Answer:
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1,
0.29a = 0.29 and 0.34a = 0.34
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f
Step-by-step explanation:
ψ(x) = 2a(√sin(πxa)
The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction
The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as
P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx
That is, integrating from 0.29a to 0.34a
ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)
∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx
∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))
= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ
- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.
Can someone please help me find the missing lengths in the following diagram?
Answer:
Step-by-step explanation:
18 a)Since line DE is parallel to line BC, it means that triangle ADE is similar to triangle ABC. Therefore,
AB/AD = AC/AE = BC/DE
AC = AE + CE = 18 + 9
AC = 27
AB = AD + DB
AB = AD + 5
Therefore,
27/18 = (AD + 5)/AD
Cross multiplying, it becomes
27 × AD = 18(AD + 5)
27AD = 18AD + 90
27AD - 18AD = 90
9AD = 90
AD = 90/9
AD = 10
b) AC = AE + EC = 13 + 3
AC = 16
To find AD,
16/13 = 24/AD
16 × AD = 13 × 24
16AD = 312
AD = 312/16
AD = 19.5
DB = 24 - 19.5
DB = 4.5
Your grandmother has been putting $ 3 comma 000$3,000 into a savings account on every birthday since your first (that is, when you turned 1). The account pays an interest rate of 7 %7%. How much money will be in the account on your 18th birthday immediately after your grandmother makes the deposit on that birthday?
Answer: there would be $102000 in the account.
Step-by-step explanation:
We would apply the future value which is expressed as
FV = C × [{(1 + r)^n - 1}/r]
Where
C represents the yearly payments on each birthday.
FV represents the amount of money
in your account at the end of 18 years.
r represents the annual rate.
n represents number of years or period.
From the information given,
r = 7% = 7/100 = 0.07
C = $3000
n = 18 years
Therefore,
FV = 3000 × [{(1 + 0.07)^18 - 1}/0.07]
FV = 3000 × [{3.38 - 1}/0.07]
FV = 3000 × 34
FV = $102000
A corporation has 11 manufacturing plants. Of these, seven are domestic and four are outside the United States. Each year a performance evaluation is conducted for four randomly selected plants. What is the probability that a performance evaluation will include at least one plant outside the United States
Answer:
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Step-by-step explanation:
Total plants = 11
Domestic plants = 7
Outside the US plants = 4
Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Here we have n=4, p=4/11 and q=7/11
P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰
= 1 - 0.16399
P(X≥1) = 0.836
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Exercise 12.a - Compute confidence interval; Find necessary sample size Find a 95% confidence interval for the mean efficiency. Round the answers to three decimal places.
Answer:
95% confidence interval for the mean efficiency is [84.483 , 85.517].
Step-by-step explanation:
We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.
So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\mu[/tex] = sample average efficiency = 85
[tex]\sigma[/tex] = sample standard deviation = 2
n = sample of motors = 60
[tex]\mu[/tex] = population mean efficiency
So, 95% confidence interval for the mean efficiency, [tex]\mu[/tex] is ;
P(-2.0009 < [tex]t_5_9[/tex] < 2.0009) = 0.95
P(-2.0009 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.0009 ) = 0.95
P( [tex]-2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P( [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]85 -2.0009 \times {\frac{2}{\sqrt{60} }[/tex] , [tex]85 +2.0009 \times {\frac{2}{\sqrt{60} }[/tex] ]
= [84.483 , 85.517]
Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].
Final answer:
The 95% confidence interval for the mean efficiency of electric motors, given a sample mean of 85, a standard deviation of 2, and a sample size of 60, is approximately (84.494, 85.506) when rounded to three decimal places.
Explanation:
To calculate the 95% confidence interval for the mean efficiency of electric motors, we use the sample mean, standard deviation, and the sample size along with the z-score for the 95% confidence level. Since the sample size is large (n > 30), we can use the z-distribution to approximate the sampling distribution of the sample mean.
The formula for a confidence interval is:
Confidence Interval = sample mean "+/-" (z-score * (standard deviation / sqrt(n)))
Given that the sample mean is 85, the standard deviation is 2, and the sample size (n) is 60, and using the z-score of approximately 1.96 for 95% confidence, we can compute the confidence interval.
Confidence Interval = 85 "+/-" (1.96 * (2 / √(60)))
After the calculation, we get the two ends of the interval:
Lower Limit = 85 - (1.96 * (2 / √(60))) = 84.494
Upper Limit = 85 + (1.96 * (2 / √(60))) = 85.506
Rounding these to three decimal places, the 95% confidence interval for the mean efficiency of electric motors is approximately (84.494, 85.506). This means we can be 95% confident that the true average efficiency of all electric motors is between 84.494% and 85.506%.
Circle your answer and justify it by showing your work. (a) T F: (b) T F: Let A be any square matrix, then ATA, AAT, and A + AT are all symmetric. If S is invertible, then ST is also invertible. If a row exchange is required to reduce matrix A into upper triangular form U, then (c) T F: A can not be factored as A = LU. (d) T F: Suppose A reduces to upper triangular U but U has a 0 in pivot position, then A has no LDU factorization. (e) T F: If A2 is not invertible, then A is not invertible. 10 10. [10points] (a) T F: All(x,y,z)∈R3 withx=y+z+1isasubspaceofR3 (b) T F: All(x,y,z)∈R3 withx+z=0isasubspaceofR3 (c) T F: All 2 × 2 symmetric matrices is a subspace of M22. (Here M22 is the vector space of all 2 × 2 matrices.) (d) T F: All polynomials of degree exactly 3 is a subspace of P5. (Here P5 is the vector space of all polynomials a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0 of degree less than or equal to 5.) (e) T F: P3 is a subspace of P5. (Here Pi is the vector space of all polynomials aixi+ai−1xi−1+ ai−2x1−2 + ... + a2x2 + a1x + a0 of degree less than or equal to i.)a. TrueB. False.
Answer:
a-True
b-True
c-True
d- False
e-True
f-True
g-True
h- False
i-True
j- False
Step-by-step explanation:
See the attached pictures.
What is the product?
Answer: the third option is correct
Step-by-step explanation:
The first matrix is a 2 × 3 matrix while the second matrix is a 3 × 2 matrix. To get the product of both matrices, we would multiply each term in each row by the terms in the corresponding column and add.
1) row 1, column 1
1×2 + 3×3 + 1×4 = 2 + 9 + 4 = 15
2) row 1, column 2
1×-2 + 3×5 + 1×1 = - 2 + 15 + 1 = 14
3) row 2, column 1
-2×-2 + 1×3 + 0×4 = - 4 + 3 + 0 = - 1
4) row 2, column 2
- 2×-2 + 1×5 + 0×1 = 4 + 5 = 9
The solution becomes
15 14
- 1 9
Here are summary statistics for randomly selected weights of newborn girls: nequals202, x overbarequals28.3 hg, sequals6.1 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very different from the confidence interval 27.8 hgless thanmuless than29.6 hg with only 17 sample values, x overbarequals28.7 hg, and sequals1.8 hg? What is the confidence interval for the population mean mu? nothing hgless thanmuless than nothing hg (Round to one decimal place as needed.) Are the results between the two confidence intervals very different? A. Yes, because the confidence interval limits are not similar. B. Yes, because one confidence interval does not contain the mean of the other confidence interval. C. No, because each confidence interval contains the mean of the other confidence interval. D. No, because the confidence interval limits are similar.
Answer:
The confidence interval is 27.5 hg less than mu less than 29.1 hg
(A) Yes, because the confidence interval limits are not similar.
Step-by-step explanation:
Confidence interval is given as mean +/- margin of error (E)
mean = 28.3 hg
sd = 6.1 hg
n = 202
degree of freedom = n-1 = 202-1 = 201
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196
E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg
Lower limit = mean - E = 28.3 0.8 = 27.5 hg
Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg
95% confidence interval is (27.5, 29.1)
When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17
The confidence intervals for the two experiments have overlapping intervals, suggesting that the results are not very different.
Explanation:The first experiment resulted in a 95% confidence interval of 3.070-3.164 g for the population mean weight of newborn girls. The second experiment had a 95% confidence interval of 3.035-3.127 g. Although the two confidence intervals are not identical, the mean for each experiment is within the confidence interval of the other experiment. This suggests that the results are not very different and there is an appreciable overlap between the two intervals. Therefore, the answer is C. No, because each confidence interval contains the mean of the other confidence interval.
A tennis club offers two payment options:
Option1: $35 monthly fee plus $4/hour for court rental
Option 2: No monthly fee but $6.50/hour for court rental.
Let x = hours per month of court rental time.
a) Write a mathematical model representing the total monthly cost, C, in terms of x for the following:
Option 1: C= _________________
Option 2: C=_________________
b) How many hours would you have to rent the court so that the monthly cost of option 1, is less than option 2. Set up an inequality and show your work algebraically using the information in part a.
Multiply the cost per hour by number of hours x and for option 1 you need to add the monthly fee:
A) option 1: C= 4x +35
Option2: C = 6.50x
B)
4x+35 < 6.50x
Subtract 4x from both sides:
35 < 2.50x
Divide both sides by 2.50 :
X < 14
You would need to rent more than 14 hours for option 1 to be cheaper.
A client arrives in the emergency department in a crisis state. The client demonstrates signs of profound anxiety and is unable to focus on anything but the object of the crisis and the impact on self. The initial data collection would focus on which information
Answer:The client's physical condition
Step-by-step explanation: The first or initial data that should be collected when a person have a symptoms of profound anxiety and is unable to focus on anything but the object of the crisis and the impact on self, the initial data to be collected would focus on the physical condition of the clients as it will give the adequate Background data from where other actions or data analysis will be done in order to evaluate and handle such a situation, this initial data will also assist the health care provider to know what is the root cause of his condition and what other steps are required.
Your body loses sodium when you sweat. Researchers sampled 38 random tennis players. The average sodium loss was 500 milligrams per pound and the standard deviation was 62 milligrams per pound. Construct and interpret a 99% confidence interval to estimate the mean loss in sodium in the population.
Answer:
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{62}{\sqrt{38}} = 25.90[/tex]
The lower end of the interval is the mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.
The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
Answer:
Option A
Step-by-step explanation:
The 99% confidence interval is (472.69, 527.31). We are 99% confident that the true population mean of sodium loss for tennis players will be between 472.69 milligrams per pound and 527.31 milligrams per pound.
Your large corporation manufactures a certain popular brand of robot lawn mower. After manufacture, all mowers go through a standard quality control check. The Portland factory, which manufactures 30% of your mowers, has a probability of .8 that each mower will pass the quality control check. The Dallas factory, which manufactures 50% of your mowers, has a probability of .7 that each mower will pass the quality control check. For the Buffalo factory,
which manufactures the remaining 20% of your mowers, you have been unable to learn what the probability is that each mower will pass inspection. But, the corporation’s annual report claims that the overall probability that one of their manufactured mowers will pass the quality control check is .72
a) What do you conclude is the probability that a mower manufactured at the Buffalo factory will pass the quality control check?
b) A customer orders one of your mowers, and of course receives one that has passed the quality control check. What is the probability it was manufactured in Dallas?
Answer:
(a) The probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.
(b) The probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.
Step-by-step explanation:
Denote the events as follows:
X = a mower is manufactured at the Portland factory
Y = a mower is manufactured at the Dallas factory
Z= a mower is manufactured at the Buffalo factory
A = a mower passes the quality check.
The information provided is:
[tex]P(X)=0.30\\P(A|X)=0.80\\P(Y)=0.50\\P(A|Y)=0.70\\P(Z)=0.20\\P(A)=0.72[/tex]
(a)
The probability that a mower manufactured at the Buffalo factory will pass the quality control check is:
P (A|Z)
Compute the value of P (A|Z) as follows:
[tex]P(A)=P(A\cap X)+P(A\cap Y) + P (A\cap Z)\\0.72=(0.80\times0.30)+(0.70\times0.50)+(0.20\times P(A|Z))\\0.20\times P(A|Z)=0.72-0.24-0.35\\P(A|Z)=\frac{0.13}{0.20}\\=0.65[/tex]
Thus, the probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.
(b)
Compute the value of P (Y|A) as follows:
[tex]P(Y|A)=\frac{P(A|Y)P(Y)}{P(A)}=\frac{0.70\times0.50}{0.72}=0.4861[/tex]
Thus, the probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.
The probability that a mower from the Buffalo factory will pass the quality control check is 13%, and if a customer receives a mower that has passed the check, there is a 49% probability that it was manufactured in Dallas.
Explanation:To find the probability that a robot lawn mower made at the Buffalo factory will pass the quality control check, we first understand that the total probability of a mower passing the check is a sum of all the probabilities from the three factories. This is given as 0.72.
The Portland factory which makes 30% of the mowers has a 0.8 probability of passing the check. Therefore, the contribution of Portland to the total probability is 0.3*0.8 = 0.24. Similarly, the Dallas factory makes 50% of the mowers and each has a passing probability of 0.7, so the Dallas contribution is 0.5*0.7 = 0.35.
Knowing this, we can subtract the total contributions of Portland and Dallas from the overall probability of 0.72 to get Buffalo's contribution, which is the Buffalo passing probability. Therefore, the Buffalo passing probability becomes 0.72 - 0.24 - 0.35 = 0.13 or 13%.
In the second part, if the customer receives a mower that has passed the quality control check, the probability that it was manufactured in Dallas is the contribution of the Dallas factory to the passing mowers i.e., 0.35 ÷ 0.72 = 0.486 or approximately 49%.
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Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:
θ1=X1+X2+......+X7 / 7
θ2= (2X1-X3+X5) / 2
a. Is either estimator unbiased?
b. Which estimator is best? In what sense is it best? Calculate the relative efficiency of the 2 estimtors.
Answer:
a) In order to check if an estimator is unbiased we need to check this condition:
[tex] E(\theta) = \mu[/tex]
And we can find the expected value of each estimator like this:
[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]
So then we conclude that [tex] \theta_1 [/tex] is unbiased.
For the second estimator we have this:
[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]
And then we conclude that [tex]\theta_2[/tex] is unbiaed too.
b) For this case first we need to find the variance of each estimator:
[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]
And for the second estimator we have this:
[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]
And the relative efficiency is given by:
[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]
Step-by-step explanation:
For this case we assume that we have a random sample given by: [tex] X_1, X_2,....,X_7[/tex] and each [tex] X_i \sim N (\mu, \sigma)[/tex]
Part a
In order to check if an estimator is unbiased we need to check this condition:
[tex] E(\theta) = \mu[/tex]
And we can find the expected value of each estimator like this:
[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]
So then we conclude that [tex] \theta_1 [/tex] is unbiased.
For the second estimator we have this:
[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]
And then we conclude that [tex]\theta_2[/tex] is unbiaed too.
Part b
For this case first we need to find the variance of each estimator:
[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]
And for the second estimator we have this:
[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]
And the relative efficiency is given by:
[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]
Both θ1 and θ2 are unbiased estimators of the population mean μ. However, θ1 is more efficient due to having a lower variance compared to θ2. The relative efficiency of θ1 compared to θ2 is 10.5.
To determine if the estimators [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased and which one is best, we follow these steps:
A. Evaluating Unbiasedness
Estimator [tex](\theta_1):\theta_1=\frac{X_1+X_2+\\cdots+X_7}{7}[/tex][tex]\(E[\theta_1]=\frac{E[X_1+X_2+\cdots+X_7]}{7}=\frac{7\\mu}{7}=\mu\)[/tex]Thus, [tex](\theta_1)[/tex] is an unbiased estimator of \\(\\mu\\).Estimator [tex](\theta_2=\frac{2X_1-X_3+X_5}{2})[/tex][tex]\(E[\theta_2]=E[\frac{2X_1-X_3+X_5}{2}]=\frac{2E[X_1]-E[X_3]+E[X_5]}{2}=\frac{2\mu-\mu+\mu}{2}=\mu\)[/tex]Thus,[tex](\theta_2)[/tex] is also an unbiased estimator of [tex](\mu)[/tex].B. Evaluating Efficiency
Variance of [tex](\theta_1):\theta_1=\frac{X_1+X_2+\cdots+X_7}{7}[/tex][tex]\(Var(\theta_1)=Var(\frac{X_1+X_2+\cdots+X_7}{7})=\frac{\sigma^2}{7}\)[/tex]Variance of[tex](\theta_2):\theta_2=\frac{2X_1-X_3+X_5}{2})[/tex][tex]\(Var(\theta_2)=Var(\frac{2X_1-X_3+X_5}{2})=\frac{4Var(X_1)+Var(X_3)+Var(X_5)}{4}=\frac{4\sigma^2+\sigma^2+\sigma^2}{4}=\frac{6\sigma^2}{4}=1.5\sigma^2)[/tex]Since[tex](Var(\theta_1)=\frac{\sigma^2}{7}) is less than \(Var(\theta_2)=1.5\sigma^2), (\theta_1)[/tex] is the more efficient estimatorThe relative efficiency of [tex](\theta_1)[/tex] to [tex](\theta_2)[/tex] is:
[tex]\(RE=\frac{Var(\theta_2)}{Var(\theta_1)}=\frac{1.5\sigma^2}{\frac{\sigma^2}{7}}=10.5\)[/tex]Both [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased estimators of [tex](\mu)[/tex], but [tex](\theta_1)[/tex] is the best in terms of efficiency since it has a lower variance.
12. Wendy sliced a loaf of bread into 12
equal slices. She used 4 of the slices to
make sandwiches. What fraction of the
loaf of bread was left?
Answer:2/3
Step-by-step explanation:
Answer: 2/3 the
loaf of bread was left.
Step-by-step explanation:
Wendy sliced a loaf of bread into 12
equal slices. This means that the portion or size of each slice is 1/12.
She used 4 of the slices to
make sandwiches. This means that the portion of the loaf of bread that she used to make the sandwiches is
4 × 1/12 = 4/12 = 1/3
Therefore, the fraction of the
loaf of bread that was left is
1 - 1/3 = 2/3
Consider the differential equation y '' − 2y ' + 37y = 0; ex cos(6x), ex sin(6x), (−[infinity], [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.The functions satisfy the differential equation and are linearly independent sincea) W(ex cos 6x, ex sin 6x) =
The question is:
Consider the differential equation y'' - 2y' + 37y = 0;
[tex]e^x cos(6x), e^x sin(6x), (-\infty, \infty).[/tex]
Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.
The functions satisfy the differential equation and are linearly independent since
a) [tex]W(e^x cos 6x, e^x sin 6x) \neq 0[/tex]
Answer:
To verify if the given functions form a fundamental set of solutions to the differential equation, we find the Wronskian of the two functions.
The Wronskian of functions [tex]y_1 $and$ y_2[/tex] is given as[tex]W(y_1, y_2) = \left|\begin{array}{cc}y_1&y_2\\y_1'&y_2'\end{array}\right|\\\\y_1= e^x cos 6x \\y_2 = e^x sin 6x \\y_1' = -6e^x sin 6x \\y_2' = 6e^x cos 6x \\W\left(e^x cos 6x, e^x sin 6x \right) = \left|\begin{array}{cc}e^x cos 6x &e^x sin 6x \\ \\ -6e^x sin 6x&6e^x cos 6x \end{array}\right|\\ \\= 6e^{2x} cos^2 6x + e^{2x} sin^2 6x \\ \\ = 6e^{2x}\left( cos^2 6x + sin^2 6x\right)\\ \\$but $cos^2 6x + sin^2 6x = 1\\ \\W\left(y_1, y_2 \right) = 6e^{2x} \neq 0[/tex]
Because the Wronskian is not zero, we say the solutions are linearly independent
The two functions form a fundamental set of solutions to the differential equation if both satisfy the differential equation and their Wronskian is non-zero, indicating they are linearly independent. This can be confirmed by computing the determinant.
Explanation:The given differential equation is a second-order homogeneous equation. To confirm if the given functions form a fundamental set of solutions, they should satisfy the differential equation and should be linearly independent.
To solve the equation, first find the Wronskian, which is a determinant used to test whether solutions are linearly independent. For this particular case:
W(ex cos 6x, ex sin 6x) = det[[ex cos(6x)]', [ex cos(6x)]'; [ex sin(6x)]', [ex cos(6x)]'].
If the resulting value does not equal zero, the solutions are indeed linearly independent and they form a fundamental set of solutions for the given differential equation over the entire real line.
To conclude, if both solutions satisfy the differential equation and W does not equal zero, then ex cos(6x) and ex sin(6x) are linearly independent and they form a fundamental set of solutions for the given differential equation.
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For a data set of weights (pounds) and highway fuel consumption amounts (mpg) of eight types of automobile, the linear correlation coefficient is found and the P-value is 0.044. Write a statement that interprets the P-value and includes a conclusion about linear correlation.
The P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is [WHAT PERCENT] which is [LOW OR HIGH] so there [IS OR IS NOT] sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
(Type an integer or a decimal. Do not round.)
Answer:
The P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is 4.4 which is LOW so there IS sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
Step-by-step explanation:
Hello!
Remember:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
Let's say that the significance level of this correlation test is α: 0.05
If the p-value is the probability of obtaining
you can express it as a percentage: 4.4%Is a very low probability. The decision rule using the p-value is:
p-value < α ⇒ Reject the null hypothesis
p-value ≥ α ⇒ Do not reject the null hypothesis.
The p-value is less than the significance level, the decision is to reject the null hypothesis.
In a linear correlation analysis the statement "there is no linear correlation between the two variables" is always in the null hypothesis, so if you reject it, you can conclude that there is a linear correlation between the variables.
I hope it helps!
In statistical analysis, a P-value of 0.044 indicates there is a 4.4% chance of obtaining a linear correlation as extreme as the observed correlation coefficient. A P-value under 0.05 provides enough evidence to reject the null hypothesis of no correlation, therefore suggesting a significant correlation. Hence, there is sufficient evidence of a linear correlation between car weight and highway fuel consumption.
Explanation:The P-value of 0.044 in this context represents the likelihood of obtaining a linear correlation coefficient for the data points in your dataset that is as extreme as, or more extreme than, the one you calculated, assuming there is no linear relationship between the two variables (weight and highway fuel consumption in automobiles). This probability is 4.4% - rather low. A common threshold for significance in many fields is 0.05, or 5%. If your P-value is below this threshold, we reject the null hypothesis that there is no correlation and conclude there may be a correlation. Therefore, as the P-value is 0.044, which is below the 0.05 threshold, there is sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
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The lengths of the sides of a rectangle are consecutive prime numbers. The area of the rectangle is represented by a three-digit number composed only of the two smallest prime digits and it will not change if we reverse it . What is the perimeter of the rectangle?
WILL MARK BRAINLIEST!
Answer:
Perimeter = 72
Step-by-step explanation:
Mathematics Puzzle
The area of a rectangle is the product of the base and the height
A=b.h
We know the base and the height are two consecutive prime numbers, not much of useful information so far.
We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.
We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.
The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.
The perimeter of that rectangle is 2*17+2*19= 72
Answer:
72 units
Step-by-step explanation:
72 units
A researcher believes that 9% of females smoke cigarettes. If the researcher is correct, what is the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%? Round your answer to four decimal places.
To calculate the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%, we can use the normal approximation to the binomial distribution. The probability is approximately 0.0030.
Explanation:To calculate the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%, we can use the normal approximation to the binomial distribution. The first step is to calculate the standard error, which is the square root of the product of the population proportion (0.09) and its complement (0.91), divided by the sample size (703). The formula for the standard error is:
SE = sqrt((p * (1-p))/n)
Substituting the values into the formula, we get:
SE = sqrt((0.09 * 0.91)/703) ≈ 0.0096
Next, we need to calculate the z-score, which measures how many standard errors the observed proportion is from the population proportion. The formula for the z-score is:
z = (observed proportion - population proportion) / standard error
Substituting the values into the formula:
z = (observed proportion - population proportion) / SE
The observed proportion is unknown in this case, but we can calculate the range within which it can fall. To differ from the population proportion by more than 3%, the observed proportion can be either 3% greater than the population proportion or 3% less than the population proportion. So the range is from (0.09 - 0.03) to (0.09 + 0.03), which is (0.06 to 0.12).
Calculating the z-score for the lower limit of the range:
z = (0.06 - 0.09) / 0.0096 ≈ -3.125
Calculating the z-score for the upper limit of the range:
z = (0.12 - 0.09) / 0.0096 ≈ 3.125
To find the probability that the proportion differs from the population proportion by more than 3%, we need to find the area under the standard normal curve beyond these z-scores. We can use a standard normal table or a calculator to find the probabilities. However, since the z-scores are symmetric, we only need to find the probability for one tail and double it. Using a standard normal table or calculator, we find that the probability of a z-score less than -3.125 or greater than 3.125 is approximately 0.001521. Doubling this probability gives us the final answer:
Probability = 2 * 0.001521 ≈ 0.003042
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derived the MOM and MLE for an exponential distribution with parameter ????. Conduct a Bootstrap simulation to compare the estimation of λ with sample sizes of n = 10, n = 100, and n = 500. Choose true value λ = 0.2 and use B = 1000. Calculate and compare the mean and standard error for each set of simulations to each other as well as their theoretical values.
Answer:
rm(list=ls(all=TRUE))
set.seed(12345)
N=c(10,100,500)
Rate=0.2
B=1000
MN=SE=rep()
for(i in 1:length(N))
{
n=N[i]
X=rexp(n,rate=Rate)
EST=1/mean(X)
ESTh=rep()
for(j in 1:B)
{
Xh=rexp(n,rate=EST)
ESTh[j]=1/mean(Xh)
}
MN[i]=mean(ESTh)
SE[i]=sd(ESTh)
}
cbind(N,Rate,MN,SE)
1. Using R, construct time series (line) plots for both stock prices and return series. R functions ts.plot or plot can create the plot. Describe the patterns and compare the plots.
Answer:
The code in R for the time series is given as below.
Step-by-step explanation:
As the complete question is not presented thus a sample code for the variables is given as below
# Libraries
library(ggplot2)
library(dplyr)
library(plotly)
library(hrbrthemes)
# Load dataset
filepath=" Your dataset path here in csv format"
data <- read.table("filepath", header=T)
data$date <- as.Date(data$date)
# plot
data %>%
ggplot( aes(x=date, y1=value1,x=date, y2=value2)) +
geom_line(color="#69b3a2") +
ylim(0,22000) +
theme_ipsum()
In R, stock prices and return series can be visualized using the ts.plot() or plot() functions in a line graph. The time series graph will then help in identifying and understanding trends and patterns in the data over time. Comparing the plots of the stock prices and the return series can provide insights into market behavior.
Explanation:To plot both stock prices and return series using R, one would use either the ts.plot or plot function. Firstly, you need to import or generate the stock prices and the return series data. The horizontal axis of the plot corresponds to the date or time increments, while the vertical axis corresponds to the values of the variable (i.e., stock prices or return series).
Load your time series data in R.Next, use the ts.plot() or plot() functions to visualize your data in a line graph.Pay attention to the patterns that emerge. Line graphs are an effective method to show the relationships between two changing variables such as time and stock price.Detail the trends and patterns you observe in your data. Are there seasonal trends, cyclical fluctuations or random variations? Is the trend increasing or decreasing over time? Comparatively analyze the plots for the stock prices and the return series.Time series graphs facilitate spotting trends and are used to graphically represent the same variable values recorded over an extended period of time. In the context of stock prices and return series, these patterns over time can provide valuable insights into market behavior and investment opportunities.
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QUESTION 7 A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them. Suppose 9 employees are selected randomly and will be interviewed about company loyalty. What is the probability that none of the 9 employees will say their company is loyal to them? g
Answer:
7.51% probability that none of the 9 employees will say their company is loyal to them.
Step-by-step explanation:
For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
25% of employees said their company is loyal to them.
This means that [tex]p = 0.25[/tex]
9 employees are selected randomly
This means that [tex]n = 9[/tex]
What is the probability that none of the 9 employees will say their company is loyal to them?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751[/tex]
7.51% probability that none of the 9 employees will say their company is loyal to them.
What is the age distribution of adult shoplifters (21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution. Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use 45.5 as the class midpoint. (Enter your answers to one decimal place.)
Age range (years) 21-30 31-40 41 and over
Number of shoplifters 280 368 247
We can estimate the mean, variance, and standard deviation by using class midpoints and the number of observations. First, find the midpoint of each age group. Then, to find the mean, multiply each midpoint by the number of shoplifters in that age group, sum those products, and divide by the total number of observations. To find the variance, subtract the mean from each midpoint, square the result, multiply by the number of shoplifters in that group, sum those products, and divide by the total number of observations minus one. The standard deviation is the square root of the variance.
Explanation:To estimate the mean age, sample variance, and sample standard deviation for the shoplifters from a random sample of 895 incidents we need to follow a few steps. Here's how to do it:
Calculate the midpoint of each class: For the age ranges 21-30 and 31-40, the midpoints are 25.5 and 35.5 respectively. The problem already provides 45.5 as the class midpoint for the range '41 and over'.Calculate the estimated mean (µ) by multiplying the midpoint of each class by the number of observations in that class, summing these values, and dividing by the total number of observations.Calculate the estimated variance (σ²) by subtracting the estimated mean from each class midpoint, squaring the result, multiplying by the number of observations in that class, summing these values, and dividing by the total number of observations - 1.Finally, calculate the standard deviation (σ) by taking the square root of the estimated variance.Learn more about Statistics here:https://brainly.com/question/31538429
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If two events are independent, then Group of answer choices the sum of their probabilities must be equal to one. they must be mutually exclusive. their intersection must be zero. None of the above.
Answer:
For this case we can define the following two events A and B.
In order to classify A and B as independent we needd to satisfy this condition:
[tex] P(A \cap B) = P(A) *P(B)[/tex]
None of the above.
True, because none of the options were correct.
See explanation below
Step-by-step explanation:
For this case we can define the following two events A and B.
In order to classify A and B as independent we need to satisfy this condition:
[tex] P(A \cap B) = P(A) *P(B)[/tex]
So let's analyze one by one the possible options:
the sum of their probabilities must be equal to one.
False, the sum of the probabilities can be <1 so this statement is not true
they must be mutually exclusive.
False when we talk about mutually exclusive events we are saying that:
[tex] P(A \cap B) =0[/tex]
But independence not always means that we have mutually exclusive events
their intersection must be zero.
False the intersection of the probabilities is 0 just if we have mutually exclusive events, not independent events
None of the above.
True, because none of the options were correct.
Independent events refer to the concept in which the outcome of one event doesn't influence the outcome of another. If two events are independent, the probability of both events taking place is the product of their individual probabilities.
Explanation:In mathematics, when we say that events are independent, it means that the outcome of one event does not affect the outcome of the other. The correct statement is: 'If two events are independent, the probability of both events happening is the product of the probabilities of each event.' When the sum of their probabilities is equal to one, they are mutually exclusive, not independent. And their intersection being zero also refers to mutually exclusive events not independent ones.
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Professor stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particulare on the the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, also independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half of the crossings encountered.
A.) Which route should he take to minimize the probability of being late to the meeting?
B.) If he tosses a fair coin to decide on a route and is late, what is the probability he took the four crossing route?
Answer:
A) He should take the route with 4 crossings
B) The probability that he took the 4 crossing route is 0.2158
Step-by-step explanation:
Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.
Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1
The probability that the Professor is on time on route 1 is equal to
P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477
On the other hand, the probability that the professor is on time on route 2 is
P(X|B = 0) = 0.9² = 0.81
Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.
B) Lets call L the event 'The professor is late'. We know that
P(L|A) = 1-0.9477 = 0.0524
P(L|B) = 1-0.81 = 0.19
Also
P(A) = P(B) = 1/2 (this only depends on the result of the coin.
For the Bayes theorem we know, therefore that
[tex]P(A|L) = \frac{P(L|A) * P(A)}{P(L|A)*P(A) + P(L|B)*P(B)} = \frac{0.0523*0.5}{0.0523*0.5 + 0.19*0.5 } = 0.2158[/tex]
Hence, the probability that he took the 4 crossing route is 0.2158.
Suppose the coffee industry claimed that the average adult drinks 1.7 cups of coffee per day. To test this claim, a random sample of 40 adults was selected, and their average coffee consumption was found to be 1.9 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.6 cups. Using alphaequals0.10, complete parts a and b below. a. Is the coffee industry's claim supported by this sample? Determine the null and alternative hypotheses.
Answer:
(a) No, the coffee industry's claim is not supported by this sample.
(b) Null hypothesis: The average adult drinks 1.7 cups of coffee per day.
Alternate hypothesis: The average adult drinks more than 1.7 cups of coffee per day.
Step-by-step explanation:
(a) Test statistic (z) = (sample mean - population mean) ÷ sd/√n
sample mean = 1.9 cups per day
population mean = 1.7 cups per day
sd = 0.6 cups per day
n = 40
z = (1.9 - 1.7) ÷ 0.6/√40 = 0.2 ÷ 0.095 = 2.11
The test is a one-tailed test. Using alpha (significance level) = 0.1, the critical value is 2.326.
Conclusion:
Reject the null hypothesis because the test statistic 2.11 falls within the rejection region of the critical value 2.326.
The coffee industry's claim is contained in the null hypothesis, hence it is not supported by the sample because the null hypothesis is rejected.
(b) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.
An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.
A certain vibrating system satisfies the equation u'' + γu' + u = 0. Find the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.
In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.
Explanation:The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.
To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:
1.9 = 1/(sqrt(1 - (γ/2)^2))
Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.
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The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.
Explanation:The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.
The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.
The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.
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Which of the following are continuous variables, and which are discrete? (a) number of traffic fatalities per year in the state of Florida continuous discrete (b) distance a golf ball travels after being hit with a driver continuous discrete (c) time required to drive from home to college on any given day continuous discrete (d) number of ships in Pearl Harbor on any given day continuous discrete (e) your weight before breakfast each morning continuous discrete
Answer:
a) Discrete
b) Continuous
c) Continuous
d) Discrete
e) Continous
Step-by-step explanation:
Continuous:
Real numbers, can be integer, decimal, etc.
Discrete:
Only integer(countable values). So can be 0,1,2...
(a) number of traffic fatalities per year in the state of Florida
You cannot have half of a traffic fatality, for example. So this is discrete
(b) distance a golf ball travels after being hit with a driver
The ball can travel 10.25m, for example, which is a decimal number. So this is continuous.
(c) time required to drive from home to college on any given day
You can take 10.5 minutes, for example, which is a decimal value. So this is continuous.
(d) number of ships in Pearl Harbor on any given day
There is no half ship, for example. So this is discrete.
(e) your weight before breakfast each morning continuous discrete
You can weigh 80.4kg, for example, which is a decimal number. So this is continuous.
a) Discrete
b) Continuous
c) Continuous
d) Discrete
e) Continuous
Discrete data is the numerical type of data which includes whole, concrete numbers that has specific and fixed data values. therefore, they can be determined by counting.
example: Number of students in the class, ect.,
Continuous type of data includes complex numbers or the varying data values that are measured over a specific time interval.
example: Height of students in a school, etc.,
(a) Number of traffic fatalities per year in the state of Florida.
This is a Discrete data, as it is a countable data and will be always a whole be number. Number of traffic fatalities per year will be 75, 150, 200, etc.
(b) distance a golf ball travels after being hit with a driver.
This is a Discrete data, as the data can be measures and will be not always be a whole number. distance a golf ball travels after being hit can be 1.25 meters, 10.9 meters, 21.1 meters, etc.
(c) time required to drive from home to college on any given day.
This is a Discrete data, as the data can be measures and will be not always be a whole number. time required to drive from home to college can be 5minutes 30 seconds, 15 minutes 26 seconds.
(d) number of ships in Pearl Harbor on any given day.
This is a Discrete data, as it is a countable data and will be always a whole number. number of ships in Pearl Harbor on any given day will be always 1, 10, 8, etc.
(e) your weight before breakfast each morning.
This is a Discrete data, as the data can be measures and will be not always be a whole number. weight can be 44.56 kgs., 52.3 kgs, etc.
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According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers; and of those not having lung disease, 25% are smokers. What is the probability that a smoker has lung disease?
Answer:
Probability that a smoker has lung disease = 0.2132
Step-by-step explanation:
Let L = event that % of population having lung disease, P(L) = 0.07
So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93
S = event that person is smoker
% of population that are smokers given they are having lung disease, P(S/L) = 0.90
% of population that are smokers given they are not having lung disease, P(S/L') = 0.25
We know that, conditional probability formula is given by;
P(S/L) = [tex]\frac{P(S\bigcap L)}{P(L)}[/tex]
[tex]P(S\bigcap L)[/tex] = P(S/L) * P(L)
= 0.90 * 0.07 = 0.063
So, [tex]P(S\bigcap L)[/tex] = 0.063 .
Now, probability that a smoker has lung disease is given by = P(L/S)
P(L/S) = [tex]\frac{P(S\bigcap L)}{P(S)}[/tex]
P(S) = P(S/L) * P(L) + P(S/L') * P(L')
= 0.90 * 0.07 + 0.25 * 0.93 = 0.2955
Therefore, P(L/S) = [tex]\frac{0.063}{0.2955}[/tex] = 0.2132
Hence, probability that a smoker has lung disease is 0.2132 .
1. You have an aluminum bar of dimensions 2cm*5cm*10 cm. You want to put it into electric circuit such a way that this bar will demonstrate the smallest possible resistance. You should connect your bar to the opposite faces with dimensions of:
Answer:
( 5 x 10 ) cm
Step-by-step explanation:
Given:
- The dimensions of the bar are:
( 2 x 5 x 10 ) cm
Find:
Which two faces with dimensions ( _x _ ) should be connected to get smallest possible resistance.
Solution:
- The electrical resistance R of any material with density ρ and corresponding dimensions is expressed as:
R = ρ*L / A
- Where, A: cross sectional Area
L: The length in between the two faces.
- We need to minimize the electrical resistance of the bar. For that the Area must be maximized and Length should be minimized.
- A_max & L_min ---- > R_min
(5*10) & ( 2) ------> R_min
Hence, the electrical resistance is minimized by connecting the face with following dimensions ( 5 x 10 ) cm
A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?
After 1 second has passed, the balloon will be 215 ft above the ground, and the automobile will be 66 ft away from its initial position directly below the balloon, forming a right triangle whose hypotenuse has length [tex]\sqrt{215^2+66^2}=\sqrt{50,581}[/tex] ft.
The distance between the balloon and automobile [tex]d[/tex] is related to the height of the balloon [tex]b[/tex] and the distance [tex]a[/tex] the automobile has moved away from its starting position by the Pythagorean formula,
[tex]d^2=a^2+b^2[/tex]
Differentiate both sides with respect to time [tex]t[/tex]:
[tex]2d\dfrac{\mathrm dd}{\mathrm dt}=2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}[/tex]
Plug in everything you know and solve for the rate you want:
[tex]2(\sqrt{50,581}\,\mathrm{ft})\dfrac{\mathrm dd}{\mathrm dt}=2(66\,\mathrm{ft})\left(66\dfrac{\rm ft}{\rm s}\right)+2(215\,\mathrm{ft})\left(15\dfrac{\rm ft}{\rm s}\right)[/tex]
[tex]\implies\dfrac{\mathrm dd}{\mathrm dt}\approx33.71\dfrac{\rm ft}{\rm s}[/tex]
To find the rate at which the distance between the balloon and the car changes after 1 second, we can use the Pythagorean theorem and differentiate the equation with respect to time.
Explanation:To find the rate at which the distance between the balloon and the car changes, we can use the Pythagorean theorem. Let's call the distance between them D. We have D^2 = (200 + 15t)^2 + (66t)^2, where t is the time in seconds. To find how fast D is changing after 1 second, we can differentiate the equation with respect to t and substitute t = 1. Finally, we can solve for the rate at which D is changing.
Learn more about Rate of change here:https://brainly.com/question/29181502
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Female III-4 is pregnant via male III-5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV-3). If you need help with how to approach calculating these probabilities, use the Hint.
Answer: the following are required for IV-3 to have condition
Explanation:
-II-4 passes an X b chromosome to III-4 (probability = 1/2).
-If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).
-III-5 passes a Y chromosome to IV-3 (probability = 1/2).
All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).
Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).