For each part below, give an example of a linear system of
equations in two variables that has the given property. In each case, draw the lines
corresponding to the solutions of the equations in the system.
(a) has no solution
(b) has exactly one solution
(c) has infinitely many solutions
(i) Add or remove equations in (b) to make an inconsistent system.
(ii) Add or remove equations in (b) to create infinitely many solutions.
(iii) Add or remove equations in (b) so that the solution space remains unchanged.
(iv) Can you add or remove equations in (b) to change the unique solution you had
to a different unique solution?
In each of (i) - (iv) justify your action in words.

Answer:

c. the concentration of 1 mg dose in the blood 2 hours after injection.

Step-by-step explanation:

We have, C = f(x, t) = t*e^-t(5-x)

then

C = f(1, 2) = 2*e^-2(5-1)

C = 2*e^-2(4)

C = 2*e^-8 OR C = 2/e^8

The concentration of a drug in blood depends on the dose and the time since administration. The given function allows calculating these concentrations, resulting in a concentration of a 1 mg dose 2 hours after injection of 8e^-2 mg/liter and a 2 mg dose 1 hour after injection of 3e^-1 mg/liter. The change in concentration cannot be determined without the previous concentration information.

To find the concentration of the drug, we need to plug the values of x and t into the given function C = f(x, t) = te^-t(5-x). For f(1,2), we replace x with 1 and t with 2. Therefore, C = 2e^-2(5-1) = 2e^-2*4 = 2*4*e^-2 = 8e^-2.

a. The change in concentration of a 1 mg dose in the blood 2 hours after injection. To find the change in concentration, we would need the previous concentration. However, as this information is not provided, we cannot determine the change in concentration for this case.

b. The amount of a 1 mg dose in the blood 2 hours after injection. The amount of the dose remains constant at 1 mg; what changes is its concentration due to distribution and elimination processes.

c. The concentration of a 1 mg dose in the blood 2 hours after injection. With the values provided, we already calculated this to be C = 8e^-2 mg/liter.

d. The concentration of a 2 mg dose in the blood 1 hour after injection. Plain the value of x = 2 and t = 1 into the equation, we get C = 1e^-1(5-2) = 3e^-1 mg/liter.

e. The change in concentration of a 2 mg dose in the blood 1 hour after injection. Similar to part a, without previous concentration information, it is not possible to determine the change in this case.

https://brainly.com/question/31595886

#SPJ3

Answer:

13/18

Step-by-step explanation:

P(A∪B)=P(A)+P(B)−P(A∩B)

P(A∪B)=2/3 + 1/6 − 1/9 = 13/18

Final answer:

To find the equation of a line perpendicular to 2x - 3y = 13 and passing through (-6, 5), first find the slope of the given line, then its negative reciprocal for the perpendicular line, and finally apply the point-slope formula with the new slope and given point.

Explanation:

To find the equation of a line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5), follow these steps:

Rewrite the given equation in slope-intercept form (y = mx + b) to find its slope.

Calculate the slope of the perpendicular line using the negative reciprocal of the given line's slope.

Use the point-slope formula with the slope from step 2 and the given point to find the equation of the perpendicular line.

Rewriting the given equation: 2x - 3y = 13 → -3y = -2x + 13 → y = (2/3)x - 13/3. The slope (m1) is 2/3. The slope of the perpendicular line (m2) is -3/2 (the negative reciprocal of m1). Using the point-slope formula with the perpendicular slope and the point (-6, 5), we get: y - 5 = (-3/2)(x + 6), which simplifies to y = (-3/2)x - 4. This is the equation of the line perpendicular to 2x - 3y = 13 that passes through (-6, 5).

Answer:

a) 14.29% probability that they pick the same number.

b) 85.71% probability that they pick the different numbers.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The outcomes in this problem is as follows:

(Student A Number, Student B Number)

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1), (7,1)

(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)

(1,3), (2,3), (3,3), (4,3), (5,3), (6,3), (7,3)

(1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (7,4)

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (7,5)

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (7,6)

(1,7), (2,7), (3,7), (4,7), (5,7), (6,7), (7,7)

So there are 49 total outcomes.

(a) What is the probability that they pick the same number?

There are 7 possible outcomes in which they pick the same number:

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7)

So there is a 7/49 = 0.1429 = 14.29% probability that they pick the same number

(b) What is the probability that they pick different numbers?

There are 49-7 = 42 possible outcomes in which they pick different numbers.

So there is a 42/49 = 0.8571 = 85.71% probability that they pick the different numbers.

Answer:

A. D = None of the above

B. B = 18

C. A = 15

D. D = None of the above

Step-by-step explanation:

Rearranging the population from the highest to the lowest.

10, 10, 12, 15, 15, 15, 15, 17, 17, 18, 18, 20, 22, 22, 24, 25, 31, 32, 34, 37

Mean = (summation of all the 20 samples)/ no of samples

Mean per year per 1000 population = 409/20

= 20.45

Median = the middle value (for odd numbered samples)

= the sum of the middle value ÷ 2 ( for even numbered samples)

Median birth per year per 1000 population = (18 + 18)/2

= 18

Mode = the sample that has the number of frequency

Mode per year per 1000 population = 15 (frequency = 4)

Range = highest value sample - lowest value sample

Range per year per 1000 population = 37 - 10

= 27

Answer:

a) 10.6; 48.7; within

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 29.6

Standard deviation = 6.35

"According to the Empirical Rule, the ages of nearly all lead actresses will be between______________and_____________ years.

Nearly all(99.7%) will be within 3 standard deviations from the mean.

So from 29.6 - 3*6.35 = 10.6 years to 29.6 + 3*6.35 = 48.7 years.

Anna was 11-years old, so she was within this range.

So the correct answer is

a) 10.6; 48.7; within

The ages of nearly all lead actresses will be between 16.9 and 42.3 years according to the Empirical Rule. Anna Paquin's age falls within this range when she won the academy award.

According to the Empirical Rule, for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations. Since the mean age is 29.6 years and the standard deviation is 6.35 years, we can calculate the range as follows:

One standard deviation: 29.6 ± 6.35 = 23.25 to 35.95

Two standard deviations: 29.6 ± (2 × 6.35) = 16.9 to 42.3

Three standard deviations: 29.6 ± (3 × 6.35) = 10.55 to 48.65

Since Anna Paquin was 11 years old when she won the academy award, her age falls within the range of two standard deviations, which is 16.9 to 42.3 years. Therefore, the correct answer is option c) 16.9: 42.3; not within.

https://brainly.com/question/33068610

#SPJ3

sinФ = 1/6

sinФ = opposite/hypotenuse

cosФ = adjacent/hypotenuse

tanФ = opposite/adjacent

SOHCAHTOA

1^(2)+b^(2)=6^(2)

1+b^(2)=36

b^(2)=35

[tex]\sqrt{35}[/tex]

cosФ = [tex]\sqrt{35}[/tex]/6

tanФ = 1/[tex]\sqrt{35}[/tex] = [tex]\sqrt{35}[/tex]/35

Answer:

Cos ø = √35/6

Tan ø = √35/35

Step-by-step explanation:

Sin ø = opposite/hypothenus

Sin ø = 1/6

Pythagoras theorem says

Hyp² = Opp² + Adj²

6² = 1² + A²

A² = 35

A = √35

Hence, cos ø = √35/6

tan ø = 1/√35 = √35/35

Answer:

Explanation:

The function that can determine the distance from a point on the bottom edge of the banner to the floor is:

[tex]f(x)=0.25x^2-x+9.5[/tex]

That function is a quadratic function which means that it is a parabola. Given that the coefficient of the quadratic term (0.25) is positive, the parabola open upwards, and the vertex is the lowest point of the parabola and it represents how high above the floor is the lowest point on the bottom edge of the banner.

So, you need to find the vertex of the parabola.

I will complete squares to find the form A(x -h)² + k, where h and k are the coordinates of the vertex (h, k).

[tex]f(x)=0.25x^2-x+9.5\\ \\ 4(f(x))=4(0.25x^2-x+9.5)\\ \\ 4f(x)=x^2-4x+38\\ \\ 4f(x)-38=x^2-4x\\ \\ 4f(x)-38+4=x^2-4x+4\\ \\ 4f(x)-34=(x-2)^2\\ \\ 4f(x)=(x-2)^2+34\\ \\ f(x)=(1/4)(x-2)^2+8.5[/tex]

Hence, the vertex (h,k) is (2, 8.5), meaning that the lowest point on the bottom edge of the banner is at 2 feet from the left edge of the banner and 8.5 feet above the floor.

The lowest point on the bottom edge of the banner which results in a height of 8.5 feet above the floor.

To find the height above the floor of the lowest point on the bottom edge of the banner using the function f(x) = 0.25x^2 - x + 9.5, we need to determine the vertex of this quadratic function. The vertex form of a quadratic function f(x) = ax^2 + bx + c gives the minimum or maximum value of the function.

Here, the coefficient a is positive (0.25), indicating that the parabola opens upwards and thus has a minimum point. The x-coordinate of the vertex can be found using the formula x = -b / (2a).

Substitute x = 2 back into the function to find the minimum height:

f(2) = 0.25(2)^2 - 2 + 9.5 = 1 - 2 + 9.5 = 8.5 feet.

Therefore, the lowest point on the bottom edge of the banner is 8.5 feet above the floor.

Answer:

1) Explanation and figure attached below.

2) For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.

Step-by-step explanation:

Part 1

If we order the data from the smallest to the largest we got:

13.1 , 16.2  ,18.3 , 21.1 , 23.4 , 24.3 , 24.5

25.1 , 25.2 , 26.4 , 26.6 ,27.3  ,28 , 28.6 ,

28.7 , 29 , 29.4 , 29.4 , 29.5 , 30 , 30.7

31.2 , 32 , 32 , 33.5 , 33.6 , 33.7 , 33.9

34.2 , 34.4 , 34.5 , 34.5 , 34.8 , 34.8

34.8 , 34.9 , 35.3 , 35.3 , 35.6 , 35.7

35.9 , 36 , 36.6 , 36.8 , 37.7 , 37.9

38.1 , 38.8 , 39.2 , 39.9

For this case if we use 10 mpg as the lower limit and with a width of 5 mpg for the intervals we have the following table:

Interval   Frequency   Midpoint

[10-15)           1                 12.5

[15-20)          2                17.5

[20,25)         4                 22.5

[25-30)         12               27.5

[30-35)          17              32.5

[35-40)          14              37.5

Total             50

So the histogram is on the figure attached. The midpoint for the class interval with the largest number of observations is 32.5.

Part 2

For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.

Final answer:

To construct the histogram, determine the class intervals and count the number of observations in each interval. The midpoint of the cell with the largest number of observations is approximately 37.45 mpg.

Explanation:

To construct a histogram with intervals of width 5 mpg, we start by determining the class intervals. Given that the lower limit of the first interval is 10 mpg, the class intervals can be calculated as follows: 10-14.9, 15-19.9, 20-24.9, 25-29.9, 30-34.9, 35-39.9. Next, we count the number of observations falling into each interval:

10-14.9: 3

15-19.9: 6

20-24.9: 5

25-29.9: 7

30-34.9: 9

35-39.9: 10

The cell with the largest number of observations is the one corresponding to the interval 35-39.9 mpg. To calculate the midpoint of this interval, we take the average of the lower and upper limits: (35 + 39.9) / 2 = 37.45. Therefore, the midpoint of the cell with the largest number of observations (the mode) is approximately 37.45 mpg.

In this exercise we have to use the knowledge of algorithm to write the correct alternative that best matches, thus we can say that:

Letter D

The computational complicatedness of K-NN increases as the extent or bulk of some dimension of the training basic document file increase and the treasure gets considerably unhurried as the number of examples and free variables increase.

Also, K-NN happen a non-parametric machine intelligence treasure and as such form no assuming possession about the working form of the question at hand.

The invention everything better accompanying information in visible form of the same scale, therefore standard the information in visible form superior to applying the invention happen urged.

See more about algorithm at brainly.com/question/22952967

All the statements provided about k-NN (scale importance, efficiency with small input variables, and no assumptions about the functional form of the problem) are true.

The question asks which of the given statements is true about the k-NN (k-Nearest Neighbors) algorithm. The correct answer would be 'D) All of the above.' This is because all three statements accurately describe different aspects of how the k-NN algorithm works:

https://brainly.com/question/27917397

#SPJ3

Answer:

60 professors

Step-by-step explanation:

for every 55 students there were 2 professors

for 1650 students, there will be (1650 x 2) / 55 =3300/55 = 60 professors

Final answer:

To find the number of professors needed for 1650 students when the ratio is 2 professors for every 55 students, you calculate 2 / 55 = x / 1650, resulting in x = 60. Therefore, 60 professors are needed for 1650 students.

Explanation:

The question asks how many professors would be needed for 1650 students if there is a ratio of 2 professors for every 55 students. To solve this, you can set up a proportion where the number of students is directly proportional to the number of professors. The proportion can be expressed as 2 professors / 55 students = x professors / 1650 students. Solving for x involves cross-multiplying and dividing.

Step 1: Cross-multiply to get 2 × 1650 = 55 × x.

Step 2: Simplify the equation: 3300 = 55x.

Step 3: Divide both sides by 55 to solve for x: x = 3300 / 55.

Step 4: Calculate the result: x = 60. Thus, you would need 60 professors for 1650 students.

Answer:

T = 72/4 = 18years

T = 20years (to the nearest tenth)

Step-by-step explanation:

Using the rule of 72, which is used to estimate the number of years for a given investment to double at a given interest rate.

Doubling time = 72/interest rate

T = 72/r

Rate r (in percentage) = 4%

Time T (in years)

T = 72/4 = 18years

T = 20years (to the nearest tenth)

The student's question relates to analyzing trends in car safety award data and understanding supply curve dynamics in the context of business or economics at the college level. It requires statistical analysis and economic reasoning.

The student's question involves analyzing data to determine if there has been a change in the distribution of cars that earned top safety picks between the years 2009 and 2013. This question lies within the field of Business or specifically within the area of business statistics, focusing on the analysis of trends over time.

Furthermore, considering the supply of cars and how it changes with price is another key aspect of this question. Using the supplied figures and tables, which is common in business studies, the student is asked to interpret how the quantity supplied of cars changes as the price increases from $20,000 to $22,000—a concept known in economics as the law of supply.

In summary, the question requires the application of statistical analysis to evaluate changes in car safety awards over time and the understanding of basic supply curve dynamics.

Your complete question is: High demand cars that are also in low supply tend to retain their value better than other cars. The data in the table below is for a car that won a resale value award. a. Write a function to represent the change in the percentage of the car's value over time. Suppose that the function is linear for the first 5 years. b. Based on your model, by what percent did the car's value drop the day it was bought and driven off the lot? c. Do you think the linear model would still be useful after 10 years? Why or why not? d. Assume you used months instead of years to write a function. How would your model change?

Answer:

Step-by-step explanation:

Check the attachment for solution

First you need to find your True Air Speed (TAS). I used the formula:

[tex]TAS=IAS\sqrt{\frac{\rho_0}{\rho}}\\[/tex]

where [tex]\rho_0[/tex] is the density at mean sea level at 15°C which is 1.225 kg/m^3 and [tex]\rho[/tex] is the density of the air in which the airplane is flying. In this case with the given altitude and air temperature is about: 1.073 kg/m^3.

So TAS is equal to: 123 kt in this case.

To get the time of arrival you need to get your ground speed. For that you need to compensate for the wind speed and direction.  With the drift angle you can get your ground speed.

In this case you get ground speed of 115 kt which means to travel 218 nautical miles you need 1 h and 53 min which means you will arrive at: 1135.

Answer:

1920 k calories is the basal metabolism.

Step-by-step explanation:

Formula to get the total basal metabolism has been given by the formula

R(t) = [tex]80-0.18cos\frac{\pi t}{12}[/tex]

Where t = time

Now to calculate the total basal metabolism we will integrate the function with respect to time from 0 to 24 hours.

[tex]\int_{0}^{24}R(t)=\int_{0}^{24}(80-0.18cos\frac{\pi t}{12})dt[/tex]

= [tex][80t-{0.18}\times \frac{sin\frac{\pi t}{12}}{\frac{\pi}{12}}]_{0}^{24}[/tex]

= [tex][(80-0)24-\frac{0.18\times 12}{\pi}(sin2\pi - sin0)][/tex]

= [(80-0)24-(sin2\pi - sin0)]

= 80×24

= 1920 k calories

Therefore, 1920 k calories is the total basal metabolism of the young man.

Answer: The mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.

Step-by-step explanation:

if there is a linear relationship between two variables x and y , then we represent this relation in the form of equation as :[tex]y= mx+c[/tex]   (*)

, where m = rate of change of y with respect to x

c= Constant or the value of y when x=0.

We assume there is a linear relationship between the curb weight of a car (in thousands of kilograms) and its combined mileage (in miles per gallon).

Let y= combined mileage (in miles per gallon)

x= curb weight of a car (in thousands of kilograms)

When x= 3 thousand kg  , y = 9 mpg

⇒ [tex]9= m(3)+c[/tex]                   [Put values in (*)]

⇒ [tex]9= 3m+c--------(1)[/tex]

When x= 1.9 thousand kg  , y = 21 mpg

⇒ [tex]21= m(1.9)+c[/tex]                   [Put values in (*)]

⇒ [tex]21= 1.9m+c--------(2)[/tex]

Eliminate the equation (1) from (2) , we get

[tex]1.9m-3m=21-9[/tex]

[tex]-1.1m=12\\\Rightarrow\ m=\dfrac{12}{-1.1}=\dfrac{-120}{11}[/tex]

Put value of m in (1) , we gte

[tex]9= 3(\dfrac{-120}{11})+c[/tex]

[tex]9= \dfrac{-360}{11}+c\\\Rightarrow\ c=9+\dfrac{360}{11}=\dfrac{459}{11}[/tex]

Substitute the value of m and c in (*) , we get

[tex]y= \dfrac{-120}{11}x+\dfrac{459}{11}[/tex]

When x= 1.1

[tex]y= \dfrac{-120}{11}(1.1)+\dfrac{459}{11}[/tex]

[tex]y= \dfrac{-120}{11}(\dfrac{11}{10})+\dfrac{459}{11}[/tex]

[tex]y=-12+\dfrac{459}{11}=\dfrac{327}{11}\approx29.72[/tex]

Hence, the mileage of a Chevy Cavalier weighing 1.1 thousand kg is 29.72 mpg.

Individual:  A single country.

Variable:   The mean density of people per square kilometer of all countries.

Population:  Set of all countries.

Sample:  Set of the 56 countries on which data is collected.

Parameter: The mean density of people per square kilometer calculated from the population.

Statistic:  The mean density of people per square kilometer calculated from the sample.

Individual: Individuals are the objects described by a set of data.

Variable: Variables are characteristics of individuals.

Population: Population is all individuals of interest.

Sample: Sample is a subset of the population.

Parameter: Parameter is a characteristic of a population.

Statistics: Statistic is a characteristic of a sample.

The individual is the country, the variable is the density of people per square kilometer, the population is all the countries in the world, the sample is the 56 countries, the parameter is the true mean density of all countries, and the statistic is the estimated mean density from the 56 sampled countries.

In the example given, the individual is each of the 56 countries from which the World Health Organization collects data. The variable is the density of people per square kilometer in each of these countries. The population consists of all the countries in the world. The sample includes the 56 countries for which data was collected. The parameter is the true mean density of people per square kilometer for all countries. Finally, the statistic is the estimated mean density of people per square kilometer based on the data from the sample of 56 countries.

https://brainly.com/question/31538429

#SPJ3

Answer:

11,875,000 common shares outstanding

Step-by-step explanation:

The number of common shares outstanding of Harper industries is determined by the company's common equity added to its MVA and then divided by the stock price per share.

If common equity is $900 million, MVA is $50 million and the price per share is $50, the number of shares outstanding is:

[tex]n=\frac{900,000,000+50,000,000}{80}\\n=11,875,000[/tex]

Answer:

N = 11,875,000

Therefore, the number of common shares that are currently outstanding is 11,875,000

Step-by-step explanation:

Oustanding shares can be defined as the company's stock currently held by all its shareholders, it also including share blocks held by institutional investors and restricted shares owned by the company’s officers and insiders. The outstanding common shares can be defined mathematically with the equation below.

MVA = (P ×N) - BV .....1

Where;

MVA = market value added

P = price per share

N = Number of outstanding shares

BV = balance sheet value of common equity

From equation 1, we can make N the subject of formula.

N = (MVA + BV)/P .....2

Since;

MVA = $50,000,000

BV = $900,000,000

P = $80

Substituting into equation 2

N = (50,000,000 + 900,000,000)/80

N = 11,875,000

Therefore, the number of common shares that are currently outstanding is 11,875,000

(x-3)(x-5)<_0

If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.  

x−3=0x-3=0  

x+5=0x+5=0  

Set the first factor equal to 00 and solve.  

x=3x=3  

Set the next factor equal to 00 and solve.  

x=−5x=-5  

Consolidate the solutions.  

x=3,−5x=3,-5  

Use each root to create test intervals.  

x<−5x<-5  

−5<x<3-5<x<3  

x>3x>3  

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.  

x<−5x<-5 False  

−5<x<3-5<x<3 True  

x>3x>3 False  

The solution consists of all of the true intervals.  

−5<x<3-5<x<3  

The result can be shown in multiple forms.  

Inequality Form:  

−5<x<3-5<x<3  

Interval Notation:  

(−5,3)

Answer:

a. 0.4

b. Not independent events

c. 0.47

Step-by-step explanation:

Let A= Practices Gluten free diet

B= Practices Dairy free diet

A and B=  Practices Both diets

P(A)=0.20

P(B)=0.15

P(A and B)=0.08

a.

[tex]P(B/A)=\frac{P(A and B)}{P(A)}[/tex]

[tex]P(B/A)=\frac{0.08}{2}[/tex]

[tex]P(B/A)=0.40[/tex]

b.

The two events are independent if P(B/A) =P(B) or P(A/B)=P(A)

As, P(B/A) ≠P(B)

0.4≠0.15

So, the event gluten free diet and dairy free diet are dependent events.

c.

[tex]P(A'/B)=\frac{P(A' and B)}{P(B)}[/tex]

[tex]P(A'and B)= P(B)-P(A and B)[/tex]

[tex]P(A' and B)=0.15-0.08=0.07[/tex]

[tex]P(A'/B)=\frac{P(A' and B)}{P(B)}[/tex]

[tex]P(A'/B)=\frac{0.07}{0.15}[/tex]

[tex]P(A'/B)=0.47[/tex]

Answer:

a. 0.4

b. Not independent

c. 0.47

Step-by-step explanation:

Answer:

A. Statistic

Step-by-step explanation:

Usually whole population is difficult to cover so the part of population is considered to assess the aspects of population. The statistic is the quantity which summarizes the sample. The problem illustrate that summary measure is calculated from sample to describe a population. As the summary measure is computed from sample then it is a statistic.

The number of ounces that can be bought for $4.60 will be 9.2 ounces.

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Conversion means to convert the same thing into different units.

If a pound of almonds costs $8.

We know that in one pound, there are 16 ounces. Then the cost of each ounce will be

⇒ 8 / 16

⇒ $0.5 per ounce

Then the number of ounces that can be bought for $4.60 will be

⇒ 4.6 / 0.5

⇒ 9.2 ounces

The number of ounces that can be bought for $4.60 will be 9.2 ounces.

More about the Algebra link is given below.

https://brainly.com/question/953809

#SPJ2

Answer:

The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.

Step-by-step explanation:

Given a function y, the average rate of change S of [tex]y=g(z)[/tex] in an interval  [tex](z_{s}, z_{f})[/tex] will be given by the following equation:

[tex]S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}}[/tex]

In this problem, we have that:

[tex]g(z) = 4z^{2} - 5[/tex]

Between the points( -2,11) and (2,11).

So [tex]z_{f} = 2, z_{s} = -2[/tex]

[tex]g(z_{f}) = g(2) = 4*(2)^{2} - 5 = 11[/tex]

[tex]g(z_{s}) = g(-2) = 4*(-2)^{2} - 5 = 11[/tex]

So

[tex]S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}} = \frac{11 - 11}{2 - (-2)} = \frac{0}{4} = 0[/tex]

The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.

Answer:

a. The mean growth of plant is same for each type of soil.

Step-by-step explanation:

The analysis of variance is statistical procedure used to assess the equality of more than two means by computing two different estimate of variances. Basically when we have more than two means to compare, we can't use t-test or z-test because these procedure will be tedious and time taking. So, the null hypothesis will be the same. Hence the null hypothesis in the given scenario is that "the mean growth of plant is same for each type of soil".

Answer:

0.3667 or 36.67%

Step-by-step explanation:

The probability of getting at least 1 defective hard disk among the 5 (P(X>0)) is equal to 100% minus the probability of getting none defectives (1-P(X=0)).

If 23 out of 25 hard disks are non-defective, the probability is:

[tex]P(X>0) = 1 -P(X=0)\\P(X>0) = 1 -\frac{23}{25}*\frac{22}{24} *\frac{21}{23} *\frac{20}{22} *\frac{19}{21}\\ P(X>0) = 0.3667=36.67\%[/tex]

The probability that there is at least 1 defective hard disk is 0.3667 or 36.67%.

Answer: (0.1, 0.9) We cannot reject the null hypothesis.

Step-by-step explanation:

First, calculate with the negative. Next, with the positive.

B1 - 2 x SE(B1)

0.5 - 2 x 0.2 = 0.1

B1 + 2 x SE(B1)

0.5 + 2 x 0.2 = 0.9

Final answer:

The student's question about hypothesis testing in linear regression focuses on whether to reject the null hypothesis based on a slope estimate, its standard error, and the p-value. Since the p-value exceeds the significance level, the null hypothesis is not rejected, indicating insufficient evidence of a nonzero slope at the 5 percent level.

Explanation:

The student is asking about hypothesis testing in the context of linear regression. Specifically, the question pertains to whether the null hypothesis, which posits no effect (i.e., a slope of zero), should be rejected based on a given slope estimate, its standard error, and a provided p-value. In hypothesis testing, if the p-value is greater than the chosen level of significance (alpha), you do not reject the null hypothesis.

In the provided scenario, if the p-value is indeed 0.2150 and the level of significance (alpha) is set to 5 percent (0.05), the decision would be not to reject the null hypothesis. This is because the p-value is greater than the alpha level (0.2150 > 0.05). The conclusion is that there is insufficient evidence at the 5 percent significance level to suggest that the true slope is different from zero.

Answer:

$1,448.66

Step-by-step explanation:

The future value of an annuity with yearly deposits 'P' at an interest rate of 'r' invested for 'n' years is determined by:

[tex]FV = P[\frac{(1+r)^n-1}{r}][/tex]

For P = $100, r = 0.08 and n = 10 years:

[tex]FV = 100[\frac{(1+0.08)^{10}-1}{0.08}]\\FV=\$1,448.66[/tex]

The amount at the end of the ten years is $1,448.66

Answer:

a) 1/3

b) 0.0658436214  

Step-by-step explanation:

Part a

Ann wins first round = (P , R) + (S , P) + (R , S) = 3 possibilities of win

Total outcomes = Wins + Losses + Ties = 9 possibilities

Hence,

P (Ann wins first round) = 3/9 = 1/3

Part b

Ann losses or ties first 4 rounds and wins 5th round

P(Ann loosing or tie in any round) = 6/9 = 2/3

Hence,

P(Ann wins 5th round only) = (2/3)^4 * (1/3) = 0.0658436214  

Part c

Final answer:

In rock-paper-scissors, Ann has a 1/3 probability of winning the first round. The probability of Ann's first win in round #5 is (2/3)^4 × (1/3), and the probability of Ann's first win coming after round #5 is 1 - (2/3)^5.

Explanation:

To answer these questions, we will calculate the probabilities based on the uniform random choices made by Ann and Bill in the game of rock-paper-scissors.

Probability Ann wins the first round: Each player has 3 choices, resulting in 9 possible outcomes. There are 3 ways Ann can win (rock beats scissors, paper beats rock, scissors beats paper), so the probability is 3/9 or 1/3.

Probability Ann's first win happens in round #5: We need 4 consecutive non-wins (losses or ties) and a win in the 5th round. The probability of a non-win per round is 2/3, making the probability of four non-wins (2/3)^4. The probability Ann wins the 5th round is 1/3. Therefore, the probability of this scenario is (2/3)^4 × (1/3).

Probability that Ann's first win comes after round #5: This is the probability of Ann not winning in the first 5 rounds. As each round is independent, we raise the non-winning probability to the power of 5, (2/3)^5, and subtract from 1 to get the desired probability.