a. the probability that a randomly selected candy weighs more than 0.8537 g is approximately 0.5062.
b. the probability that a sample of 458 candies will have a mean weight of at least 0.8537 g is approximately 0.4920.
a. To find the probability that a randomly selected candy weighs more than 0.8537 g, we can use the z-score and the standard normal distribution.
Given:
Mean weight (μ) = 0.8545 g
Standard deviation (σ) = 0.0517 g
We need to find the probability P(X > 0.8537), where X is the weight of a randomly selected candy.
First, let's calculate the z-score for 0.8537 g:
z = (x - μ) / σ
z = (0.8537 - 0.8545) / 0.0517
z ≈ -0.0155
Using the standard normal distribution table or a calculator, we can find the probability corresponding to a z-score of -0.0155, which is approximately 0.5062.
Therefore, the probability that a randomly selected candy weighs more than 0.8537 g is approximately 0.5062.
b. To find the probability that a sample of 458 candies will have a mean weight of at least 0.8537 g, we need to calculate the sampling distribution of the sample mean.
Given:
Sample size (n) = 458
Mean weight (μ) = 0.8545 g
Standard deviation (σ) = 0.0517 g
The sample mean follows a normal distribution with the same mean as the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σ/√n).
Standard deviation of the sample mean (σ/√n) = 0.0517 / √458 ≈ 0.002415
To find the probability P([tex]\bar{X}[/tex] ≥ 0.8537), where [tex]\bar{X}[/tex] is the mean weight of the sample of 458 candies:
Using the z-score formula:
z = ([tex]\bar{X}[/tex] - μ) / (σ/√n)
z = (0.8537 - 0.8545) / 0.002415
z ≈ -0.0331
Using the standard normal distribution table or a calculator, the probability corresponding to a z-score of -0.0331 is approximately 0.4920.
Therefore, the probability that a sample of 458 candies will have a mean weight of at least 0.8537 g is approximately 0.4920.
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Find the area of the surface.
The helicoid (or spiral ramp) with vector equation r(u, v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ π
To find the area of the surface, we can use the surface area formula for a parametric surface given by r(u, v):
A = ∬√[ (∂r/∂u)² + (∂r/∂v)² + 1 ] dA
where ∂r/∂u and ∂r/∂v are the partial derivatives of the vector function r(u, v) with respect to u and v, and dA is the area element in the u-v coordinate system.
In this case, the vector equation of the helicoid is r(u, v) = u cos(v) i + u sin(v) j + v k, with the given parameter ranges 0 ≤ u ≤ 1 and 0 ≤ v ≤ π.
Taking the partial derivatives, we have:
∂r/∂u = cos(v) i + sin(v) j + 0 k
∂r/∂v = -u sin(v) i + u cos(v) j + 1 k
Plugging these values into the surface area formula and integrating over the given ranges, we can calculate the surface area of the helicoid. However, this process involves numerical calculations and may not yield a simple closed-form expression.
Hence, the exact value of the surface area of the helicoid in this case would require numerical evaluation using appropriate numerical methods or software.
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If X ∼ t(p), prove that X2 ∼ F(1,p)
The t-distribution is defined as the distribution of a random variable T that is obtained by dividing a standard normal variable Z by the square root of a chi-square random variable χ^2 with p degrees of freedom, that is:
T = Z / √(χ^2 / p)
If X ∼ t_p with p degrees of freedom, then:
X = Z / √(χ^2 / p)
We can square this expression to obtain:
X^2 = Z^2 / (χ^2 / p)
Since the numerator is a chi-square variable with one degree of freedom and the denominator is a chi-square variable with p degrees of freedom, we have:
X^2 ∼ F(1,p)
Therefore, if X ∼ t_p, then X^2 ∼ F(1,p).
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a cone with volume 5000 m³ is dilated by a scale factor of 15. what is the volume of the resulting cone? enter your answer in the box.
When a cone with a volume of 5000 m³ is dilated by a scale factor of 15, the volume of the resulting cone is 3375000 m³.
The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius of the base and h is the height. Since the scale factor of 15 applies to all dimensions of the cone, the new radius and height will be 15 times the original values. Let's assume the original cone has radius r and height h.
After dilation, the new cone will have a radius of 15r and a height of 15h. Plugging these values into the volume formula, we get
V' = (1/3)π(15r)²(15h) = (1/3)π(15²)(r²)(h) = 3375V.
Given that the original cone has a volume of 5000 m³, we can calculate the volume of the resulting cone by multiplying 5000 by 3375:
V' = 5000× 3375 = 3375000 m³.
Therefore, the volume of the resulting cone, after being dilated by a scale factor of 15, is 3375000 m³.
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Suppose that the intervals between car accidents at a known accident blackspot can be modelled by an exponential distribution with unknown parameter θ, θ > 0. The p.d.f. of this distribution is f(x; θ) = θ e−θx, x> 0. The four most recent intervals between accidents (in weeks) are x1 = 4.5, x2 = 1.5, x3 = 6, x4 = 4.4; these values are to be treated as a random sample from the exponential distribution. (a) Show that the likelihood of θ based on these data is given by L(θ) = θ4 e−16.4
The likelihood of θ based on these data is given by L(θ) = θ⁴ e^(-16.4).
The likelihood function for the given data is given by
L(θ) = f(x1;θ).f(x2;θ).f(x3;θ).f(x4;θ)= θ.e^(-θx1).θ.e^(-θx2).θ.e^(-θx3).θ.e^(-θx4)= θ⁴ e^(-θ(x1+x2+x3+x4))= θ⁴ e^(-θ(16.4))
Therefore, the likelihood of θ based on these data is given by L(θ) = θ⁴ e^(-16.4).
Hence, the required result is obtained.
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Data Analysis (20 points)
Dependent Variable: Y Method: Least Squares
Date: 12/19/2013 Time: 21:40 Sample: 1989 2011
Included observations:23
Variable Coefficient Std. Error t-Statistic Prob.
C 3000 2000 （ ） 0.1139
X1 2.2 0.110002 20 0.0000
X2 4.0 1.282402 3.159680 0.0102
R-squared ( ) Mean dependent var 6992
Adjusted R-square S.D. dependent var 2500.
S.E. of regression （ ） Akaike info criterion 19.
Sum squared resid 2.00E+07 Schwarz criterion 21
Log likelihood -121 F-statistic （ ）
Durbin-Watson stat 0.4 Prob(F-statistic) 0.001300
Using above E-views results：:
Put correct numbers in above parentheses（with computation process）
(12 points)
(2)How is DW statistic defined? What is its range? (6 points)
（3） What does DW=0.4means? (2 points)
The correct numbers are to be inserted in the blanks (with calculation process) using the given E-views results above are given below: (1) Variable Coefficient Std. Error t-Statistic Prob.
C. 3000 2000 1.50 0.1139X1 2.2 0.110002 20 0.0000X2 4.0 1.282402 3.159680 0.0102R-squared 0.9900 Mean dependent var 6992. Adjusted R-square 0.9856 S.D. dependent var 2500. S.E. of regression 78.49 Akaike info criterion 19. Sum squared redid 2.00E+07 Schwarz criterion 21 Log likelihood -121 F-statistic 249.9965 Durbin-Watson stat 0.4 Prob(F-statistic) 0.0013 (2)DW (Durbin-Watson) statistic is defined as a test
statistic that determines the existence of autocorrelation (positive or negative) in the residual sequence. Its range is between 0 and 4, where a value of 2 indicates no autocorrelation. (3) DW = 0.4 means there is a positive autocorrelation in the residual sequence, since the value is less than 2. This means that the error term of the model is correlated with its previous error term.
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characterize the likely shape of a histogram of the distribution of scores on a midterm exam in a graduate statistics course.
The shape of a histogram of the distribution of scores on a midterm exam in a graduate statistics course is likely to be bell-shaped, symmetrical, and normally distributed. The bell curve, or the normal distribution, is a common pattern that emerges in many natural and social phenomena, including test scores.
The mean, median, and mode coincide in a normal distribution, making the data symmetrical on both sides of the central peak.In a graduate statistics course, it is reasonable to assume that students have a good understanding of the subject matter, and as a result, their scores will be evenly distributed around the average, with a few outliers at both ends of the spectrum.The histogram of the distribution of scores will have an approximately normal curve that is bell-shaped, with most of the scores falling in the middle of the range and fewer scores falling at the extremes.
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probability of an event B in relationship to an event A, is defined as the probability that event B occurs after event A has already occurred. O a. Empirical Ob. Unconditional Oc. Conditional Od. Samp
The answer is option (c) Conditional. The probability of an event B in relationship to an event A, is defined as the conditional probability that event B occurs after event A has already occurred.
Explanation: The probability of an event A occurring given that event B has already occurred is known as a conditional probability. P(A|B) = Probability of A given that B has occurredP(B|A) = Probability of B given that A has occurred.If B is the occurrence of one event and A is the occurrence of another event, then we can say that the probability of event B happening given that event A has already happened is known as a conditional probability.
Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about the likelihood of an event happening, or how likely it is. Probability can range from 0 to 1, with 0 denoting an impossibility and 1 denoting a certainty.
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Researchers studying the learning of speech often compare measurements made on the recorded speech of adults and children. One variable of interest is called the voice onset time (VOT). Here are the results for 6-year-old children and adults asked to pronounce the word "bees". The VOT is measured in milliseconds and can be either positive or negative.
Group n x-bar s
Children 10 -3.67 33.89
Adults 20 -23.17 50.74
Give a P-value and state your conclusions.
P is between _______ and ________.
The Pvalue of the study described in the problem given is between 0.1 and 0.2.
Calculating the standard error values using the relation:
SE = √(s1²/n1 + s2²/n2)
SE = √(33.89²/10 + 50.74²/20) = 17.81
Obtaining the test statistic :
t = (x1 - x2)/SE
Where x1 and x2 are the mean values of the samples
t = (-3.67 - (-23.17))/17.81 = 0.62
Looking up the t-statistic in a t-table. The t-table will tell us the P-value associated with a t-statistic of 0.62.
The P-value is 0.156.
Hence, the Pvalue is between 0.1 and 0.2
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b) In a sample of 25 observations from a normal distribution with mean 98.6 and standard deviation 17.2 (10) (i) What is P(92
Given that there are 25 observations in a sample taken from a normal distribution having mean 98.6 and standard deviation 17.2.
We are to find [tex]P(92 < X < 100)[/tex], where X is the random variable with a normal distribution having mean 98.6 and standard deviation 17.2.
We know that the distribution is normal, and we have the mean and standard deviation. Let Z be the standard normal variable such that [tex]Z = (X - μ) / σ[/tex], where [tex]μ[/tex] and [tex]σ[/tex] are the mean and standard deviation of the normal distribution.
The probability that X lies between two given values a and b is the probability that Z lies between the corresponding standard scores.
(ii) To find [tex]P(X > 105)[/tex], we use the same standard normal variable Z.
[tex]P(X > 105) = P(Z > (105 - 98.6) / 17.2)= P(Z > 0.372)[/tex]. From standard normal tables, the area under the curve to the right of z = 0.372 is 0.3520.
P(X > 105) = 0.3520. This implies that the probability that the value of X is greater than 105 is 0.3520.
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You pick a card at random. 5 6 7 What is P(less than 7)? Write your answer as a fraction or whole number.
The probability of picking a card less than 7 from a standard deck of cards is 2/13.
To find the probability (P) of picking a card less than 7, we need to determine the number of favorable outcomes (cards less than 7) and divide it by the total number of possible outcomes.
In this case, the favorable outcomes are the cards with values less than 7, which are 5 and 6. The total number of possible outcomes is the total number of cards in the deck, which depends on the specific deck being used.
Assuming a standard deck of 52 cards, there are four 5s and four 6s, making a total of eight favorable outcomes.
Therefore, P(less than 7) = favorable outcomes / total outcomes = 8 / 52.
Simplifying the fraction, we find that P(less than 7) = 2 / 13.
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(4 points) Saab, a Swedish car manufacturer, is interested in estimating average monthly sales in the US, using the following sales figures from a sample of 5 months: 793, 565, 630, 649, 351 Using thi
Saab, a Swedish car manufacturer, is interested in estimating average monthly sales in the US, using the following sales figures from a sample of 5 months: 793, 565, 630, 649, and 351.
Saab can calculate the average monthly sales by adding up the sales figures for each month and dividing by the number of months in the sample. This is called the sample mean. The formula for the sample mean is: Sample mean = (sum of values) / (number of values)Using the sales figures given above, Saab can calculate the sample mean as follows: Sample mean = (793 + 565 + 630 + 649 + 351) / 5= 2988 / 5= 597.6Therefore, Saab can estimate the average monthly sales in the US to be $597.6. Saab, a Swedish car manufacturer, is trying to estimate the average monthly sales in the US. To do this, Saab uses sales figures from a sample of 5 months. The sample mean is a statistic that measures the central tendency of the data. It is calculated by summing up all the values in the sample and dividing by the number of values. In this case, the sample mean is calculated by adding up the sales figures for each month and dividing by the number of months in the sample, which is 5. Using this formula, Saab can estimate the average monthly sales in the US to be $597.6 based on the sales figures given above. The sample mean is a useful tool for estimating the population mean, which is the average sales figure for all months in the US. However, the sample mean may not always be an accurate estimate of the population mean. To obtain a more accurate estimate of the population mean, Saab would need to increase the sample size and use statistical techniques such as hypothesis testing and confidence intervals.
Saab can estimate the average monthly sales in the US to be $597.6 based on the sales figures from a sample of 5 months. The sample mean is a useful tool for estimating the population mean, but it may not always be an accurate estimate. Saab could obtain a more accurate estimate of the population mean by increasing the sample size and using statistical techniques such as hypothesis testing and confidence intervals.
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what is the probability that a randomly selected patient is 0-12 years old given that the patient suffers from knee pain?
To find the probability that a randomly selected patient is 0-12 years old given that the patient suffers from knee pain,
we need to use conditional probability.The conditional probability formula is:P(A|B) = P(A ∩ B) / P(B)where P(A|B) denotes the probability of A given that B has occurred. Here, A is "patient is 0-12 years old" and B is "patient suffers from knee pain".
Thus, the probability that a randomly selected patient is 0-12 years old given that the patient suffers from knee pain can be expressed as:P(0-12 years old | knee pain) = P(0-12 years old ∩ knee pain) / P(knee pain)The joint probability of 0-12 years old and knee pain can be busing the multiplication rule:P(0-12 years old ∩ knee pain) = P(0-12 years old) × P(knee pain|0-12 years old)where P(knee pain|0-12 years old) is the probability of knee pain given that the patient is 0-12 years old. We are not given this value, so we cannot calculate the joint probability.
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You measure 28 randomly selected textbooks' weights, and find
they have a mean weight of 70 ounces. Assume the population
standard deviation is 8.6 ounces. Based on this, construct a 90%
confidence in
The 90% confidence interval for the mean weight of the population of textbooks is [tex]$(68.08, 71.92)$[/tex] ounces.
To construct a 90% confidence interval for the mean weight of 28 randomly selected textbooks, we will use the following formula: [tex]$$\bar{X} \pm z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right)$$[/tex].
Where: [tex]$\bar{X}$[/tex] is the sample mean weight, [tex]$\sigma$[/tex] is the population standard deviation, [tex]$n$[/tex] is the sample size, [tex]$z_{\frac{\alpha}{2}}$[/tex] is the critical value obtained from the z-table or calculator, and [tex]$\alpha$[/tex] is the significance level which is equal to 1 - confidence level.
So, let's plug in the given values and solve:
Sample size, [tex]$n = 28$[/tex]
Sample mean weight, [tex]$\bar{X} = 70$[/tex]
Population standard deviation, [tex]$\sigma = 8.6$[/tex]
Confidence level, [tex]$C = 90\%$[/tex], which means [tex]$\alpha = 1 - C = 0.1$[/tex].
Therefore, the critical value from the z-table for a 90% confidence level is [tex]$z_{\frac{\alpha}{2}}= 1.645$[/tex].
Now, we can calculate the confidence interval:[tex]$$\begin{aligned}\bar{X} \pm z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) &= 70 \pm 1.645\left(\frac{8.6}{\sqrt{28}}\right) \\&= 70 \pm 1.915 \\&= (68.08, 71.92)\end{aligned}$$[/tex].
Therefore, the 90% confidence interval for the mean weight of the population of textbooks is [tex]$(68.08, 71.92)$[/tex] ounces.
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Find the missing value required to create a probability
distribution. Round to the nearest hundredth.
x / P(x)
0 / 0.18
1 / 0.11
2 / 0.13
3 / 4 / 0.12
The missing value to create a probability distribution is 0.46.
To find the missing value required to create a probability distribution, we need to add the probabilities and subtract from 1.
This is because the sum of all the probabilities in a probability distribution must be equal to 1.
Here is the given probability distribution:x / P(x)0 / 0.181 / 0.112 / 0.133 / 4 / 0.12
Let's add up the probabilities:
0.18 + 0.11 + 0.13 + 0.12 + P(4) = 1
Simplifying, we get:0.54 + P(4) = 1
Subtracting 0.54 from both sides, we get
:P(4) = 1 - 0.54P(4)
= 0.46
Therefore, the missing value to create a probability distribution is 0.46.
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Ryan is practicing his shot put throw. The path of the ball is given approximately by the function H(x) = -0.01x² + .66x +5.5, where H is measured in feet above the ground and is the horizontal dista
Answer:
The highest peak when x= 33 and the ball reaches a height of 16.39 ft
Step-by-step explanation:
The peak is obtained at x= -b/2a
x= -0.66 / 2(-0.01)
x = -0.66/(-0.02)
x= 33
The highest is reached when x=33 :
-0.01(33)^2 +0.66*33 +5.5 = 16.39 ft
The ball finally touches the ground again when -0.01x² + .66x +5.5 = 0
and the starting height is when x=0 => -0.01(0)^2 + 0.66(0) +5.5 = 5.5 ft
He started his throw at 5.5 ft
The normal monthly precipitation (in inches) for August is
listed for 20 different U.S. cities. Construct a boxplot for the
data set. Enter the maximum value.
3.7, 3.5, 4.4, 1.9, 2.8, 6.5, 1.5, 5.5, 2
The maximum value from the given data set is 6.5 A box plot is a pictorial representation of data that demonstrates the center, spread, and distribution of a data set, as well as any potential outliers.
It is made up of a rectangle that encompasses the interquartile range (IQR), which encompasses the middle 50% of the data, with whiskers extending to the minimum and maximum values or the maximum value that falls within a particular range.
To create a boxplot, first find the minimum, maximum, median, and quartiles of the data set. The quartiles divide the data into four equal parts, each containing 25% of the data.
The median is the middle value of the data set once it has been sorted in ascending or descending order. The interquartile range (IQR) is the difference between the upper and lower quartiles. It is also known as the middle 50% of the data. It's now time to construct the boxplot after obtaining the necessary statistics. A rectangular box is used to represent the IQR. The whiskers are represented by two lines extending out from the top and bottom of the box. The maximum and minimum values are represented by the whiskers. Any points that are further from the box than the whiskers are considered outliers. As we can see the maximum value from the given data set is 6.5, so that will be the answer.
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The Arnolds want to install a 3 ft tall circular pool with a 15 ft diameter in their rectangular patio. The patio will be surrounded
by new fencing, and the patio area surrounding the pool will be covered with new tiles.
How many square feet of ground will be covered in tiles? Round to the nearest tenth's place.
Suppose that a random variable X follows a binomial distribution with n=30, p=0.35. Which of the following is correct about the probability distribution? Select all that apply.
a) The SD is 0.087
b) The SD is 2.61
c) The distribution is skewed
d) The distribution is roughy asymmetric
e) The distribution is roughly symmetric
f) The expected value is 10.5
If a random variable X follows a binomial distribution with n=30, p=0.35.
a) The SD is 0.087
c) The distribution is skewed
f) The expected value is 10.5
The Correct options are a), c), and f).
For a binomial distribution with parameters n and p, the standard deviation (SD) is calculated as √(n * p * (1 - p)). In this case, the SD would be √(30 * 0.35 * (1 - 0.35)) ≈ 2.61. Therefore, option b) is incorrect.
Binomial distributions are generally skewed, especially when the probability of success (p) is far from 0.5. In this case, p = 0.35, so the distribution would be skewed. Thus, option c) is correct.
The expected value (mean) of a binomial distribution is given by n * p. Therefore, for n = 30 and p = 0.35, the expected value would be 30 * 0.35 = 10.5. Hence, option f) is correct.
Regarding the symmetry of the distribution, binomial distributions are typically roughly symmetric when p is close to 0.5. However, in this case, p = 0.35, which indicates some asymmetry. Thus, option e) is incorrect.
Therefore, the correct options are a), c), and f) for this probability distribution.
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i
need help
Score on last try: 0 of 3 pts. See Details for more. > Next question Get a similar question You can retry this question below Assume that a sample is used to estimate a population proportion p. Find t
The formula is t = (p - P) / (s / √n) where p is the sample proportion, P is the population proportion, s is the standard error of the sample proportion, and n is the sample size.
It is not possible to find t just by knowing that a sample is used to estimate a population proportion p.
The value of t depends on several factors such as the sample size, the level of confidence, and the standard error of the sample mean or proportion.
However, if you have information on these factors, then you can use the formula for the t-statistic to find t.
The formula is t = (p - P) / (s / √n) where p is the sample proportion, P is the population proportion, s is the standard error of the sample proportion, and n is the sample size.
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The area of a square can be represented by the expression x10 Which monomial represents a side of the square?
The monomial that represents a side of the square is 10x.
In the given expression, "x10" represents the area of a square. To find the monomial that represents a side of the square, we need to rewrite the expression in a form that represents the side length.
In the expression "x10," the variable "x" represents the unknown side length of the square, and the coefficient "10" represents the area. We know that the area of a square is equal to the side length squared (A = s^2), so we need to rearrange the expression to isolate the side length.
To do this, we divide both sides of the equation by 10 to get the side length alone. Dividing "x10" by 10 yields "x." Therefore, "x" represents the side length of the square.
In summary, the monomial that represents a side of the square is 10x. The coefficient "10" represents the area, and the variable "x" represents the side length.
The expression "x10" represents the area of a square. However, to find the monomial that represents the side length, we need to rearrange the expression to isolate the side length. By dividing both sides of the equation by 10, we obtain the side length alone, represented by the variable "x." This means that "x" is the monomial that represents a side of the square. By understanding the relationship between area and side length in a square, we can determine the correct monomial representation.
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online homework manager Course Messages Forums Calendar Gradebook Home > MAT120 43550 Spring2022 > Assessment Homework Week 7 Score: 14/32 11/16 answered X Question 6 < > Score on last try: 0 of 2 pts
Answer:.
Step-by-step explanation:
The approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61, using the empirical rule, is 81.5%.
The empirical rule, also known as the 68-95-99.7 rule, states that for a bell-shaped distribution (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.
In this case, the mean of the distribution is 49, and the standard deviation is 4. To find the percentage of 1-mile long roadways with potholes numbering between 41 and 61, we need to calculate the percentage of data within one standard deviation of the mean.
Since the range from 41 to 61 is within one standard deviation of the mean (49 ± 4), we can apply the empirical rule to estimate the percentage. According to the rule, approximately 68% of the data falls within this range.
Therefore, the approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61 is 68%.
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Complete Question:
online homework manager Course Messages Forums Calendar Gradebook Home > MAT120 43550 Spring2022 > Assessment Homework Week 7 Score: 14/32 11/16 answered X Question 6 < > Score on last try: 0 of 2 pts. See Details for more. > Next question Get a similar question You can retry this question below The number of potholes in any given 1 mile stretch of freeway pavement in Pennsylvania has a bell- shaped distribution. This distribution has a mean of 49 and a standard deviation of 4. Using the empirical rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61? Do not enter the percent symbol. ans = 81.5 % Question Help: Post to forum Calculator Submit Question ! % & 5 B tab caps Hock A N 2 W S X #3 E D C $ 54 R T G 6 Y H 7 00 * 8 J K 1
in the matrix equation below, what are the values of x and y? x = –1, y = 2 x = 3, y = 8 x = 5, y = –4 x = 5, y = 2
The value of x is 5 and value of y is 2 in the matrix equation.
From the matrix equation let us find the value of y:
1/2(8)-3(y+1)=-5
4-3y-3=-5
Isolate the variable y:
-3y=-5-1
-3y=-6
Divide both sides by -3:
y=2
Now let us solve for x:
1/2(x+3)-3(-1)=7
x/2+3/2+3=7
x/2+3/2=4
x/2=4-3/2
x/2=5/2
x=5
Hence, the value of x is 5 and value of y is 2.
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The values of x and y in the matrix equation below are x = 5, y = 2, the correct option is D.
We are given the matrix equation;
[tex]1/2\left[\begin{array}{ccc}4&8\\x+ 3&-4\end{array}\right] - 3\left[\begin{array}{ccc}1&y+ 1\\-1&-2\end{array}\right] = \left[\begin{array}{ccc}-1&-5\\7&4\end{array}\right][/tex]
Now, solving
[tex]\left[\begin{array}{ccc}2&4\\x+ 3/2&-2\end{array}\right] - \left[\begin{array}{ccc}3&3y+ 3\\-3&-6\end{array}\right] = \left[\begin{array}{ccc}-1&-5\\7&4\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}2-3&4-3y+ 3\\x+ 3/2 + 3&-2+6\end{array}\right] = \left[\begin{array}{ccc}-1&-5\\7&4\end{array}\right][/tex]
Two matrices are equal if their entries are equal.
1-3y = -5
⇒ -3y = -5-1
⇒ -3y = -6
y = 2
Thus, x+9 = 14
x = 14-9
x = 5
Therefore, by the equation answer will be x = 5, y = 2.
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Calculate partial derivatives of second order and mixed partial derivatives of f (x, y) = arctan (x + y/l - x y).
The partial derivatives of the second order and mixed partial derivatives of f(x, y) = arctan(x + y/l - x*y) are:
f_xx = -2y/(l^2*(1 + (x + y/l - xy)^2))
f_yy = -2x/(l^2(1 + (x + y/l - xy)^2))
f_xy = -1/(l^2(1 + (x + y/l - x*y)^2))
To find the partial derivatives of f(x, y), we differentiate the function with respect to each variable while holding the other variable constant.
Partial derivatives of the first order:
Differentiating f(x, y) with respect to x, we get:
f_x = (1 + y/l - y) / (1 + (x + y/l - x*y)^2)
Differentiating f(x, y) with respect to y, we get:
f_y = x/(l(1 + (x + y/l - x*y)^2))
Partial derivatives of second order:
Differentiating f_x with respect to x, we get:
f_xx = -2y/(l^2*(1 + (x + y/l - x*y)^2))
Differentiating f_y with respect to y, we get:
f_yy = -2x/(l^2*(1 + (x + y/l - x*y)^2))
Mixed partial derivative:
Differentiating f_x with respect to y, we get:
f_xy = -1/(l^2*(1 + (x + y/l - x*y)^2))
These formulas represent the partial derivatives of the second order and mixed partial derivatives of f(x, y) with respect to x and y.
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A doctor brings coins, which have a 50% chance of coming up "heads". In the last ten minutes of a session, he has all the patients flip the coins until the end of class and then ask them to report the numbers of heads they have during the time. Which of the following conditions for use of the binomial model is NOT satisfied?
a) fixed number of trials
b) each trial has two possible outcomes
c) all conditions are satisfied
d) the trials are independent
e) the probability of 'success' is same in each trial
The correct answer is (a) fixed number of trials because there is no fixed number of trials in this case.
The doctor has the patients flip the coins until the end of the session, and then asks them to report the number of heads they got. Which of the following conditions for using the binomial model is not satisfied?The doctor has coins with a 50% chance of coming up heads. The doctor has patients flip the coins until the end of the session. The patients will then report how many heads they got. Which of the following conditions for using the binomial model is not met?The condition that is not satisfied for the use of the binomial model is a fixed number of trials. Since there is no fixed number of trials, the doctor may have to flip the coins several times. It is essential that the number of trials is fixed so that the binomial model can be used properly.In a binomial experiment, there are a fixed number of trials, each trial has two possible outcomes, the trials are independent, and the probability of success is the same for each trial. If any of these conditions are not met, the binomial model cannot be used. Therefore, the correct answer is (a) fixed number of trials because there is no fixed number of trials in this case.
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the figure shows level curves of a function . (a) draw gradient vectors at and . is longer than, shorter than, or the same length as ?
The level curves represent the function. The gradient of the function at a point is perpendicular to the level curve through that point, and the direction of the gradient is the direction of the steepest ascent (maximum increase) of the function.
The gradient of the function points in the direction of the greatest rate of increase of the function. In this case, we have to draw gradient vectors at and .At , a gradient vector can be drawn perpendicular to the level curve, such that it points to the higher values of the function. The magnitude of the gradient vector can be determined by the rate of change of the function, which is given by the slope of the tangent line to the level curve at .
The gradient vector at is: At , a gradient vector can be drawn perpendicular to the level curve, such that it points to the higher values of the function. The magnitude of the gradient vector can be determined by the rate of change of the function, which is given by the slope of the tangent line to the level curve at .
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In a fish hatchery tank, there are 25 fish of which 5 are tagged. A sample of 6 fish is randomly selected from the tank. (a) What is the probability that exactly two of the selected fish are tagged? (
The probability that exactly two of the selected fish are tagged is 0.278
Probability of selecting 2 tagged fish and 4 non-tagged fishP(2T and 4NT) = (5C2 × 20C4) / 25C6= (10 × 4845) / 177100 = 0.2727
Probability of selecting 3 tagged fish and 3 non-tagged fishP(3T and 3NT) = (5C3 × 20C3) / 25C6= (10 × 1140) / 177100 = 0.0652
Probability of selecting 4 tagged fish and 2 non-tagged fishP(4T and 2NT) = (5C4 × 20C2) / 25C6= (5 × 190) / 177100 = 0.0054
Probability of selecting 5 tagged fish and 1 non-tagged fishP(5T and 1NT) = (5C5 × 20C1) / 25C6= (1 × 20) / 177100 = 0.0001
Probability of selecting 6 tagged fish and 0 non-tagged fishP(6T and 0NT) = (5C6 × 20C0) / 25C6= (1 × 1) / 177100 = 0.0000
Therefore, the probability that exactly two of the selected fish are tagged isP(2T) = 0.2727
We can summarise the above findings in a tabular form as shown below:No of tagged fish No of non-tagged fish Probability2 4 0.27273 3 0.06524 2 0.00545 1 0.00016 0 0.0000Total probability = Σ P(X) = 0.3439 ≈ 0.34
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3. Five number summary results of diastolic blood pressures (dbp) of a large group of adults are displayed below: L= = 60, Q₂ = 64, Q₂=75, Q₁ =79, H=84 What percentage of survey adults have dbp
The percentage of survey adults have dbp is 62.5%.
To determine what percentage of survey adults have diastolic blood pressure (dbp) using the given five-number summary results below:
L = 60Q₁ = 64Q₂ = 75Q₃ = 79H = 84
The five-number summary results above contain information about a dataset of diastolic blood pressures (dbp) of a large group of adults.
The minimum value, quartile 1 (Q₁), median (Q₂), quartile 3 (Q₃), and maximum value provide information about the spread and central tendency of the dataset.
To find the percentage of adults who have dbp, follow the steps below;
Step 1: Find the interquartile range (IQR).
IQR = Q₃ - Q₁
IQR = 79 - 64
IQR = 15
Step 2: Determine the lower and upper limits for outliers.
Lower limit = Q₁ - 1.5 (IQR)
Lower limit = 64 - 1.5(15)
Lower limit = 42.5
Upper limit = Q₃ + 1.5 (IQR)
Upper limit = 79 + 1.5(15)
Upper limit = 100.5
The limits show that any dbp values below 42.5 or above 100.5 are outliers and not part of the dataset.
Step 3: Determine the range of the values in the dataset.
Range = H - L
Range = 84 - 60
Range = 24
Step 4: Determine what percentage of survey adults have dbp between the limits.
Percent dbp = 100( Q₃ - Q₁) ÷ Range
Percent dbp = 100(79 - 64) ÷ 24
Percent dbp = 62.5%
Therefore, approximately 62.5% of survey adults have dbp within the lower and upper limits (between 64 and 79).
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Q. Let X have the following density function: fx(x) = { otherwise 3x5 for 0 < x < 2 • Find the çdf of X. Find P(X>1). Find E[X] Define a new random variable by setting Y=X2. Find the density of
The cumulative distribution function (cdf) of a random variable is defined as the probability that the random variable X takes on a value less than or equal to x. The cdf is therefore defined as F(x) = P(X ≤ x). Now, we have to find the cdf of X. Therefore, we must calculate the probability that X is less than or equal to some value x.
Using the density function, we can calculate the cdf of X as follows:
For 0 ≤ x < 2,
fx(x) = 3x5
So,F(x) = ∫0x3t5dt= x6.
For x < 0, F(x) = 0.
For x > 2,
F(x) = 1.
Finding P(X > 1)P(X > 1)
= 1 - P(X ≤ 1)
= 1 - F(1)
= 1 - (1/6)
= 5/6
Finding E[X]:
The expected value of X is given by:
E(X) = ∫(-∞)∞x.f(x)dx.
For 0 ≤ x < 2,
fx(x) = 3x5
∴ E(X) = ∫0∞x.3x5dx
= 3x716∣∣∣∣02
= 310
Thus, the çdf of X is F(x) = { 0 for x ≤ 0;
x6 for 0 < x < 2 ;
1 for x ≥ 2 }
The probability that X > 1 is 5/6. The expected value of X is E(X) = 310. The density function of Y is
fy(y) = 1√yfx(√y) = { otherwise 15y32 for 0 < y < 4, 0 elsewhere }.
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for each 7-subset of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}, give the next largest 7-subset or indicate that the 7-subset is the last one in lexicographic order.
To answer the question, we must list the subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} in lexicographic function order.
The first 7-subset is {1, 2, 3, 4, 5, 6, 7}. The next 7-subset is obtained by replacing the last element 7 by the smallest element greater than 7, which is 8. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 8}.The next 7-subset is obtained by replacing the last element 8 by the smallest element greater than 8, which is 9. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 9}.The next 7-subset is obtained by replacing the last element 9 by the smallest element greater than 9, which is 10. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 10}.
The next 7-subset is obtained by replacing the last element 10 by the smallest element greater than 10, which is 11. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 11}.The next 7-subset is obtained by replacing the last element 11 by the smallest element greater than 11, which is 12. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 12}.The next 7-subset is obtained by replacing the last element 12 by the smallest element greater than 12, which is 13. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 13}.The next 7-subset is obtained by replacing the last element 13 by the smallest element greater than 13, which is 14. Thus, the next largest 7-subset is {1, 2, 3, 4, 5, 6, 14}.Finally, the last 7-subset is {8, 9, 10, 11, 12, 13, 14}.
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when choosing a sample size for a population proportion a practitioner is considering either setting p* = 0.5, or, using a preliminary estimate of p-hat = 0.7. which is true?
When choosing a sample size for a population proportion, a practitioner may consider setting p* = 0.5 or using a preliminary estimate of p-hat = 0.7.
The practitioner must consider the amount of content loaded when choosing a sample size for a population.
The amount of content loaded when choosing a sample size for a population refers to the fact that the size of the sample that is chosen should be large enough to get meaningful results, but it should not be too large as to waste time, effort, and resources.
The sample size should be determined such that the confidence interval is not too wide, but at the same time, it should not be too narrow that the estimate is not reliable enough.
Preliminary estimates of the population parameter can be based on historical data, expert opinion, or even small samples.
Using a preliminary estimate of p-hat = 0.7 provides better results than using p* = 0.5 because it is closer to the population proportion.
By using a preliminary estimate of p-hat = 0.7, the sample size can be determined more accurately, and the results will be more reliable as compared to using a default value like p* = 0.5.In conclusion, using a preliminary estimate of p-hat = 0.7 is more appropriate than setting p* = 0.5.
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