.05.2021 13:44

Mathematics

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft2). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y). Temp. 174 176 177 178 178 179 180 181 Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74 Temp. 184 184 184 184 184 185 185 186 Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94 Temp. 186 186 186 188 188 189 190 192 Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16 (a) Determine the equation of the estimated regression line.

Step-by-step answer

P Answered by PhD

Kindly check explanation

Step-by-step explanation:

Given the data:

Temp. 174 176 177 178 178 179 180 181

Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74

Temp. 184 184 184 184 184 185 185 186

Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94

Temp. 186 186 186 188 188 189 190 192

Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16

A)

Using the online linear regression calculator, the lie of best fit which models the data above is :

ŷ = 0.09386X - 15.55523

Where ;

X = independent variable

ŷ = predicted or dependent variable

- 15.55523 = intercept

0.09386 = gradient / slope

B)

Point estimate when tank temperature is 186

ŷ = 0.09386(186) - 15.55523

ŷ = 17.45796 - 15.55523

ŷ = 1.90273

C)

Residual error (y - ŷ), ŷ = 1.90273 when x = 186

(0.94 - 1.90273) = −0.96273

(1.85 - 1.90273) = −0.05273

(2.02 - 1.90273) = 0.11727

(2.64 - 1.90273) = 0.73727

D)

To determine the proportion of observed variation in efficiency ratio, we find the Coefficient of determination R^2, which can be found using the online Coefficient of determination calculator : the r^2 value obtained is 0.4433.

Mathematics

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft2). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y). Temp. 174 176 177 178 178 179 180 181 Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74 Temp. 184 184 184 184 184 185 185 186 Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94 Temp. 186 186 186 188 188 189 190 192 Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16 (a) Determine the equation of the estimated regression line.

Step-by-step answer

P Answered by PhD

Kindly check explanation

Step-by-step explanation:

Given the data:

Temp. 174 176 177 178 178 179 180 181

Ratio 0.86 1.31 1.42 1.01 1.15 1.02 1.00 1.74

Temp. 184 184 184 184 184 185 185 186

Ratio 1.43 1.70 1.57 2.13 2.25 0.76 1.37 0.94

Temp. 186 186 186 188 188 189 190 192

Ratio 1.85 2.02 2.64 1.53 2.48 2.90 1.79 3.16

A)

Using the online linear regression calculator, the lie of best fit which models the data above is :

ŷ = 0.09386X - 15.55523

Where ;

X = independent variable

ŷ = predicted or dependent variable

- 15.55523 = intercept

0.09386 = gradient / slope

B)

Point estimate when tank temperature is 186

ŷ = 0.09386(186) - 15.55523

ŷ = 17.45796 - 15.55523

ŷ = 1.90273

C)

Residual error (y - ŷ), ŷ = 1.90273 when x = 186

(0.94 - 1.90273) = −0.96273

(1.85 - 1.90273) = −0.05273

(2.02 - 1.90273) = 0.11727

(2.64 - 1.90273) = 0.73727

D)

To determine the proportion of observed variation in efficiency ratio, we find the Coefficient of determination R^2, which can be found using the online Coefficient of determination calculator : the r^2 value obtained is 0.4433.

Mathematics

Experimental data are collected as: x[100]={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}; y[100]={-92.3973,14.0677,98.9626,49.6151,111.88,-38.7967,52.1535,-101.325,65.6111,125.129,151.762, -123.132,-41.8744,-39.3356,-99.7856,-128.881,58.5477,-111.692,-75.3057,4.4723,-100.051,169.051,

Step-by-step answer

P Answered by Master

y = 1.00114x + 1.75243

Step-by-step explanation:

Given

The x and y values

Required

The regression line equation

Because of the length of the given data, I will run the analysis using online tools, then analyze the result.

From the analysis, we have:

$\sum x = 5050$

$\sum y = 5231.1011$

$\bar x = 50.5$

$\bar y = 52.311$

SSX = 83325 --- Sum of squares

SP = 83419.7626 --- Sum of products

The regression equation is calculated as:

y = ax + b

Where:

$a = \frac{SP}{SSX}$

So, we have:

$a = \frac{83419.76}{83325}$

a = 1.00114

$b = \bar y - a * \bar x$

b = 52.31 - (1.00114*50.5)

b = 1.75243

So:

y = ax + b becomes

y = 1.00114x + 1.75243

Mathematics

Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and find the proportion of presidents who were taller than their opponents. Use that result to construct a? 95% confidence interval estimate of the population percentage. Based on the? result, does it appear that greater height is an advantage for presidential? candidates? Why or why? not? F. Roosevelt 188 182 Harding 183 178 Polk 173 185 Clinton 188 188 J. Adams 170 189 Truman 175 173 J. Q. Adams 171 191 Eisenhower 179 178 Harrison 168 180 G. H. W.

Step-by-step answer

P Answered by PhD

1

1.) 33% < p < 66%

As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.

If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.

Step-by-step explanation:

Out of this sample, we have 11 presidents, out of 20, that were taller than their oponent.

Then, the proportion of presindents that were taller than their oponent can be calculated as:

p=X/n=11/20=0.55

We can calculate now the standard error of the proportion as:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.55*0.45}{20}}=\sqrt{0.012375}=0.11$

For a 95% confidence interval, the z-value is z=1.96 (we can loook up this value in the standarized normal distribution table).

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z\cdot \sigma_p=0.55-1.96*0.11=0.55-0.22=0.33\\\\UL=p+z\cdot \sigma_p=0.55+1.96*0.11=0.55+0.22=0.66$

As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.

If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.

Mathematics

Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and find the proportion of presidents who were taller than their opponents. Use that result to construct a? 95% confidence interval estimate of the population percentage. Based on the? result, does it appear that greater height is an advantage for presidential? candidates? Why or why? not? F. Roosevelt 188 182 Harding 183 178 Polk 173 185 Clinton 188 188 J. Adams 170 189 Truman 175 173 J. Q. Adams 171 191 Eisenhower 179 178 Harrison 168 180 G. H. W.

Step-by-step answer

P Answered by PhD

1

1.) 33% < p < 66%

As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.

If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.

Step-by-step explanation:

Out of this sample, we have 11 presidents, out of 20, that were taller than their oponent.

Then, the proportion of presindents that were taller than their oponent can be calculated as:

p=X/n=11/20=0.55

We can calculate now the standard error of the proportion as:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.55*0.45}{20}}=\sqrt{0.012375}=0.11$

For a 95% confidence interval, the z-value is z=1.96 (we can loook up this value in the standarized normal distribution table).

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z\cdot \sigma_p=0.55-1.96*0.11=0.55-0.22=0.33\\\\UL=p+z\cdot \sigma_p=0.55+1.96*0.11=0.55+0.22=0.66$

As the confidence interval includes values under 50% and over 50%, it doesn't appear that greater height is an advantage for presidential.

If the lower bound of the confidence interval were over 50%, one could interpret that greater height is an advantage for presidential, but it is not the case for this sample.

Mathematics

Experimental data are collected as: x[100]={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}; y[100]={-92.3973,14.0677,98.9626,49.6151,111.88,-38.7967,52.1535,-101.325,65.6111,125.129,151.762, -123.132,-41.8744,-39.3356,-99.7856,-128.881,58.5477,-111.692,-75.3057,4.4723,-100.051,169.051,

Step-by-step answer

P Answered by Specialist

y = 1.754 + 1.001 x

Step-by-step explanation:

The equation of Regression equation is of the form of:

y = a + bx

where, a is intercept and b is slope

The formula for a and b is given by,

$a = \frac{\sum Y \sum X^2- \sum X \sum XY}{n\sum X^2-(\sum X)^2} \\and, b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2-(\sum X)^2}$

Thus, a = 1.754

and b = 1.001

Thus, Regression Line is given by,

y = 1.754 + 1.001 x

Now plotting these line:

Experimental data are collected as: x[100]={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,

Business

The trend analysis report of Marswell, Inc. is given below (in millions) 2019 2018 2017 2016 2015 Net income $690 $604 $470 $404 $398 Trend percentages 173% 152% 118% 102% 100% Which of the following is a correct conclusion from the above analysis? A) Net income for 2017 has decreased by 1 18% over that for 2015. B) Net income for 2017 has increased by 1 18% over that for 2015 C) Net income for 2017 has decreased by 18% over that for 2015. D) Net income for 2017 has increased by 18% over that for 2015

Step-by-step answer

P Answered by PhD

D) Net income for 2017 has increased by 18% over that for 2015

Explanation:

Trend Analysis shows the difference in the value of a variable overtime. In the analysis below, the base year is 2015 and so has a trend percentage of 100%.

The increases or decrease in Net Income in subsequent years can be inferred by the different in the trend percentages of the various years. For instance, the increase (decrease) in net income in 2019 over 2015 is;

= 173 - 100

= 73%

This means that income in 2019 is 73% higher than it was in 2015.

The same goes for 2017 and 2015;

= 118 - 100

= 18%

Income in 2017 has increased by 18% since 2015.

Mathematics

Experimental data are collected as: x[100]={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}; y[100]={-92.3973,14.0677,98.9626,49.6151,111.88,-38.7967,52.1535,-101.325,65.6111,125.129,151.762, -123.132,-41.8744,-39.3356,-99.7856,-128.881,58.5477,-111.692,-75.3057,4.4723,-100.051,169.051,

Step-by-step answer

P Answered by Master

y = 1.754 + 1.001 x

Step-by-step explanation:

The equation of Regression equation is of the form of:

y = a + bx

where, a is intercept and b is slope

The formula for a and b is given by,

$a = \frac{\sum Y \sum X^2- \sum X \sum XY}{n\sum X^2-(\sum X)^2} \\and, b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2-(\sum X)^2}$

Thus, a = 1.754

and b = 1.001

Thus, Regression Line is given by,

y = 1.754 + 1.001 x

Now plotting these line:

Business

The trend analysis report of Marswell, Inc. is given below (in millions) 2019 2018 2017 2016 2015 Net income $690 $604 $470 $404 $398 Trend percentages 173% 152% 118% 102% 100% Which of the following is a correct conclusion from the above analysis? A) Net income for 2017 has decreased by 1 18% over that for 2015. B) Net income for 2017 has increased by 1 18% over that for 2015 C) Net income for 2017 has decreased by 18% over that for 2015. D) Net income for 2017 has increased by 18% over that for 2015

Step-by-step answer

P Answered by PhD

D) Net income for 2017 has increased by 18% over that for 2015

Explanation:

Trend Analysis shows the difference in the value of a variable overtime. In the analysis below, the base year is 2015 and so has a trend percentage of 100%.

The increases or decrease in Net Income in subsequent years can be inferred by the different in the trend percentages of the various years. For instance, the increase (decrease) in net income in 2019 over 2015 is;

= 173 - 100

= 73%

This means that income in 2019 is 73% higher than it was in 2015.

The same goes for 2017 and 2015;

= 118 - 100

= 18%

Income in 2017 has increased by 18% since 2015.

Mathematics

The vertices of a rectangle are given in the columns of the matrix (picture 1). if (picture 2) is found to perform a transformation, what are the coordinates of the transformed rectangle? (0, 0), (0, 18), (18, 18), (18, 0) (0, 0), (18, 0), (18, –18), (0, –18) (0, 0), (–18, 0), (–18, –18), (0, –18) (0, 0), (18, 0), (18, 18), (0, 18)

Step-by-step answer

P Answered by PhD

75

(0 , 0) , (18 , 0) , (18 , -18) , (0 , -18) ⇒ the second answer

Step-by-step explanation:

∵ The vertices of the rectangles are:

(0 , 0) , (0 , 6) , (6 , 6) , (6 , 0)

∵ 3 × $\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$ × R

∴ That is mean The rectangle rotate 270° around the origin

(270° anti-clockwise or 90° clockwise)

and enlargement by scale factor 3

I need help with number 18 and 20

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