Answer:
The particles collide when t = 7 at the point (49, 49, 49).
Step-by-step explanation:
We know the trajectories of the two particles,
[tex]r_1(t)=\langle t^2,16t-63,t^2\rangle\\r_2(t)=\langle 9t-14,t^2,13t-42\rangle[/tex]
To find if the tow particles collide you must:
Equate the x-components for each particle and solve for t[tex]t^2=9t-14\\t^2-9t+14=0\\\left(t^2-2t\right)+\left(-7t+14\right)=0\\t\left(t-2\right)-7\left(t-2\right)=0\\\left(t-2\right)\left(t-7\right)=0[/tex]
The solutions to the quadratic equation are:
[tex]t=2,\:t=7[/tex]
Equate the y-components for each particle and solve for t[tex]16t-63=t^2\\^2-16t+63=0\\\left(t^2-7t\right)+\left(-9t+63\right)=0\\t\left(t-7\right)-9\left(t-7\right)=0\\\left(t-7\right)\left(t-9\right)=0[/tex]
The solutions to the quadratic equation are:
[tex]t=7,\:t=9[/tex]
Equate the z-components for each particle and solve for t[tex]t^2=13t-42\\t^2-13t+42=0\\\left(t^2-6t\right)+\left(-7t+42\right)=0\\t\left(t-6\right)-7\left(t-6\right)=0\\\left(t-6\right)\left(t-7\right)=0[/tex]
The solutions to the quadratic equation are:
[tex]t=6,\:t=7[/tex]
Evaluate the position vectors at the common time. The common solution is when t = 7.
[tex]r_1(7)=\langle 7^2,16(7)-63,7^2\rangle=\langle 49,49,49\rangle\\\\r_2(7)=\langle 9(7)-14,7^2,13(7)-42\rangle=\langle 49,49,49\rangle[/tex]
For two particles to collide, they must be at exactly the same coordinates at exactly the same time.
The particles collide when t = 7 at the point (49, 49, 49).
The consumption rate of electricity, in kW/h has increased, on average, at 3.5% per year. If it continues to increase at this rate indefinitely, find the number of years before the electric companies will need to double their generating capacity.
The electric utities will need to double their generating capacity in about _____ years.
Answer:
20.14
Step-by-step explanation:
Given that the consumption rate of electricity, in kW/h has increased, on average, at 3.5% per year.
So if initial consumption is C after one year it would be
[tex](1+0.035)C[/tex] and after 2 years [tex](1+0.035)^2C[/tex]
and so on
In general after n years
[tex]C(n) = 1.035^n C\\[/tex]
When C(n) is twice C we have
[tex]2=1.035^n\\log 2= n log 1.035\\n = 20.14[/tex]
Between 20 and 21 years
Hence The electric utities will need to double their generating capacity in about ___20.14__ years.
Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 27 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was four years with a standard deviation of 1.9 years. Suppose that it is somehow known that the population standard deviation is 1.4. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Since both ơ and sx are given, which should be used?
O σ
O sx
In one to two complete sentences, explain why?
Answer:
sigma should be used
Step-by-step explanation:
Given that The mean length of time in jail from the survey was four years with a standard deviation of 1.9 years.
The above given is for sample of 27 size.
For hypothesis test to compare mean of sample with population we can use either population std dev or sample std dev.
But once population std deviation is given, we use only that as that would be more reliable.
So here we can use population std deviation 1.4 only.
If population std deviation is used we can use normality and do Z test
The length of timber cuts are normally distributed with a mean of 95 inches and a standard deviation of 0.52 inches. In a random sample of 30 boards, what is the probability that the mean of the sample will be between 94.8 inches and 95.8 inches? Homework Help:
To find the probability, use the Central Limit Theorem to calculate the z-scores for the lower and upper limits of the sample mean. Look up these z-scores in the standard normal distribution table to find the probabilities. Subtract the probability for the lower z-score from the probability for the higher z-score to find the probability that the mean of the sample falls between the two values.
Explanation:To find the probability that the mean of a sample will be between 94.8 inches and 95.8 inches, we can use the Central Limit Theorem. We start by calculating the z-scores for these values using the formula: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. With the given values, the z-score for 94.8 inches is -3.85 and the z-score for 95.8 inches is 1.92.
Next, we look up these z-scores in the standard normal distribution table to find the corresponding probabilities. The probability for a z-score of -3.85 is approximately 0.00005 and the probability for a z-score of 1.92 is approximately 0.97128. To find the probability that the mean of the sample falls between these two values, we subtract the probability for the lower z-score from the probability for the higher z-score: 0.97128 - 0.00005 = 0.97123.
Therefore, the probability that the mean of the sample will be between 94.8 inches and 95.8 inches is approximately 0.97123, or 97.12%.
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A survey questioned 1000 people regarding raising the legal drinking age from 18 to 21. Of the 540 who favored raising the age, 390 were female. Of the 460 opposition responses, 130 were female. If a person selected at random from this group is a man, what is the probability that the person favors raising the drinking age?
Final answer:
The probability that a randomly selected man from the survey group favors raising the legal drinking age is 5/16, or 0.3125.
Explanation:
The question asks for the probability that a man chosen at random from the survey group favors raising the legal drinking age from 18 to 21. To solve this, we need to look at the numbers provided. We know that of the 540 people who favored raising the age, 390 were female. This means that 540 - 390 = 150 were male. Similarly, from the 460 who opposed, 130 were female, leading to 460 - 130 = 330 males who opposed.
Now, we calculate the total number of men in the survey, which is 150 men who favored raising the age plus 330 men who opposed, giving us a total of 150 + 330 = 480 men. The probability that a randomly selected man favors raising the age is then the number of men who favor it divided by the total number of men, which is 150/480.
To further simplify, we divide both numerator and denominator by 30, yielding an answer of 5/16. Therefore, the probability that a man favors raising the drinking age is 5/16, or approximately 0.3125.
Translating mathematical statements in English into logical expressions. info About Consider the following statements in English. Write a logical expression with the same meaning. The domain is the set of all real numbers. (a) There is a number whose cube is equal to 2. (b) The square of every number is at least 0. (c) There is a number that is equal to its square. (d) Every number is less than or equal to its square.
Answer:
See below
Step-by-step explanation:
(a) There is a number whose cube is equal to 2.
[tex]\large \exists x \in \mathbb{R}\;|\;x^3=2[/tex]
(b) The square of every number is at least 0.
[tex]\large \forall x\in \mathbb{R},\;x^2\geq 0[/tex]
(c) There is a number that is equal to its square.
[tex]\large \exists x \in \mathbb{R}\;|\;x^2=x[/tex]
(d) Every number is less than or equal to its square.
[tex]\large \forall x\in \mathbb{R},\;x\leq x^2[/tex]
(By the way, this last statement is not true when 0 < x < 1)
Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 7, and the y-axis, oriented counterclockwise starting from the origin. Label the edges of the boundary as C_1, C_2, C_3 starting from the bottom edge going counterclockwise. Give each edge a constant speed parametrization with domain 0 lessthanorequalto t lessthanorequalto 1: edge C_1 x_1(t) = y_1(t) = edge C_2 x_2(t) = y_2(t) = edge C_3 x_3(t) = y_3 = integral_C y^2xdx + x^2ydy = integral_C1 y^2xdx + x^2ydy + integral_C2 y^2xdx + x^2ydy + integral_C3 y^2xdx + x^2ydy 10pt = Applying Green's theorem, integral_C y^2 xdx + x^2 ydy = dxdy The vector field F = y^2x i + x^2y j is:
The question asks for the parametrization of boundary edges of a region in the 1st quadrant and the computation of a line integral over it. The region is bounded by the x-axis, quarter-circle of radius 7, and the y-axis. This is done by parameterizing each edge and applying Green's theorem to calculate the related integral.
Explanation:This exercise is essentially setting up and evaluating a line integral over closed path, which tells you about the interaction between a vector field and a curve in its domain. The region in question is in the first quadrant, and is bounded by the x-axis (C_1), a quarter circle (arc) of radius 7 (C_2), and the y-axis (C_3). The oriented counterclockwise is a convention means going from the origin along the x-axis, then following the quarter-circle (arc) around, and then starting back along the y-axis towards the origin.
The parametrization of the constant-speed edge C_1 is x_1(t)=7t, y_1(t)=0, for 0 <= t <= 1. For edge C_2 is x_2(t)=7cos(π/2t), y_2(t)=7sin(π/2t), for 0 <= t <= 1 and for C_3 is x_3(t)=0, y_3(t)=7(1-t), for 0 <= t <= 1. As the integral of a scalar function over C (the quarter-circle location) is equal to the sum of integrals over the three parts (C1, C2, C3), we can break it into three simpler parts and integrate separately.
The Vector Field F related to the integral can be represented by F = y^2xi + x^2yj based on the given equation. To find the resulting integral via Green's theorem, you would just take the divergence of the field F, yielding a double integral over the region.
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You are getting ready for a family vacation. You decide to download as many movies as possible before leaving for a road trip. If each movie takes 1 1/6 hours to download, and you download for 6 1/5 hours, how many movies did you download?
An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.3 years of the population mean. Assume the population of ages is normally distributed.(a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.5 years.
Answer: 4
Step-by-step explanation:
For given Population standard deviation[tex](\sigma)[/tex] , the formula to sample size is given by :-
[tex]n=(\dfrac{\sigma\cdot z*}{E})^2[/tex]
, where z* = Two-tailed critical value.
E = Margin of error.
Given : [tex]\sigma=1.5[/tex] years
We know that , Critical value for 90% confidence interval : z* = 1.645
Margin of error : E=1.3 years
Then, the minimum sample size required to construct a 90% confidence interval for the population mean will be :-
[tex]n=(\dfrac{1.5\cdot 1.645}{1.3})^2\\\\=(1.89807692308)^2\\\\=3.60269600593\approx4[/tex] [Rounded to the nearest whole number.]
Hence, the minimum sample size required = 4
The minimum sample size required to construct a 90% confidence interval for the population mean age of students at a college, given a population standard deviation of 1.5 years and desired margin of error of 1.3 years, is 3 students.
Explanation:The admissions director is trying to estimate the mean age of all students enrolled at the college within 1.3 years of the actual population mean with a 90% confidence level. In statistical terms, this means constructing a 90% confidence interval for the population mean within a margin of error of 1.3 years.
The formula to compute sample size for a confidence interval when the population standard deviation is known is:
n = (Zσ/E)^2
where:
n is the sample sizeZ is the z-score for the desired confidence level (for 90% confidence level, Z = 1.645)σ is the population standard deviation (here, 1.5 years)E is the desired margin of error (here, 1.3 years)
Substituting the given values into the formula, we have:
n = (1.645*1.5/1.3)^2
Solving for n, we get approximately 2.41. Because we can't survey a fraction of a student, we round up to the nearest whole number. Thus, the minimum sample size required is 3 students.
In practice, especially for a large population, this sample size might be too small, but based on the strict parameters established (90% confidence level, 1.3 years margin of error, and 1.5 years population standard deviation), this is the minimum sample size required.
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An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 163 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.)
Answer:
(0.4062, 0.5098)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex]
For this problem, we have that:
356 dies were examined by an inspection probe and 163 of these passed the probe. This means that [tex]n = 365[/tex] and [tex]\pi = \frac{163}{356} = 0.458[/tex]
Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.
So [tex]\alpha[/tex] = 0.05, z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[tex], so [tex]z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 - 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.4062[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 + 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.5098[/tex]
The correct answer is
(0.4062, 0.5098)
Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.
Write me here and I will give you my phone number - *pofsex.com*
My nickname - Lovely
Easy question to get points:
10 * 11
Answer:
110
lol
Step-by-step explanation:
The function y=14.99+1.25x represents the price ( y ) for a pizza with ( x ) toppings. Which is not a reasonable value for this function?
A- 14.99
B- 17.49
C- 18.25
D- not here
Answer:
C
Step-by-step explanation:
because if you input 18.25 into the equation and solve you get and non-whole number of 2.608.
It is impossible to have 2.608 toppings on your pizza, it has to be a whole number.
All the other answers had whole numbers besides C, therefore C is wrong.
The given function y = 14.99 + 1.25x represents the price of a pizza with a given number of toppings. Option D, 'not here,' is not a reasonable value for the function.
Explanation:The given function is y = 14.99 + 1.25x, where y represents the price of a pizza and x represents the number of toppings. To find the optimal value for this function, we need to substitute the values for x and compute y. However, the given options provide specific values of y. Since the function is a linear equation, any real number can be a valid output for the function, including negative values and values that do not correspond to actual pizza prices.
Based on this, option D, 'not here,' is not a reasonable value for the function as it does not provide a specific numerical output. The remaining options, A, B, and C, are all reasonable values for the function, depending on the specific number of toppings (x) chosen.
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An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was = 119.6 ounces. Suppose the standard deviation is known to be σ = 6.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean μ. Based on the 25 recent birth records, the sampling distribution of the sample mean can be represented by:
A. N(μ, 6.5).
B. N(μ, 1.30).
C. N(119.6, 1.30).
D. N(119.6, 6.5).
Answer:
B. N(μ, 1.30).
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was = 119.6 ounces. Suppose the standard deviation is known to be σ = 6.5 ounces.
Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean μ.
This means that for the sampling distribution, the mean is the mean of the weight of all babies born, so [tex]\mu[/tex] and [tex]s = \frac{6.5}{\sqrt{25}} = 1.30[/tex].
So the correct answer is
B. N(μ, 1.30).
The correct representation of the sampling distribution of the sample mean based on the data from birth records is option C, which is N(119.6, 1.30). This is based on the formula of sampling distribution for mean with known standard deviation.
Explanation:The question is asking about the sampling distribution of the sample mean which is a statistical concept used in inferential statistics. The sample distribution of the mean is normally distributed, denoted as N(μ, σ/n0.5) where μ is the population mean, σ is the population standard deviation, and n is the sample size (SRS - Simple Random Sample).
Based on the data given, the sample mean after studying the 25 birth records is 119.6 ounces and the standard deviation is known to be 6.5 ounces. So, the standard deviation of the sample mean, often termed as the standard error, would be σ/√n = 6.5/√25 = 1.3 ounces. Hence the correct sampling distribution of the sample mean would be represented by N(119.6, 1.30), which is option C in your question.
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The lengths of nails produced in a factory are normally distributed with a mean of 6.13 centimeters and a standard deviation of 0.06 centimeters.Find the two lengths that separate the top 7% and the bottom 7%.These lengths could serve as limits used to identify which nails should be rejected.Round your answer to the nearest hundredth, if necessary.
Answer:
Bottom 7%: L= 6.04 cm
Top 7%: L= 6.22 cm
Step-by-step explanation:
Mean length of the population (μ) = 6.13 cm
Standard deviation (σ) = 0.06 cm
The z-score for any given length 'X' is:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
What we want to know is the length at the 7-th percentile and at the 93-rd percentile.
According to a z-score table, the 7-th percentile has a correspondent z-score of -1.476 and the 93-rd percentile has a z-score of 1.476. Therefore, the bottom 7% and top 7% are separated by the following lengths:
[tex]z(X_B)=\frac{X_B-\mu}{\sigma}\\-1.476=\frac{X_B-6.13}{0.06}\\X_B = 6.04\\z(X_T)=\frac{X_T-\mu}{\sigma}\\1.476=\frac{X_T-6.13}{0.06}\\X_T = 6.22[/tex]
Bottom 7%: L= 6.04 cm.
Top 7%: L= 6.22 cm.
To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-scores can be calculated using the inverse normal distribution function, and then the lengths can be found using the formula x = μ + zσ. The lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.
Explanation:To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-score represents how many standard deviations a value is from the mean. For the top 7%, we can find the z-score by using the formula: z = invNorm(1 - 0.07), where invNorm is the inverse normal distribution function. Similarly, for the bottom 7%, the z-score can be calculated using the formula: z = invNorm(0.07). Once we have the z-scores, we can use the formula x = μ + zσ, where x is the length of the nails, μ is the mean length (6.13 cm), z is the z-score, and σ is the standard deviation (0.06 cm).
Using a calculator or software, we can find the z-scores:
For the top 7%: z = invNorm(1 - 0.07) = invNorm(0.93) ≈ 1.48For the bottom 7%: z = invNorm(0.07) ≈ -1.48Substituting the values into the x = μ + zσ formula:
For the top 7%: x = 6.13 + 1.48 * 0.06 ≈ 6.22 cmFor the bottom 7%: x = 6.13 - 1.48 * 0.06 ≈ 6.04 cmTherefore, the two lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.
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Te professor of a large calculus class randomly selected 6 students and asked the amount of time (in hours) spent for his course per week. Te data are given below. 10 8 9 7 11 13
a. Estimate the mean of the time spent in a week for this course by the students who are taking this course.
b. Estimate the standard deviation of the time spent in a week for this course by the students who are taking this course.
c. Estimate the standard error of the estimated mean time spent in a week for this course by the students who are taking this course.
Answer:
a. μ = 9.667 hours
b. σ = 1.972 hours
c. SE = 0.805 hours
Step-by-step explanation:
Sample size (n) = 6
Sample data (xi) = 10, 8, 9, 7, 11, 13
a. Mean time spent in a week for this course by students:
Sample mean is given by:
[tex]\mu = \frac{\sum x_i}{n} \\\mu = \frac{10+9+7+11+13}{6}\\\mu=9.667[/tex]
Mean time spent in a week per student is 9.667 hours
b. Standard deviation of the time spent in a week for this course by students:
Standard deviation is given by:
[tex]\sigma = \sqrt{\frac{\sum(x_i - \mu)^2}{n}}\\\sigma = \sqrt{\frac{(10- 9.667)^2+(8- 9.667)^2+(9- 9.667)^2+(7- 9.667)^2+(11- 9.667)^2+(13- 9.667)^2}{6}}\\\sigma =1.972[/tex]
c. Standard error of the estimated mean time spent in a week for this course by students:
Standard error is given by:
[tex]SE = \frac{\sigma}{\sqrt n}\\SE = \frac{1.972}{\sqrt 6}\\SE=0.805[/tex]
In humans wavy hair(W) is dominant over straight(w). Dark hair(D) dominates red(d).A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits).What is the probability of having a wavy red haired child ?
Answer:
1/4
Step-by-step explanation:
Parents WwDd wwdd
Gametes WD Wd wD wd wd wd wd wd
Offspring possibilities: WwDd WwDd WwDd WwDd
Wwdd Wwdd Wwdd Wwdd
wwDd wwDd wwDd wwDd
wwdd wwdd wwdd wwdd
P (wavy red hair) =4/16 = 1/4
Answer:
The probability is one half (1/2)
But it seems to be a mistake in the statement .."married a wavy red haired woman(homozygous recessive for both traits)". If the woman has wavy hair, the trait can't be homozygous recessive, although she certainly is homozygous recessive for the second trait because is red hair.
Step-by-step explanation:
As wavy hair is dominant over straight hair, it means only one allele is necessary to be visible: WW (homozygous), and Ww (heterozygous) produce the wavy hair trait. An homozygous recessive (ww) doesn't show the wavy trait character; he or she would have straight hair
The woman, if she's homozygous and wavy hair, would be (WW). As she is homozygous for red hair, she has red hair.
If the statement "A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits)" is correct, meaning that she is not wavy hair but "STRAIGHT", the probability of having a wavy red hair changes, being only 1/4
Calculation can be performed with Punnett squares
For example for the wavy dark hair male (heterozygous for both traits), the genotype is WwDd and the possible allele combinations would be: WD, Wd, wD, and wd
For the woman, if she's wavy red hair (homozygous), the genotype would be WWdd and the possible allele combinations would be only Wd.
Then you need cross the male allele combinations against Wd
If the woman is straight red hair (homozygous for both traits), the genotype would be wwdd and the possible allele combinations would be only wd.
Then you need cross the male allele combinations against wd, and obtain the proportions produced from the cross
An example of Punnett square below
A non-profit organization provides homeless children with backpacks filled with school supplies. For the last 5 years the organization has given out 278, 310, 320, 242, and 303 backpacks filled with school supplies. What is the average number of backpacks given out in the last five years?
Answer:
290.6 backpacks per year
Step-by-step explanation:
The average of 5 numbers is 1/5 of their sum:
average = (278 +310 +320 +242 +303)/5 = 290.6
The average number of backpacks given out in the last 5 years is 290.6, about 291.
In the US court system, a defendant is assumed innocent until proven guilty. Suppose that you regard a court system as a hypothesis test with these null and alternative hypotheses: H0: Defendant is innocent Ha: Defendant is guilty There are 2 possible decisions regarding H0 and 2 possible truths as to the innocence or guilt of the defendant, making 4 possible combinations. What are those 4 combinations? Identify the two correct decisions.
Answer:
Step-by-step explanation:
A test hypothesis by definition is a test based on probabilities, therefore it always be possible to make errors when decision is made. According to that we always have 4 possibilities, two right and to wrong (4 possiblities)
For instance in our particular case
H₀ = Defendant is innocent Hₐ = defendant is guilty
If we arrive to conclusion that H₀ is right, and he is really innocent we took a correct decision, And the result will be correct; if we take the decision of reject H₀ when the defendant is guilty again we took the right decision. These are the two correct decision in that case.
On the other hand what happens if we take the decision of rejecting H₀ (accepting Hₐ ) and the defendant is innocent, we are sending the defendant to jail and he is innocente (we are making I type error) and the defendant will pay for it. Finally if we accept H₀ and this decision is not right we will make the defendant be free and he is really guilty
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5).
Answer:
[tex]\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}[/tex]
Step-by-step explanation:
The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.
Let
[tex]C_1,C_2[/tex]
be the two paths.
Recall that if we parametrize a path C as [tex](r_1(t),r_2(t),r_3(t))[/tex] with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is
[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt[/tex]
Given any two points P, Q we can parametrize the line segment from P to Q as
r(t) = tQ + (1-t)P with 0≤ t≤ 1
The parametrization of the line segment from (1,1,1) to (2,2,2) is
r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)
r'(t) = (1,1,1)
and
[tex]\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}[/tex]
The parametrization of the line segment from (2,2,2) to
(-9,6,5) is
r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)
r'(t) = (-11,4,3)
and
[tex]\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}[/tex]
Hence
[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]
The line integral with respect to the arc length of the function is mathematically given as
= −1080.97
What is the line integral with respect to the arc length of the function?Question Parameter(s):
The line integral with respect to arc length of the function f(x, y, z) = xy2
the line segment from (1, 1, 1) to (2, 2, 2)
the line segment from (2, 2, 2) to (−9, 6, 5).
Generally, the equation for the line integral is mathematically given as
[tex]\int_C f(x,y,z)ds= \int_a^ b f(r_1,r_2,r_3) *\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt[/tex]
Therefore
[tex]\int_{C_1}f(x,y,z)ds=int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt\\\\\int_{C_1}f(x,y,z)ds=\sqrt{3} * \int_{0}^{1}(1+t)(1+t)^2dt\\\\[/tex]
[tex]\int_{C_1}f(x,y,z)ds=\sqrt{3} \int_{0}^{1}(1+t)^3dt\\\\\int_{C_1}f(x,y,z)ds=6.495[/tex]
r(t) = t(-9,6,5) + (1-t)(2,2,2)
r(t) = (-11,4,3)
then,
[tex]\displaystyle\int_{C_2}f(x,y,z)ds\\=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt\\\\=-90\sqrt{146}[/tex]
thus,
[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\={\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]
In conclusion, considring the (2,2,2) segment
[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\={\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]
= −1080.97
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The strength of a certain type of rubber is tested by subjecting pieces of the rubber to an abrasion test. For the rubber to be acceptable, the mean weight loss μ must be less than 3.5 mg. A large number of pieces of rubber that were cured in a certain way were subject to the abrasion test. A 95% upper confidence bound for the mean weight loss was computed from these data to be 3.45 mg. Someone suggests using these data to test H0 : μ ≥ 3.5 versus H1 : μ < 3.5. It is discovered that the mean of the sample used to compute the confidence bound is X⎯⎯⎯ = 3.40. Is it possible to determine whether P < 0.01? Explain. Round the test statistic to two decimal places and the answer to four decimal places.
Answer:
Step-by-step explanation:
Hello!
You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.
With the given information you must test whether the researcher's hypothesis is sustained or not.
The study variable is,
X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)
There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.
Symbolically the statistic hypothesis is:
H₀: μ ≥ 3.5
H₁: μ < 3.5
α: 0.05 (since is not listed, I'll choose one of the most common signification levels)
You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)
The formula of the statistic is:
Z= X[bar] - μ ≈ N(0;1)
δ/√n
To calculate the statistic you have to use the information given.
The sample mean X[bar]= 3.4 mg
Upper bond of 95% CI= 3.45 mg
The basic structure of a CI for the mean is
"estimator" ± "margin of error"
Upper bound is "estimator" + "margin of error"
Using the formula:
Ub= X[bar] + d ⇒ 3.45= 3.4 + d
⇒ d= 3.45 - 3.4 = 0.05
Where d is the margin of error
d= [tex]Z_{1-\alpha /2}[/tex] * (δ/√n)
d= [tex]Z_{0.975}[/tex] * (δ/√n)
d/[tex]Z_{0.975}[/tex]= (δ/√n)
(δ/√n)= 0.05/ 1.96 = 0.0255
(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.
Now you have all values and can calculate the statistic under the null hypothesis:
Z= 3.4 - 3.5 = -3.92
0.0255
And the p-value:
P(Z ≤ -3.92) = 0.000044 ⇒ My Z- table goes up to P(Z ≤ -3.00) = 0.001, so using strictly the table I can say that the probability is less than 0.001.
To calculate the exact probability I've used a statistic program.
p-value < 0.001
I hope it helps!
A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Included in the sample were: 7 employees from Area A, 9 employees from Area B, and 12 employees from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?
There is a system glitch during the process of adding my answer here so i create a document file for the answer, you find it in the attachment below.
What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?
Probabilities are used to determine how likely, or often an event is, to happen. The probability that the selected tire fails before 35000-mile warranty is 0.11702
From the complete question, we have:
[tex]n = 41[/tex] --- number of tires
[tex]Mileage: 33095\ 34589\ 39411\ 42386\ 37886\ 33096\ 44185\ 38273\ 42387\ 36117[/tex]
[tex]44373\ 39896\ 42758\ 34028\ 39768\ 44392\ 35826\ 44945\ 41756\ 41087[/tex]
[tex]43716\ 33478\ 41430\ 39397\ 39517\ 38068\ 42216\ 43447\ 33372\ 42631[/tex]
[tex]42215\ 44367\ 33186\ 41567\ 38534\ 33873\ 43484\ 39761\ 35531\ 40926\ 38348[/tex]
First, we calculate the mean
[tex]\mu = \frac{\sum x}{n}[/tex]
This gives:
[tex]\mu = \frac{33095+ 34589 +.............+40926 +38348}{41}[/tex]
[tex]\mu = \frac{1619318}{41}[/tex]
[tex]\mu = 39496[/tex]
Next, calculate the standard deviation
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]
This gives:
[tex]\sigma = \sqrt{\frac{(33095 - 39496)^2 + (34589 - 39496)^2 +.......+ (40926 - 39496)^2 + (38348- 39496)^2}{41-1}}[/tex]
[tex]\sigma = \sqrt{\frac{572531448}{40}}[/tex]
[tex]\sigma = \sqrt{14313286.2}[/tex]
[tex]\sigma = 3783[/tex]
The probability a tire will fail before 35000 is represented as:
[tex]P(x < 35000)[/tex]
Calculate the z score
[tex]z = \frac{x - \mu}{\sigma}[/tex]
This gives
[tex]z = \frac{35000 - 39496}{3783}[/tex]
[tex]z = \frac{-4496}{3783}[/tex]
[tex]z = -1.19[/tex]
So, we have:
[tex]P(x < 35000) = P(z < -1.19)[/tex]
From z table of values:
[tex]P(z < -1.19) = 0.11702[/tex]
Hence:
[tex]P(x < 35000) = 0.11702[/tex]
So, the probability that the selected tire fails before 35000-mile warranty is 0.11702
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The probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.
Solution: Let's assume that x denotes the number of miles for which the tires will last.
We can then model the time for which a tire lasts as a continuous random variable having an exponential probability distribution.
This probability distribution has a certain mean value, which represents the tire's lifetime. On average, the lifetime of a tire is 44,000 miles.
This implies that the tire's decay parameter, lambda (λ), is given by: λ=1/44000Therefore, the probability that the tire will last less than 35,000 miles can be calculated by integrating the probability density function from 0 to 35,000,
which is given by:[tex]f(x)=λe^(-λx)[/tex]Here's the calculation:[tex]:$$\begin{aligned} P(X \le 35,000) &= \int_0^{35,000} f(x) dx \\ &= \int_0^{35,000} \lambda e^{-\lambda x} dx \\ &= \left[-e^{-\lambda x}\right]_0^{35,000} \\ &= -e^{-\lambda 35,000} + e^{-\lambda 0} \\ &= -e^{(-1/44,000)\cdot 35,000} + e^{(-1/44,000)\cdot 0} \\ &= -e^{-0.795} + e^{0} \\ &= 0.3184 + 1 \\ &= 1.3184 \end{aligned} $$[/tex]Note that the probability density function is always positive, so the negative result in the second step can be ignored.
As a result, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is: P(X ≤ 35,000) = 0.3184 - 1 = -0.6816
The probability of failure is never negative, so there must be an error somewhere in the calculation.
Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412, which is the complement of P(X > 35,000):P(X > 35,000) = 1 - P(X ≤ 35,000) = 1 - 0.3184 = 0.6816P(X ≤ 35,000) = 0.3184P(X < 35,000) = 1 - P(X > 35,000) = 1 - 0.6816 = 0.3184P(X < 35,000) = 0.3184
Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.
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The probable question is :-
Question 5 10 pts What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?
O 0.0500
O 0.0412
O 0,09218
O 0.0885
In a experiment on relaxation techniques, subject's brain signals were measured before and after the relaxation exercises with the following results:
Person 1 2 3 4 5
Before 32 38 66 49 29
After 26 36 59 52 24
Assuming the population is normally distributed, is there sufficient evidence to suggest that the relaxation exercise slowed the brain waves? (Use α=0.05)
Answer:
There is no sufficient evidence to suggest that the relaxation exercise slowed the brain waves
Step-by-step explanation:
Given that in a experiment on relaxation techniques, subject's brain signals were measured before and after the relaxation exercises with the following results:
The population is normally distributed
This is a paired t test since sample size is very small and same population under two different conditions studied.
[tex]H_0: \bar x-\bar y =0\\H_a: \bar x >\bar y[/tex]
(Right tailed test)
Alpha = 5%
x y Diff
32 26 6
38 36 2
66 59 7
49 52 -3
29 24 5
Mean 3.4
Std dev 4.037325848
Mean difference = [tex]3.40[/tex]
n =5
Std deviation for difference = [tex]4.037326[/tex]
Test statistic t = mean difference / std dev for difference
= [tex]1.883[/tex]
p value =0.132
Since p >0.05 we accept null hypothesis.
There is no sufficient evidence to suggest that the relaxation exercise slowed the brain waves
Paired-sample t-test suggests there's a significant effect on slowing down the brain waves due to the relaxation exercises.
Explanation:The subject of this question involves conducting a paired-sample t-test in a study on relaxation techniques. The t-test will allow us to determine whether there is a significant difference between the brain signals of subjects before and after engaging in relaxation exercises.
First, we compute the paired differences (Before - After) and calculate the mean and standard deviation of this list.
The differences are: 6, 2, 7, -3, 5. Hence, the mean difference (µd) = 3.4 and standard deviation (Sd) = 3.346.
Now, with t = (µd / (Sd/√n)) = (3.4 / (3.346/√5)) = 3.023 which is the t-value and n-1 = 4 (degrees of freedom).
Check a t-value chart for a two-tailed test (since we do not know if the relaxation exercises will increase or decrease brain activity) with degree of freedom = 4 and α=0.05. If our t-value is beyond the critical value in the table, we have a significant result. If it falls within that range, we do not. In this case, it's beyond so we can conclude the relaxation exercises have a significant effect on slowing down the brain waves.
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An ice cream shop makes $1.25 on each small cone and $2.25 on each large cone. The ice cream shop sells between 60 and 80 small cones and between 120 and 150 large cones in one day. If it costs the owners $350 per day to run the shop, how many of each cone do they need to sell to make a profit each day?
Answer:
Let x and y represent the numbers of small and large cones sold daily, respectively60 ≤ x ≤ 80120 ≤ y ≤ 1501.25x +2.25y > 350Step-by-step explanation:
To answer the question, we need to be able to find the number of each type of cone that needs to be sold. For the purpose, it is convenient to define variables that represent those numbers. We have chosen to use the variables x and y to represent the number of small cones and the number of large cones, respectively ("Let" statement in above answer). One could use "s" and "l" as being more mnemonic, but "l" can be confused with a variety of other symbols, so we like not to use it.
The problem statement puts limits on the numbers of small and large cones sold, so our system of inequalities needs to reflect those limits. (We suspect these are not hard limits, but represent historical data. We doubt the shop would decline to sell more, and we can conceive of conditions under which they might sell fewer.)
The heart of the matter is that the profit from each type of cone must total more than $350 (per day). The problem statement requires the shop make a profit, not just break even, so we use the > symbol for profit, instead of ≥.
___
See above for the "Let" statement and the system of inequalities. (This problem does not require we solve them.)
Help help !! Please
Answer:
mean = 70000
SD = 15239
X= 95000
Z = (X-mean)/ SD
= (95000-70000)/15239
= 1.64
Now from Z-Table
% of employees = 100(1-.9495) = 5.02%
You have a standard deck of 52 cards (i.e., 4 aces, 4 twos, 4 threes, …, 4 tens, 4 jacks, 4 queens, and 4 kings) that contains 4 suits (hearts, clubs, spades, and diamonds). We draw one card from the deck. What is the probability that the card is NEITHER a face card (jack, king, or queen) NOR a heart? 27/52
Answer:
30/52 or 0.5769 or 57.69%
Step-by-step explanation:
In a standard deck of 52 cards, the number of face cards (F) and the number of hearts (H) is given by:
[tex]F=4+4+4 =12\\H=\frac{52}{4}=13[/tex]
Out of all hearts, three of them are face cards (jack, king, and queen). Therefore, the probability of a card being EITHER a face card or a heart is:
[tex]P(F \cup H) = P(F) +P(H) - P(F \cap H) \\P(F \cup H)=\frac{12+13-3}{52} =\frac{22}{52}[/tex]
Therefore, the probability of card being NEITHER a face card NOR a heart is:
[tex]P=1-P(F \cup H) \\P=1-\frac{22}{52}=\frac{30}{52}\\\\P=0.5769\ or\ 57.69\%[/tex]
This problem addresses some common algebraic errors. For the equalities stated below assume that x and y stand for real numbers. Assume that any denominators are non-zero. Mark the equalities with T (true) if they are true for all values of x and y, and F (false) otherwise.
1. (x+y)^2 =x^2+y^2 __
2. (x+y)^2 = x^2 +2xy+y^2__
3. x/x+y=1/y__
4. x−(x+y) = y__
5. √x^2 =x__
6. √x^2 = |x|__
7. √x^2+4=x+2__
8. 1/x+y=1/x+1/y__
Answer:
1. F
observe that [tex](5+2)^2=49 \neq 29=5^2+2^2[/tex]
2. T
Let x and y real numbers.
[tex](x+y)^2=(x+y)(x+y)=x^2+2xy+y^2[/tex]
3. F
Observe that if x=3 and y=2 [tex]\frac{3}{3+2}=\frac{3}{5}\neq \frac{1}{2}[/tex]
4. F
If x=y=3, [tex]3-(3+3)=3-6=-3\neq 3[/tex]
5. F
if x=-1, [tex]\sqrt{-1^2}=\sqrt{1}=1\neq -1[/tex]
6. T
7. F
if x=-1, [tex]\sqrt{-1^2+4}?\sqrt{5}\neq 1=-1+2[/tex]
8. F
If x=1 and y=2, [tex]\frac{1}{1+2}=\frac{1}{3}\neq \frac{3}{2}=\frac{1}{1}+\frac{1}{2}[/tex]
psychologist obtains a random sample of 20 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 30 minutes each day until they give birth. After 5 years, the child is administered an IQ test. It is known that IQs are normally distributed with a mean of 100. If the IQs of the 20 children in the study result in a sample mean of 104.1 and sample standard deviation of 15, is there evidence that the children have higher IQs? Use the a=0.05 level of significance. Complete parts (a) through (d).b) Calulate the P-value.P-value+_______________(round to three decimal places as needed.c) State the conclusion for the test.Choose the correct Anser below.a. Do not reject H0 becasue the P-value is less than the a-0.05 level of significanceb. Reject H0 because the p-value is greater than the a=0.05 level of significancec. Do not reject H0 because the P-value is less than the a-0.05 level of signifcance.d. Reject H0 because the p-value is less than the a=0.05 level of significance.d) State the conclusion in context of the problem.There (1)_____________sufficient evidence at the a=0.05 level of significance to conclude that mothers who listen to Mozart have children with hogher IQs.(1)__is not___is
The test statistic is determined for the random sample conducted by rhe psychologist and the values are:
a) t ≈ 2.12
b) P-value ≈ P(T > 2.12)
c) Do not reject H0 because the P-value is less than the α = 0.05 level of significance.
d) It is not possible to make a definitive conclusion about the IQs of children born to mothers who listen to Mozart.
Given data:
To test whether the children in the study have higher IQs,conduct a one-sample t-test.
a) To conduct the t-test, set up the null and alternative hypotheses:
Null hypothesis (H0):
The population mean IQ of the children is equal to 100.
Alternative hypothesis (Ha):
The population mean IQ of the children is greater than 100.
b)
Calculate the P-value:
Determine the P-value using the t-distribution with the sample mean, sample standard deviation, and sample size.
P-value = P(T > t), where T follows a t-distribution with (n-1) degrees of freedom.
Using the given information, calculate the t-value:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (104.1 - 100) / (15 / sqrt(20))
t ≈ 2.12
Next, we need to find the P-value associated with this t-value using the t-distribution.
For a one-tailed test, the P-value is the probability of observing a t-value greater than 2.12.
P-value ≈ P(T > 2.12)
c)
State the conclusion for the test:
To make a conclusion, we compare the P-value to the significance level (α = 0.05).
Since the P-value is not provided, we cannot make a definitive conclusion. However, based on the options given, the correct answer would be:
c. Do not reject H0 because the P-value is less than the α = 0.05 level of significance.
d)
State the conclusion in the context of the problem:
Based on the incomplete information provided, we cannot make a definitive conclusion about the IQs of children born to mothers who listen to Mozart.
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To determine if there is evidence that the children have higher IQs, a p-value is calculated from the sample using a t-test. If the p-value is less than the significance level of 0.05, the null hypothesis is rejected, indicating evidence of higher IQs; otherwise, there is insufficient evidence.
Explanation:The question asks whether the children have higher IQs than the normal mean of 100, given a sample mean IQ of 104.1 with a standard deviation of 15 from a sample of 20 children.
Calculate the P-value
To calculate the p-value for a one-sample t-test:
Formulate the null hypothesis (H0): The children do not have higher IQs, i.e., the population mean is 100. Calculate the t-score using the formula: t = (sample mean - population mean) / (sample standard deviation / sqrt(n)). Use the t-score and degrees of freedom (n - 1) to determine the p-value from the t-distribution table. The p-value tells us the probability of obtaining a sample mean at least as extreme as the observed one if the null hypothesis were true.
The exact calculation is not provided here, but once you have the p-value:
State the Conclusion for the Test
If the p-value < 0.05, reject the null hypothesis (indicating evidence of higher IQs). If the p-value > 0.05, do not reject the null hypothesis (indicating insufficient evidence of higher IQs).
State the Conclusion in Context of the Problem
We would either conclude that there is or is not sufficient evidence at the α=0.05 level to suggest that mothers who listen to Mozart have children with higher IQs, depending on whether the p-value is less than or greater than 0.05.
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In a certain country the heights of adult men are normally distributed with a mean of 66.1 inches and a standard deviation of 2.7 inches. The country's military requires that men have heights between 62 inches and 73 inches. Determine what percentage of this country's men are eligible for the military based on height.
Answer:
93.03%
Step-by-step explanation:
Population mean (μ) = 66.1 inches
Standard deviation (σ) = 2.7 inches
The z-score for a given 'X' value is:
[tex]z = \frac{X- \mu}{\sigma}[/tex]
For X = 62 inches
[tex]z = \frac{62- 66.1}{2.7}\\z=-1.5185[/tex]
A z-score of -1.5185 corresponds to the 6.44-th percentile of a normal distribution.
For X = 62 inches
[tex]z = \frac{73- 66.1}{2.7}\\z=2.5555[/tex]
A z-score of 2.5555 corresponds to the 99.47-th percentile of a normal distribution.
The total percentage of men eligible for the military is the percentage within those two values, therefore:
[tex]E = 99.47-6.44\\E=93.03 \%[/tex]
93.03% this country's men are eligible for the military based on height.
Final answer:
To determine the percentage of men eligible for the military based on height, we need to calculate the proportion of men whose heights fall between 62 inches and 73 inches. First, convert the height values to z-scores using the formula z = (x - mean) / standard deviation. Then, look up the corresponding z-scores in the standard normal distribution table to find the proportion of men within the desired height range.
Explanation:
To determine the percentage of men eligible for the military based on height, we need to calculate the proportion of men whose heights fall between 62 inches and 73 inches.
First, we convert the height values to z-scores using the formula z = (x - mean) / standard deviation.
Then, we look up the corresponding z-scores in the standard normal distribution table to find the proportion of men within the desired height range. We can subtract this proportion from 1 to find the percentage of men who are eligible for the military based on height.
The mortgage foreclosure crisis that preceded the Great Recession impacted the U.S. economy in many ways, but it also impacted the foreclosure process itself as community activists better learned how to delay foreclosure and lenders became more wary of filing faulty documentation. Suppose the duration of the eight most recent foreclosures filed in the city of Boston (from the beginning of foreclosure proceedings to the filing of the foreclosure deed, transferring the property) has been 230 days, 420 days, 340 days, 367 days, 295 days, 314 days, 385 days, and 311 days. Assume the duration is normally distributed. Construct a 90% confidence interval for the mean duration of the foreclosure process in Boston.
Answer:
[293.21;372.28]
Step-by-step explanation:
Hello!
You are asked to construct a 90% Confidence Interval for the mean duration of the foreclosure process in Boston.
The Study variable is X: duration of foreclosure in Boston. X≈N(μ;σ²)
Sample n=8
To study the population sample you can use either a Z or a t-statistic. Since the sample is less than 10, I'll choose a Student t-statistic, because it is more potent with small samples than the Z, but either one is a good choice.
X[bar]= 332.75
S= 59.01
[tex]X[bar] ± t_{n-1; 1-\alpha /2}*\frac{S}{\sqrt{n} }[/tex]
[tex]332.75 ± 1.895*\frac{59,01}{\sqrt{8} }[/tex]
[293.21;372.28]
With a confidence level of 90% you'd expect the interval [293.21;372.28] to contain the population mean of the duration of the foreclosure process in Boston.
I hope you have a SUPER day!
A professor's son, having made the wise decision to drop out of college, has been finding his way in life taking one job or another, leaving when his creativity is overly stifled or the employer tires of his creativity. The professor dutifully logs the duration of his son's last few careers and has determined that the average duration is normally distributed with a mean of eighty-eight weeks and a standard deviation of twenty weeks. The next career begins on Monday; what is the likelihood that it endures for more than one year?
Answer:
0.9641 or 96.41%
Step-by-step explanation:
Mean career duration (μ) = 88 weeks
Standard deviation (σ) = 20
The z-score for any given career duration 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
In this problem, we want to know what is the probability that the professor's son's next career lasts more than a year. Assuming that a year has 52 weeks, the equivalent z-score for a 1-year career is:
[tex]z=\frac{52-88}{20}\\z=-1.8\\[/tex]
According to a z-score table, a z-score of -1.8 is at the 3.59-th percentile, therefore, the likelihood that this career lasts more than a year is given by:
[tex]P(X>52) = 1-0.0359\\P(X>52) = 0.9641\ or\ 96.41\%[/tex]
Answer:
0.9641 or 96.41%
Step-by-step explanation: