The correct option is (a) [7.53, 77.41].
To construct a 99% confidence interval for the true variance (σ²) of the increase in pulse rate of astronauts performing a given task, we can use the Chi-Square distribution.
The formula for the confidence interval for the variance is:
[ (n-1) * s² / χ²_upper , (n-1) * s² / χ²_lower ]
Where:
n is the sample size
s² is the sample variance
χ²_upper and χ²_lower are the upper and lower critical values from the Chi-Square distribution, respectively, based on the desired confidence level and degrees of freedom (n-1).
In this case, we have:
n = 12 (number of astronauts)
s² = (standard deviation)² = 4.28² = 18.2984
degrees of freedom = n - 1 = 12 - 1 = 11
critical values from the Chi-Square distribution for a 99% confidence level are χ²_upper = 26.759 and χ²_lower = 2.179
Now we can substitute these values into the formula to calculate the confidence interval:
[ (11 * 18.2984) / 26.759 , (11 * 18.2984) / 2.179 ]
Simplifying:
[ 7.531 , 77.414 ]
Therefore, the 99% confidence interval for the true variance (σ²) of the increase in the pulse rate of astronauts performing the given task is approximately [7.53, 77.41].
The correct option is (a) [7.53, 77.41].
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Which of the following is a true statement about the first movies made in hollywood?
A. Music was recorded as part of ghe film B. They were silent films C. they only lasted 30 minutes
D. they were filmed in color
The correct statement about the first movies made in Hollywood is: B. They were silent films.
During the early days of Hollywood, which refers to the late 19th and early 20th centuries, movies were primarily silent films. This means that there was no synchronized sound accompanying the visuals on screen. The technology for recording and reproducing sound in movies had not yet been developed.
Instead of recorded sound, music was often performed live in theaters during the screenings of these silent films. Musicians would play instruments or provide live vocal accompaniment to enhance the viewing experience. However, this music was not recorded as part of the film itself.
Additionally, during this time, color film technology was still in its early stages of development. Most films were shot and presented in black and white, as color film processes were not yet widely available or affordable.
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pls
help im confused on how to add/subtract them
A = 4x +-39 B = 6x +-59 Č= -9x+6y Complete each vector sum. A+B+C= A-B+C= 24 + A+B-C- A-B-C= 2+ 2+
Final vector sum would be : A + B - C= x - 2 + 6y.
Let's calculate each vector sum one by one.
A + B + C= (4x + (-39)) + (6x + (-59)) + (-9x + 6y)
= x - 53 + 6yA - B + C= (4x + (-39)) - (6x + (-59)) + (-9x + 6y)
= -11x + 98 + 6yA - B - C= (4x + (-39)) - (6x + (-59)) - (-9x + 6y)
= 7x - 22
Let's calculate the values of
24 + A + B - C, A - B + C, and 2A + 2B - 2C one by one.
24 + A + B - C = 24 + (4x + (-39)) + (6x + (-59)) - (-9x + 6y)
= x - 2 + 6yA - B + C = (4x + (-39)) - (6x + (-59)) + (-9x + 6y)
= -11x + 98 + 6y2A + 2B - 2C
= 2(4x + (-39)) + 2(6x + (-59)) - 2(-9x + 6y)
= -10x - 44
Let's put all the results together,
A + B + C= x - 53 + 6y
A - B + C= -11x + 98 + 6y
A - B - C= 7x - 22
A + B - C= x - 2 + 6y
Hence, the solutions are:
A + B + C= x - 53 + 6y
A - B + C= -11x + 98 + 6y
A - B - C= 7x - 22
A + B - C= x - 2 + 6y.
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How many ways can a group of 20, including six boys and fourteen girls, be formed into two ten-person volleyball teams with no restrictions?
(b) How many ways can a group of 20, including six boys and fourteen girls, be formed into two ten-person volleyball teams so that each team has three of the boys?
(c) How many ways can a group of 20, including six boys and fourteen girls, be formed into two ten-person volleyball teams so that all of the boys are on the same team?
a) The number of ways a group of 20, including six boys and fourteen girls, can be formed into two ten-person volleyball teams with no restrictions is given by the combination formula. Since the order of selection doesn't matter in this case, we can use the combination formula to calculate the total number of combinations.
The formula for combination is: nCr = n! / (r!(n-r)!)
Where n is the total number of individuals and r is the number of individuals in each team.
In this scenario, we have 20 individuals in total, and we need to form two teams of ten individuals each. Therefore, the number of ways to form the teams without any restrictions is:
20C10 = 20! / (10!(20-10)!) = 184,756 ways.
(b) In this case, we want each team to have three boys. Since we have six boys in total, we need to select three boys for each team. The remaining slots will be filled by the girls.
The number of ways to select three boys from six is given by the combination formula: 6C3 = 6! / (3!(6-3)!) = 20 ways.
After selecting the boys, we have 14 girls remaining, and we need to select seven girls for each team. The number of ways to select seven girls from 14 is: 14C7 = 14! / (7!(14-7)!) = 3432 ways.
To calculate the total number of ways to form the teams, we multiply the number of ways to select the boys and the number of ways to select the girls:
20 ways (boys) * 3432 ways (girls) = 68,640 ways.
(c) In this case, we want all of the boys to be on the same team. We need to select all six boys and distribute the remaining slots among the girls.
The number of ways to select six boys from six is 6C6 = 6! / (6!(6-6)!) = 1 way.
After selecting the boys, we have 14 girls remaining, and we need to select four girls for each team. The number of ways to select four girls from 14 is: 14C4 = 14! / (4!(14-4)!) = 1001 ways.
To calculate the total number of ways to form the teams, we multiply the number of ways to select the boys and the number of ways to select the girls:
1 way (boys) * 1001 ways (girls) = 1001 ways.
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Need help answering questions 5 and 6
Practice Problems for Chapter four 1. Calculate the following range of scores for a continuous variable: 9, 8, 7,6,5,4,3,2. Use upper and lower real limits to calculate your answer. 2. Calculate the f
5. The continuous variable in the range 2, 3, 4, 5, 6, 7, 8, 9 has a lower real limit of 1.5 and an upper real limit of 9.5.a) The width of each interval is equal to: [tex]$$\frac{9.5-1.5}{5}[/tex] = 2$$$$\text{ Width of each interval is }2.$$b)
Since the interval from 2 to 4 has 2 as its lower real limit and its width is 2, its upper real limit is equal to $2+2=4$. Therefore, the upper real limits of the following intervals will be $4, 6, 8,$ and $10$.c) The frequency of the first interval is 2 and the frequency of the second interval is 1. Hence, the relative frequency of the first interval is [tex]$\frac{2}{3}$[/tex]and the relative frequency of the second interval is[tex]$\frac{1}{3}$.6[/tex]. The continuous variable is in the range 2, 3, 4, 5, 6, 7, 8, 9 has a lower real limit of 1.5 and an upper real limit of 9.5. Since the range is continuous, the frequency polygon will be a line that connects the midpoints of the intervals.The width of each interval is equal to $2$. The midpoint of the first interval is[tex]$\frac{2+4}{2}=3$[/tex]. The midpoint of the second interval is[tex]$\frac{4+6}{2}=5$[/tex]. The midpoint of the third interval is [tex]$\frac{6+8}{2}=7$[/tex]. The midpoint of the fourth interval is [tex]$\frac{8+10}{2}=9$[/tex]. Hence, the frequency polygon will connect the points [tex]$(3, \frac{2}{8}), (5, \frac{1}{8}), (7, 0),$ and $(9, 0)$[/tex]. Therefore, the final answer is shown in the image below.
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The range of scores using upper and lower real limits for the given data is:2: 1.5 - 2.53: 2.5 - 3.54: 3.5 - 4.55: 4.5 - 5.56: 5.5 - 6.57: 6.5 - 7.58: 7.5 - 8.59: 8.5 - 9.5.
The median is the middle value of a set of data. When the data has an odd number of scores, the median is the middle score, which is easy to find. However, when there is an even number of scores, the middle two scores must be averaged. Therefore, to find the median of the following data, we first have to order the numbers:
60, 70, 80, 90, 100, 110
The median is the middle number, which is 85.
Finding the mean: We sum all the numbers and divide by the total number of numbers:
60 + 70 + 80 + 90 + 100 + 110 = 5106 numbers
Sum of numbers = 510
Mean of the data = Sum of numbers / Number of scores
= 510/6
= 85
f= mean/median
= 85/85
= 1
The upper and lower real limits of 2 is 1.5 and 2.5. The upper and lower real limits of 3 is 2.5 and 3.5. The upper and lower real limits of 4 is 3.5 and 4.5. The upper and lower real limits of 5 is 4.5 and 5.5. The upper and lower real limits of 6 is 5.5 and 6.5. The upper and lower real limits of 7 is 6.5 and 7.5. The upper and lower real limits of 8 is 7.5 and 8.5. The upper and lower real limits of 9 is 8.5 and 9.5.
Therefore, the range of scores using upper and lower real limits is:
2: 1.5 - 2.53: 2.5 - 3.54: 3.5 - 4.55: 4.5 - 5.56: 5.5 - 6.57: 6.5 - 7.58: 7.5 - 8.59: 8.5 - 9.5.
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use induction to prove that fn ≥ 2 0.5n for n ≥ 6
The inequality above can be simplified to f(k+1) ≥ 2 0.5(k+1). Thus, fn ≥ 2 0.5n for n ≥ 6.
Let us prove that fn ≥ 2 0.5n for n ≥ 6 using induction.
Base case: When
n = 6, we have f6 = 8 and 2(0.5)6 = 8.
Since f6 = 8 ≥ 8 = 2(0.5)6, the base case is true.
Assume that fn ≥ 2 0.5n for n = k where k ≥ 6.
Now we must show that f(k+1) ≥ 2 0.5(k+1).
Since f(k+1) = f(k) + f(k-5), we can use the assumption to get f(k+1) ≥ 2 0.5k + 2 0.5(k-5)
The inequality above can be simplified to f(k+1) ≥ 2 0.5(k+1).Thus, fn ≥ 2 0.5n for n ≥ 6.
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A single channel queuing system has an average service time of 8 minutes and an average time between arrivals of 10 minutes. What is the arrival rate? A. 8 per hour B. 6 per hour C. 2 per hour D. 5 per hour
Answer:
B. 6 per hour
Step-by-step explanation:
You want to know the arrival rate if the average time between arrivals is 10 minutes.
RateThe rate is the inverse of the period.
(1 arrival)/(10 minutes) = (1 arrival)/(1/6 h) = 6 arrivals/h
The arrival rate is 6 per hour.
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Therefore, the arrival rate is 6 per hour. Only option B has the same value as calculated, that is, 6 per hour.
A single-channel queuing system has an average service time of 8 minutes and an average time between arrivals of 10 minutes.
The arrival rate can be determined using the following formula:λ=1/twhere,λ is the arrival rate and t is the average time between arrivals. Substitute t=10 in the above equation, we getλ=1/10=0.1Now, let’s check which of the given options is equal to 0.1.5 per hour is equal to 5/60 per minute=1/12 per minute≠0.1.8 per hour is equal to 8/60 per minute=2/15 per minute≠0.1.6 per hour is equal to 6/60 per minute=1/10 per minute=0.1 (Correct)2 per hour is equal to 2/60 per minute=1/30 per minute≠0.1. Therefore, the correct answer is option B, 6 per hour. Explanation: Arrival rate=λ=1/tWhere t is the average time between arrivals. Given, the average time between arrivals =10 minutes, therefore,λ=1/10=0.1For the given options, only option B has the same value as calculated, that is, 6 per hour.
Therefore, the arrival rate is 6 per hour. Only option B has the same value as calculated, that is, 6 per hour.
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which function in vertex form is equivalent to f(x) = x2 8 – 16x?f(x) = (x – 8)2 – 56f(x) = (x – 4)2 0f(x) = (x 8)2 – 72f(x) = (x 4)2 – 32
The given function f(x) = x² - 8x can be rewritten in vertex form using the process of completing the square. The vertex form of a quadratic function is f(x) = a(x - h)² + k, where (h, k) is the vertex of the parabola. The process of completing the square involves adding and subtracting a constant term to the expression in such a way that it becomes a perfect square trinomial.
So, f(x) = x² - 8x = (x² - 8x + 16) - 16 = (x - 4)² - 16. Therefore, the function f(x) = x² - 8x is equivalent to f(x) = (x - 4)² - 16 in vertex form. Now, we need to check which function in vertex form is equivalent to f(x) = x² - 8x from the given options:Option A: f(x) = (x - 8)² - 56Comparing it with the vertex form f(x) = a(x - h)² + k, we can see that h = 8, which is not equal to -4. So, this function is not equivalent to f(x) = x² - 8x.
Option B: f(x) = (x - 4)² + 0Comparing it with the vertex form f(x) = a(x - h)² + k, we can see that h = 4, which is equal to -(-4). So, this function is equivalent to f(x) = x² - 8x.Option C: f(x) = (x + 8)² - 72Comparing it with the vertex form f(x) = a(x - h)² + k, we can see that h = -8, which is not equal to -4. So, this function is not equivalent to f(x) = x² - 8x.Option D: f(x) = (x + 4)² - 32Comparing it with the vertex form f(x) = a(x - h)² + k, we can see that h = -4, which is equal to -(-4). So, this function is equivalent to f(x) = x² - 8x.Therefore, the function in vertex form equivalent to f(x) = x² - 8x is f(x) = (x - 4)² - 16.
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Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with = 36.1 ft and o- 6.8 ft. You intend to measure a random sample of n = 81 trees. What is the mean of the distribution of sample means? the What is the standard deviation of the distribution of sample means (i.e., the standard error in estimating the mean)? (Report answer accurate to 4 decimal places.) σ= Tip: Use the Desmos calculator...
The standard deviation of the distribution of sample means, or the standard error in estimating the mean, is approximately 0.7569 ft, rounded to 4 decimal places.
To find the mean of the distribution of sample means, we use the formula:
Mean of sample means = Mean of the population
In this case, the mean of the population is given as μ = 36.1 ft.
Therefore, the mean of the distribution of sample means is also 36.1 ft.
To find the standard deviation of the distribution of sample means, also known as the standard error, we use the formula:
Standard error = Standard deviation of the population / √(Sample size)
In this case, the standard deviation of the population is given as σ = 6.8 ft, and the sample size is n = 81.
Plugging in these values into the formula, we have:
Standard error = 6.8 / √(81)
Calculating this expression, we find:
Standard error ≈ 0.7569
Therefore, the standard deviation of the distribution of sample means, or the standard error in estimating the mean, is approximately 0.7569 ft, rounded to 4 decimal places.
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(1) using the method of data linearization , find the least
sqaures function y = D/x+C that fits to the following data
points
Xk
1.0
The method of data linearization is used to make non-linear data fit a linear model. This method is useful for cases in which a known nonlinear equation is suspected but there is no straightforward way of solving for the variables. It transforms data from a nonlinear relationship to a linear relationship.
The equation of the curve is y = D/x + C. We need to fit this equation to the data points. The first step is to rewrite the equation in a linear form as follows: y = D/x + C => y = C + D/x => 1/y = 1/C + D/(Cx)
The above equation is in a linear form y = a + bx, where a = 1/C and b = D/C. The data can be tabulated as shown below: xy 1.0 0.8
The sum of xy = (1.0) (1.25) + (0.8) (1.5625) = 2.03125
The sum of x = 2
The sum of y = 2.05
The sum of x² = 2
The equation is in the form of y = a + bx, where a = 1/C and b = D/C.
The least squares method is used to find the values of a and b that minimize the sum of the squared residuals, that is the difference between the predicted value and the actual value. The equation of the least squares regression line is given by: y = a + bx, where b = (nΣxy - ΣxΣy) / (nΣx² - (Σx)²)and a = (Σy - bΣx) / n, where n is the number of data points.
The values of b and a can be calculated as follows: b = [(2)(2.03125) - (2)(2.05)] / [(2)(2) - (2)²] = -0.2265625a = (2.05 - (-0.2265625)(2)) / 2 = 1.15625
Therefore, the equation of the least squares regression line is: y = 1.15625 - 0.2265625x
The equation of the curve is y = D/x + C.
D = -0.2265625 C = 1/1.15625
D = -0.2625 C = 0.865
We can therefore rewrite the equation of the curve as: y = -0.2625/x + 0.865
Therefore, the least squares function y = -0.2625/x + 0.865 fits the data points.
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The accompanying table shows students' scores for the final exam in a history course. Scores Cumulative Frequency 50 up to 60 14 60 up to 70 32 70 up to 80 67 80 up to 90 92 90 up to 100 100 How many of the students scored at least 70 but less than 90? Multiple Choice 29 36 60 93 O O O
25 students scored at least 70 but less than 90.
To find the number of students who scored at least 70 but less than 90, we need to sum up the frequencies in the corresponding cumulative frequency interval. Looking at the table, we can see that the cumulative frequency for the interval "70 up to 80" is 67, and the cumulative frequency for the interval "80 up to 90" is 92.
To calculate the number of students in the desired range, we subtract the cumulative frequency of the lower interval from the cumulative frequency of the upper interval:
Number of students = Cumulative frequency (80 up to 90) - Cumulative frequency (70 up to 80)
= 92 - 67
= 25
Therefore, 25 students scored at least 70 but less than 90.
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in a randomly generated list of numbers from 0 to 9, what is the probability that each number will occur?
The probability that each number will occur in a randomly generated list of numbers from 0 to 9 is 1 in 3,628,800.
To understand the probability, let's consider the total number of possible outcomes in the randomly generated list. In this case, we have 10 possible numbers (0 to 9) and the list length is also 10. So, the total number of possible outcomes is given by 10 factorial (10!).
The formula for factorial is n! = n * (n-1) * (n-2) * ... * 2 * 1. Therefore, 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800.
Now, let's determine the number of favorable outcomes, which is the number of ways each number can occur exactly once in the list. Since the list is randomly generated, the occurrence of each number is equally likely.
To calculate the number of favorable outcomes, we can use the concept of permutations. The first number in the list can be any of the 10 available numbers, the second number can be any of the remaining 9 numbers, the third number can be any of the remaining 8 numbers, and so on.
Using the formula for permutations, the number of favorable outcomes is given by 10! / (10-10)! = 10!.
So, the probability that each number will occur in the randomly generated list is the number of favorable outcomes divided by the total number of possible outcomes, which is 10! / 10! = 1 in 3,628,800.
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please help with question 5 and 6
DETAILS ASK YOUR TEACHER Verify the identity. (Simplify at each step.) sin(+ x) = (cos(x) + √3 sin(x)) sin + = sin + = 40 ))+( ==(cos(x) + √3 sin(x)) Need Help? Read It 6. [-/1 Points] DETAILS 5.
The value of sin(x/2) is −(3√10/10).
Answer: −(3√10/10).
The identity that we need to verify is sin(π/3 + x) = cos(x) + √3 sin(x). Simplifying at each step:
We can use the following identities:
sin(A + B) = sinA cosB + cosA sinB
cos(A + B) = cosA cosB − sinA sinB
cos(π/3) = 1/2, sin(π/3) = √3/2
sin(π/3 + x) = sin(π/3) cos(x) + cos(π/3) sin(x) = (√3/2) cos(x) + (1/2) sin(x)
By rearranging, we have: sin(π/3 + x) = cos(x) + √3 sin(x).
Hence, we have verified the given identity. Therefore, the value of sin(π/3 + x) is cos(x) + √3 sin(x).
Answer: cos(x) + √3 sin(x). 6. We are to find the value of sin(x/2) if cos(x) = -4/5 and π/2 < x < π.We can start by drawing the unit circle for angles between 90° and 180°.
We can see that the y-coordinate of the point is negative, which means that sin(x/2) is also negative.
To find the value of sin(x/2), we can use the following identity:
sin(x/2) = ±√[(1 − cos(x))/2]
Since sin(x/2) is negative in this case, we can take the negative square root:
sin(x/2) = −√[(1 − cos(x))/2]
= −√[(1 + 4/5)/2] = −√[9/10]
= −(3/√10) × (√10/√10) = −(3√10/10)
Therefore, the value of sin(x/2) is −(3√10/10).
Answer: −(3√10/10).
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How do you find the average value of
f(x)=√x as x varies between [0,4]?
To find the average value of a function f(x) over a given interval [a, b], you can use the following formula:
Average value of f(x) = (1 / (b - a)) * ∫[a to b] f(x) dx
In this case, we want to find the average value of f(x) = √x over the interval [0, 4]. Applying the formula, we have:
Average value of √x = (1 / (4 - 0)) * ∫[0 to 4] √x dx
Now, we can integrate the function √x with respect to x over the interval [0, 4]:
∫√x dx = (2/3) * x^(3/2) evaluated from 0 to 4
= (2/3) * (4^(3/2)) - (2/3) * (0^(3/2))
= (2/3) * 8 - 0
= 16/3
Substituting this value back into the formula, we get:
Average value of √x = (1 / (4 - 0)) * (16/3)
= (1/4) * (16/3)
= 4/3
Therefore, the average value of f(x) = √x as x varies between [0, 4] is 4/3.
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.The diagram shows a cone and its axis of rotation. If a plane passes through the axis of rotation, which type of cross section will be formed?
A: a circle
B: an isosceles triangle
C: a parabola
D: an oval
A: a circle
Is the cross section formed by a plane passing through the axis of rotation of a cone a circle?
When a plane passes through the axis of rotation of a cone, the resulting cross section will be a circle. This is because a cone is a three-dimensional geometric shape that tapers from a circular base to a single point called the apex. The axis of rotation is the line passing through the apex and the center of the circular base. When a plane intersects the cone along this axis, it cuts through the cone's curved surface, resulting in a circular cross section.
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What is the common ratio for the geometric sequence?
24,−6,32,−38,...
the common ratio of the geometric sequence 24, −6, 32, −38, ... is -1.5.
The common ratio for the geometric sequence 24, −6, 32, −38, ... is -1.5.What is a geometric sequence?A geometric sequence is a sequence in which each term after the first is found by multiplying the preceding term by a fixed number. It is a sequence in which each term is obtained by multiplying the previous term by a constant value or ratio.In a geometric sequence, the ratio between any two consecutive terms is the same. The nth term of a geometric sequence can be represented as an = a1rn-1, where a1 is the first term, r is the common ratio, and n is the number of terms.Using the given terms 24, −6, 32, −38, ...The ratio between the second term and the first term is given as : (-6)/24 = -1/4Similarly, the ratio between the third term and the second term is given as: 32/(-6) = -16/3The ratio between the fourth term and the third term is given as: (-38)/32 = -19/16So, the sequence is not a constant ratio because the ratios are not the same for all of the terms.However, if you observe the ratios, you'll find that the ratio between any two consecutive terms is obtained by dividing the second term by the first term and it's the same as the ratio between the third term and the second term, and it's also the same as the ratio between the fourth term and the third term.
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Problem 7: Let X(t) = A sin πt, where A is a continuous random variable with the pdf f₁(a)= 1= {201 [2a, 0
Problem : Let X(t) = A sin πt, where A is a continuous random variable with the pdf f₁(a)= 1= {201 [2a, 0 < a < 1/2 0, elsewhereWhere X(t) is continuous?
Continuous random variable: It is a random variable that can take on any value over a continuous range of possible values.
X(t) is continuous because it can take on any value over a continuous range of possible values. Because A can be any value between 0 and 1, the possible range of values for X(t) is between -π/2 and π/2. The sine function is continuous over this range, therefore X(t) is continuous.
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The linear transformation L defined by : \(L(p(x)) = p^{'}(x) + p(0) \) maps P3into P2. Find the matrix representation of L with respect to the ordered Bases [x^2, x, 1] and [2, 1-x]. For each of the following vectors p(x) in P3, find the coordinates of L(p(x)) with respect to the ordered basis [2, 1-x].
a) x^2 + 2x -3
b) x^2 + 1
c) 3x
d)4x^2 + 2x
To find the matrix representation of the linear transformation [tex]\(L\)[/tex] with respect to the given bases, we need to find the images of the basis vectors [tex]\([x^2, x, 1]\)[/tex] under [tex]\(L\)[/tex] and express them as linear combinations of the basis vectors [tex]\([2, 1-x]\).[/tex]
Let's start by finding the image of [tex]\(x^2\)[/tex] under [tex]\(L\):[/tex]
[tex]\(L(x^2) = (x^2)' + (x^2)(0) = 2x\)[/tex]
We can express [tex]\(2x\)[/tex] as a linear combination of the basis vectors [tex]\([2, 1-x]\):\(2x = 2(2) + 0(1-x)\)[/tex]
Next, let's find the image of [tex]\(x\)[/tex] under [tex]\(L\):[/tex]
[tex]\(L(x) = (x)' + (x)(0) = 1\)[/tex]
We can express [tex]\(1\)[/tex] as a linear combination of the basis vectors [tex]\([2, 1-x]\):\(1 = 0(2) + 1(1-x)\)[/tex]
Finally, let's find the image of the constant term [tex]\(1\)[/tex] under [tex]\(L\):[/tex]
[tex]\(L(1) = (1)' + (1)(0) = 0\)[/tex]
We can express [tex]\(0\)[/tex] as a linear combination of the basis vectors [tex]\([2, 1-x]\):\(0 = 0(2) + 0(1-x)\)[/tex]
Now, we can arrange the coefficients of the linear combinations in a matrix to obtain the matrix representation of [tex]\(L\)[/tex] with respect to the given bases:
[tex]\[\begin{bmatrix}2 & 0 & 0 \\0 & 1 & 0 \\2 & 1 & 0\end{bmatrix}\][/tex]
To find the coordinates of [tex]\(L(p(x))\)[/tex] with respect to the ordered basis [tex]\([2, 1-x]\)[/tex], we can simply multiply the matrix representation of [tex]\(L\)[/tex] by the coordinate vector of [tex]\(p(x)\)[/tex] with respect to the ordered basis [tex]\([x^2, x, 1]\).[/tex]
Let's calculate the coordinates for each given vector [tex]\(p(x)\):[/tex]
a) [tex]\(p(x) = x^2 + 2x - 3\)[/tex]
The coordinate vector of [tex]\(p(x)\)[/tex] with respect to [tex]\([x^2, x, 1]\) is \([1, 2, -3]\).[/tex] Multiplying the matrix representation of [tex]\(L\)[/tex] by this coordinate vector:
[tex]\[\begin{bmatrix}2 & 0 & 0 \\0 & 1 & 0 \\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1 \\2 \\-3\end{bmatrix}= \begin{bmatrix}2 \\2 \\-4\end{bmatrix}\][/tex]
So, the coordinates of [tex]\(L(p(x))\)[/tex] with respect to [tex]\([2, 1-x]\) are \([2, 2, -4]\).[/tex]
b) [tex]\(p(x) = x^2 + 1\)[/tex]
The coordinate vector of [tex]\(p(x)\)[/tex] with respect to [tex]\([x^2, x, 1]\) is \([1, 0, 1]\).[/tex]
Multiplying the matrix representation of [tex]\(L\)[/tex] by this coordinate vector:
[tex]\[\begin{bmatrix}2 & 0 & 0 \\0 & 1 & 0 \\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1 \\0 \\1\end{bmatrix}= \begin{bmatrix}2 \\0 \\2\end{bmatrix}\][/tex]
So, the coordinates of [tex]\(L(p(x))\)[/tex] with respect to [tex]\([2, 1-x]\) are \([2, 0, 2]\).[/tex]
c) [tex]\(p(x) = 3x\)[/tex]
The coordinate vector of [tex]\(p(x)\)[/tex] with respect to [tex]\([x^2, x, 1]\) is \([0, 3, 0]\).[/tex]
Multiplying the matrix representation of [tex]\(L\)[/tex] by this coordinate vector:
[tex]\[\begin{bmatrix}2 & 0 & 0 \\0 & 1 & 0 \\2 & 1 & 0\end{bmatrix}\begin{bmatrix}0 \\3 \\0\end{bmatrix}= \begin{bmatrix}0 \\3 \\0\end{bmatrix}\][/tex]
So, the coordinates of [tex]\(L(p(x))\)[/tex] with respect to [tex]\([2, 1-x]\) are \([0, 3, 0]\).[/tex]
d) [tex]\(p(x) = 4x^2 + 2x\)[/tex]
The coordinate vector of [tex]\(p(x)\)[/tex] with respect to [tex]\([x^2, x, 1]\) is \([4, 2, 0]\).[/tex]
Multiplying the matrix representation of [tex]\(L\)[/tex] by this coordinate vector:
[tex]\[\begin{bmatrix}2 & 0 & 0 \\0 & 1 & 0 \\2 & 1 & 0\end{bmatrix}\begin{bmatrix}4 \\2 \\0\end{bmatrix}= \begin{bmatrix}8 \\2 \\8\end{bmatrix}\][/tex]
So, the coordinates of [tex]\(L(p(x))\)[/tex] with respect to \([2, 1-x]\) are \([8, 2, 8]\).
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Solve dydx=(y?1)(y+1) if the solution passes through the point (x,y)=(2,0). Graph the solution.y(x)=??
To graph the solution, plot the function y(x) over the specified interval.
Solve the differential equation dy/dx = (y-1)(y+1) with the initial condition y(2) = 0 and graph the solution.To solve the given differential equation, we can use separation of variables. Let's proceed with the solution:
dy/dx = (y-1)(y+1)We can rewrite the equation as:
dy/(y-1)(y+1) = dxNow, we integrate both sides:
∫(dy/(y-1)(y+1)) = ∫dxUsing partial fraction decomposition, we can express the integrand as:
1/2 * (∫(1/(y-1))dy - ∫(1/(y+1))dy)Integrating each term separately:
1/2 * (ln|y-1| - ln|y+1|) = x + CApplying the initial condition (x,y) = (2,0):1/2 * (ln|-1| - ln|1|) = 2 + Cln(1) - ln(1) = 4 + 2C0 = 4 + 2CC = -2Substituting C back into the equation:
1/2 * (ln|y-1| - ln|y+1|) = x - 2ln|y-1| - ln|y+1| = 2x - 4Taking the exponential of both sides:
|y-1| / |y+1| = e^(2x-4)Considering the positive and negative cases separately:
y - 1 = ± (y + 1) * e^(2x-4)Now, solving for y in both cases:
y - 1 = (y + 1) * e^(2x-4)Simplifying the equation:
y - y*e^(2x-4) = 1 + e^(2x-4)Factoring out y:y(1 - e^(2x-4)) = 1 + e^(2x-4)Dividing both sides by (1 - e^(2x-4)):
y = (1 + e^(2x-4)) / (1 - e^(2x-4)) y - 1 = - (y + 1) * e^(2x-4)Simplifying the equation:
y + y*e^(2x-4) = 1 - e^(2x-4)Factoring out y:y(1 + e^(2x-4)) = 1 - e^(2x-4)Learn more about specified
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Kevin was asked to solve the following system of inequali-
ties using graphing and then identify a point in the solution
set.
Kevin said (2, 5) is a point in the solution set. Kevin's point
is not in the solution set.
Look at Kevin's graph to determine his mistake and how to
fix it.
2. Kevin was asked to solve the following system of
inequalities using graphing and then identify a
point in the solution set.
(y> 2x-1
lys-x+5
Kevin said (2, 5) is a point in the solution set.
Kevin's p
's point i
int is not in the solution set.
Look at Kevin's graph to determine his mistake and
how to fix it.
Kevin's mistake was...
He can fix this by...
Given statement solution is :- Kevin's point (2, 5) is not in the solution set. To fix his mistake, Kevin needs to correctly identify a point in the solution set. By observing the shaded region in the graph where the two inequalities overlap, he can select any point within that region as a valid solution. He should choose a point that lies within the overlapping region, such as (1, 4), (0, 3), or any other point that satisfies both inequalities.
Kevin's mistake was incorrectly identifying (2, 5) as a point in the solution set of the system of inequalities. To determine his mistake and how to fix it, let's examine the given system of inequalities:
y > 2x - 1
y ≤ x + 5
To graph these inequalities, we need to plot their corresponding boundary lines and determine the regions that satisfy the given conditions.
For inequality 1, y > 2x - 1, we draw a dashed line with a slope of 2 passing through the point (0, -1). This line separates the plane into two regions: the region above the line satisfies y > 2x - 1, and the region below does not.
For inequality 2, y ≤ x + 5, we draw a solid line with a slope of 1 passing through the point (0, 5). This line separates the plane into two regions: the region below the line satisfies y ≤ x + 5, and the region above does not.
Now, we need to determine the overlapping region that satisfies both inequalities. In this case, we shade the region below the solid line (inequality 2) and above the dashed line (inequality 1). The overlapping region is the region that satisfies both conditions.
Upon examining the graph, we can see that the point (2, 5) lies above the dashed line (inequality 1), which means it does not satisfy the condition y > 2x - 1. Therefore, Kevin's point (2, 5) is not in the solution set.
To fix his mistake, Kevin needs to correctly identify a point in the solution set. By observing the shaded region in the graph where the two inequalities overlap, he can select any point within that region as a valid solution. He should choose a point that lies within the overlapping region, such as (1, 4), (0, 3), or any other point that satisfies both inequalities.
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3, 7, 8, 5, 6, 4, 9, 10, 7, 8, 6, 5 Using the previous question 's scores, If three points were added to every score in the distribution as a population, what would be the new mean? If three points we
The new mean of the distribution would be 8.6667.
The given data set is as follows: 3, 7, 8, 5, 6, 4, 9, 10, 7, 8, 6, 5.
The mean is calculated by adding all the values of a data set and dividing the sum by the total number of values in the data set. Therefore, the mean (μ) can be calculated as follows:
μ = (3 + 7 + 8 + 5 + 6 + 4 + 9 + 10 + 7 + 8 + 6 + 5) / 12
μ = 70 / 12
μ = 5.8333
If three points are added to each score, the new data set will be as follows: 6, 10, 11, 8, 9, 7, 12, 13, 10, 11, 9, 8.
The mean of the new data set can be calculated as follows:
μ' = (6 + 10 + 11 + 8 + 9 + 7 + 12 + 13 + 10 + 11 + 9 + 8) / 12
μ' = 104 / 12
μ' = 8.6667
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Suppose that the functions q and r are defined as follows. q(x)=-4x+1 r(x) = 3x-2 Find the following. (gor)(1) = 0 (rog) (1) = 0 x 6 ?
The value of (gor)(1) is 0, indicating that the composition of the functions g and r, evaluated at x = 1, results in an output of 0. Similarly, the value of (rog)(1) is also 0, indicating that the composition of the functions r and g, evaluated at x = 1, also gives an output of 0.
The composition of two functions, denoted as (fog)(x), is obtained by substituting the output of the function g into the input of the function f. In this case, we have two functions q(x) = -4x + 1 and r(x) = 3x - 2. To evaluate (gor)(1), we first evaluate the inner composition (or the composition of g and r) by substituting x = 1 into r(x). This gives us r(1) = 3(1) - 2 = 1. Next, we substitute this result into q(x), obtaining q(r(1)) = q(1) = -4(1) + 1 = -3. Therefore, (gor)(1) = -3.
Similarly, to evaluate (rog)(1), we first evaluate the inner composition (or the composition of r and g) by substituting x = 1 into g(x). This gives us g(1) = -4(1) + 1 = -3. Next, we substitute this result into r(x), obtaining r(g(1)) = r(-3) = 3(-3) - 2 = -11. Therefore, (rog)(1) = -11.
Since the given task asks to find when the compositions of the functions are equal to 0, neither (gor)(1) nor (rog)(1) is equal to 0.
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What kind and how much polygons do you see in the net of the triangular prism?
The net of a triangular prism consists of two triangles and three rectangles.
In the net of a triangular prism, we can observe two types of polygons: triangles and rectangles.
First, let's discuss the triangles.
A triangular prism has two triangular faces, which are congruent to each other.
These triangles are equilateral triangles, meaning they have three equal sides and three equal angles.
Each of these triangles contributes two polygons to the net, one for each face.
Next, we have the rectangles.
A triangular prism has three rectangular faces that connect the corresponding sides of the triangular bases.
These rectangles have opposite sides that are parallel and equal in length.
Each rectangle contributes one polygon to the net, resulting in a total of three rectangles.
To summarize, the net of a triangular prism consists of two equilateral triangles and three rectangles.
The triangles represent the bases of the prism, while the rectangles form the lateral faces connecting the bases.
Altogether, there are five polygons in the net of a triangular prism.
It's important to note that the dimensions of the polygons may vary depending on the specific size and proportions of the triangular prism, but the basic shape and number of polygons remain the same.
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Customers arrive at the CVS Pharmacy drive-thru at an average rate of 5 per hour. What is the probability that more than 6 customers will arrive at the drive-thru during a randomly chosen hour? 0.146
The probability that more than 6 customers will arrive at the drive-thru during a randomly chosen hour is approximately 0.2374 or 0.24 (rounded to two decimal places).
The Poisson distribution formula is used for probability problems that involve counting the number of events that happen in a certain period of time or space. It is given as:P(X = x) = (e^-λ) (λ^x) / x!
Where:X is the number of eventsλ is the average rate at which events occur.
e is a constant with a value of approximately 2.71828x is the number of events that occur in a specific period of time or spacex! = x * (x - 1) * (x - 2) * ... * 2 * 1 is the factorial of xIn the given problem, the average rate at which customers arrive at the CVS Pharmacy drive-thru is 5 per hour, and we need to find the probability that more than 6 customers will arrive at the drive-thru during a randomly chosen hour.
P(X > 6) = 1 - P(X ≤ 6)For calculating P(X ≤ 6), we can use the Poisson distribution formula as:
P(X ≤ 6) = (e^-5) (5^0) / 0! + (e^-5) (5^1) / 1! + (e^-5) (5^2) / 2! + (e^-5) (5^3) / 3! + (e^-5) (5^4) / 4! + (e^-5) (5^5) / 5! + (e^-5) (5^6) / 6!P(X ≤ 6) ≈ 0.7626
Substituting this value in the previous equation, we get:
P(X > 6) = 1 - P(X ≤ 6)
≈ 1 - 0.7626
= 0.2374
Hence, the probability that more than 6 customers will arrive at the drive-thru during a randomly chosen hour is approximately 0.2374 or 0.24 (rounded to two decimal places).
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Consider a consumer whose utility function is:U(x1, x2) = log(x₁) + log(x₂) X1 ≤ 0.5 Suppose that p₂ = 1, m = 1, and p1 is unknown. There is rationing such that ** Part a. (5 marks) Find the minimal p₁, denoted by pi, such that the if P1 > Pi, then the consumer consumes x₁ strictly less than 0.5. ** Part b. (10 marks) Now suppose increases. mathematically show that whether the threshold on you found in Part a increases/decreases/stays the same.
Part a)Given, utility function of the consumer as:U(x1, x2) = log(x1) + log(x2)X1 ≤ 0.5Let p2 = 1 and m = 1, and p1 is unknown. The consumer has a budget constraint as: p1x1 + p2x2 = m = 1Now we have to find the minimal p1 such that the consumer consumes x1 strictly less than 0.5.
We need to find the value of p1 such that the consumer spends the entire budget (m = 1) on the two goods, but purchases only less than 0.5 units of the first good. In other words, the consumer spends all his money on the two goods, but still cannot afford more than 0.5 units of good 1.
Mathematically we can represent this as:
p1x1 + p2x2 = 1......(1)Where, x1 < 0.5, p2 = 1 and m = 1
Substituting the given value of p2 in (1), we get:
p1x1 + x2 = 1x1 = (1 - x2) / p1Given, x1 < 0.5 => (1 - x2) / p1 < 0.5 => 1 - x2 < 0.5p1 => p1 > (1 - x2) / 0.5
Now we know, 0 < x2 < 1.So, we will maximize the expression (1 - x2) / 0.5 for x2 ∈ (0,1) which gives the minimum value of p1 such that x1 < 0.5.On differentiating the expression w.r.t x2, we get:d/dx2 [(1-x2)/0.5] = -1/0.5 = -2
Therefore, (1-x2) / 0.5 is maximum at x2 = 0.
Now, substituting the value of x2 = 0 in the above equation, we get:p1 > 1/0.5 = 2So, the minimal value of p1 is 2.Part b)Now, we have to show mathematically that whether the threshold on p1 found in Part a increases/decreases/stays the same when p2 increases.
That is, if p2 increases then the minimum value of p1 will increase/decrease/stay the same.Since p2 = 1, the consumer’s budget constraint is given by:
p1x1 + x2 = m = 1Suppose that p2 increases to p2′.
The consumer’s new budget constraint is:
p1x1 + p2′x2 = m = 1.
Now we will find the minimal p1 denoted by pi, such that the consumer purchases less than 0.5 units of good 1. This can be expressed as:
p1x1 + p2′x2 = 1Where, x1 < 0.5
The budget constraint is the same as that in Part a, except that p2 has been replaced by p2′. Now, using the same argument as in Part a, the minimum value of p1 is given by:
p1 > (1 - x2) / 0.5.
We need to maximize (1 - x2) / 0.5 w.r.t x2.
As discussed in Part a, this occurs when x2 = 0.Therefore, minimal value of p1 is:
pi > 1/0.5 = 2
This value of pi is independent of the value of p2′.
Hence, the threshold on p1 found in Part a stays the same when p2 increases.
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Bank Will Sell The Bond For A Commission Of 2.1%. The Market Yield Is Currently 7.6% On Twenty-Year Zero-Coupon Bonds. If Rawlings Wants To Issue A Zero-Coupon Bond, How Many Bonds Will It Need To Sell To Raise The $37,100,000? Assume That The bond is semiannual and issued at a per value of $1,000?
Rawlings will need to sell approximately 46,678 zero-coupon bonds to raise $37,100,000.
To calculate the number of bonds Rawlings needs to sell, we can use the formula for the present value of a bond. The formula is:
PV = (FV / [tex](1 + r)^n[/tex])
Where PV is the present value (the amount Rawlings wants to raise), FV is the future value (the face value of the bonds), r is the market yield, and n is the number of periods.
Given that Rawlings wants to raise $37,100,000, the face value of each bond is $1,000 (per value), and the market yield is 7.6% (or 0.076 as a decimal), we can rearrange the formula to solve for n:
n = ln(FV / PV) / ln(1 + r)
Substituting the values, we get:
n = ln(1000 / 37100000) / ln(1 + 0.076)
Using a financial calculator or spreadsheet software, we can calculate n, which comes out to be approximately 46,678. This means that Rawlings will need to sell around 46,678 zero-coupon bonds to raise the desired amount of $37,100,000.
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for a random variable z, its mean and variance are defined as e[z] and e[(z − e[z])2 ], respectively.
For a random variable z, its mean and variance are defined as e[z] and e[(z − e[z])2 ], respectively.
What is a random variable?
A random variable is a set of all possible values for which a probability distribution is defined. It is a numerical value assigned to all potential outcomes of a statistical experiment.
What is the mean of a random variable?
The mean, sometimes referred to as the expected value, is the sum of the product of each possible value multiplied by its probability, giving the value that summarizes or represents the center of the distribution of a set of data.
What is the variance of a random variable?
The variance is the expected value of the squared deviation of a random variable from its expected value. It determines how much the values of a variable deviate from the expected value.What is the formula for the mean of a random variable?
The formula for the mean of a random variable is:E(X) = ∑ xi * P(xi)
What is the formula for the variance of a random variable?
The formula for the variance of a random variable is:Variance(X) = ∑ ( xi - mean )² * P(xi)where 'mean' is the expected value.
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Given the general form of the circle 3x^2 − 24x + 3y^2 + 36y = −141
a.) Write the equation of the circle in standard (center-radius) form (x−h)^2+(y−k)^2=r^2
=
b.) The center of the circle is at the point ( , )
a) The standard form of the given circle is (x − 4)² + (y + 6)²/9 = 0
b) the center of the circle is at (h, k) = (4, -6).
The given equation of the circle is: 3x² − 24x + 3y² + 36y = −141
a.) Write the equation of circle in standard (center-radius) form (x−h)² + (y−k)² = r²
General equation of a circle is given as:x² + y² + 2gx + 2fy + c = 0
Comparing the above equation with the given circle equation, we have:
3x² − 24x + 3y² + 36y = −1413x² − 24x + 36y + 3y² = −141
Rearranging the above equation, we get:
3x² − 24x + 36y + 3y² + 141
= 03(x² − 8x + 16) + 3(y² + 12y + 36)
= 03(x − 4)² + 3(y + 6)² = 0
Comparing the above equation with (x−h)² + (y−k)² = r²,
we get:(x − 4)² + (y + 6)²/3² = 0
Hence, the standard form of the given circle is (x − 4)² + (y + 6)²/9 = 0
b.) The center of the circle is at the point (4, −6).
Hence, the center of the circle is at (h, k) = (4, -6).
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what is true of the data in the dot plot? check all that apply. number of minutes shelly spent waiting for the bus each morning
A dot plot is a graphical method that is used to represent data. The plot provides an overview of the data’s distribution, measures of central tendency, and any outliers.
From the provided question, we are supposed to determine what is true of the data in the dot plot. Below are the correct statements that apply: There is no data value that occurs more frequently than any other value in the set. This means that there are no modes in the data set. We can note that the data set is bimodal if there were two points with dots above them.
The data in the set is roughly symmetrical since it is distributed evenly around the middle. There are equal numbers of dots on either side of the middle point, and the plot is roughly symmetrical about a vertical line passing through the middle point. All data points in the data set lie within a range of 5 to 20. We can see that there are no dots below 5 or above 20.
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triangle d has been dilated to create triangle d 4, 3, 1/3, 1/4
Triangle D has been dilated to create Triangle D' with scale factors of 4, 9, and 4/3 for the corresponding sides.
To understand the dilation of Triangle D to create Triangle D', we can examine the ratio of corresponding sides.
Given that the corresponding sides of Triangle D and Triangle D' are in the ratio of 4:1, 3:1/3, and 1/3:1/4, we can determine the scale factor of dilation for each side.
The scale factor for the first side is 4:1, indicating that Triangle D' is four times larger than Triangle D in terms of that side.
For the second side, the ratio is 3:1/3. To simplify this ratio, we can multiply both sides by 3, resulting in a ratio of 9:1. This means that Triangle D' is nine times larger than Triangle D in terms of the second side.
Finally, the ratio for the third side is 1/3:1/4. To simplify this ratio, we can multiply both sides by 12, resulting in a ratio of 4:3. This means that Triangle D' is four-thirds the size of Triangle D in terms of the third side.
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The radius of the wheel on a bike is 21 inches. If the wheel is revolving at 154 revolutions per minute, what is the linear speed of the bike, in miles per hour? Round your answer to the nearest tenth, and do not include units in your answer.
Answer:
19.2 mph
Step-by-step explanation:
Given a bike wheel with a radius of 21 inches turning at 154 rpm, you want to know the speed of the bike in miles per hour.
DistanceA wheel with a radius of 21 inches will have a diameter of 42 inches, or 3.5 feet. In one turn, it will travel ...
C = πd
C = π(3.5 ft) . . . . per revolution
In one minute, the bike travels this distance 154 times, so a distance of ...
(3.5π ft/rev)(154 rev/min) = 1693.318 ft
SpeedThe speed is the distance divided by the time:
(1693.318 ft)/(1/60 h) × (1 mi)/(5280 ft) ≈ 19.2 mi/h
__
Additional comment
We could use the conversion factor 88 ft/min = 1 mi/h.
Bike wheel diameters are typically 26 inches or less, perhaps 29 inches for road racing. A 42-inch wheel would be unusually large.
On the other hand, the chainless "penny farthing" bicycle has a wheel diameter typically 44-60 inches. It would be real work to pedal that at 154 RPM.
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the speed of the bike in miles per hour is;[51408π/63360]/[1/60] mph= 30.9 mph (approx)Hence, the linear speed of the bike, in miles per hour, is 30.9 mph.
To find the linear speed of the bike, in miles per hour, given the radius of the wheel of the bike as 21 inches and the wheel revolving at 154 revolutions per minute, we can use the formula for the circumference of a circle as;C = 2πrWhere r is the radius of the circle and C is the circumference of the circle.From the given information, we can find the circumference of the wheel as;C = 2π(21) inches= 132π inchesTo find the distance traveled by the bike per minute, we can multiply the circumference of the wheel by the number of revolutions per minute;Distance traveled per minute = 154 × 132π inches= 51408π inchesTo find the speed of the bike in miles per hour, we need to convert the units of distance from inches to miles and the units of time from minutes to hours as;1 inch = 1/63360 miles (approx) and1 minute = 1/60 hours (approx)Therefore, the speed of the bike in miles per hour is;[51408π/63360]/[1/60] mph= 30.9 mph (approx)Hence, the linear speed of the bike, in miles per hour, is 30.9 mph.
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