Answer:
(a) The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b) The value of P (A₂ | A₃) is 0.7143.
(c) The events A₂ and A₃ are not independent.
(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Step-by-step explanation:
The events are defined as follows:
A₁ = an individual like vehicle #1
A₂ = an individual like vehicle #2
A₃ = an individual like vehicle #3
The information provided is:
[tex]P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88\\[/tex]
(a)
Compute the probability that the individual likes both vehicle #1 and vehicle #2 as follows:
[tex]P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40[/tex]
Thus, the probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b)
Compute the value of P (A₂ | A₃) as follows:
[tex]P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143[/tex]
Thus, the value of P (A₂ | A₃) is 0.7143.
(c)
If two events X and Y are independent then,
[tex]P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)[/tex]
The value of P (A₂ ∩ A₃) is 0.50.
The product of the probabilities, P (A₂) and P (A₃) is:
[tex]P(A_{2})\times P(A_{3})=0.65\times0.70=0.455[/tex]
Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)
The value of P (A₂ | A₃) is 0.7143.
The value of P (A₂) is 0.65.
Thus, P (A₂ | A₃) ≠ P (A₂).
The events A₂ and A₃ are not independent.
(d)
Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:
[tex]P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333[/tex]
Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Final answer:
The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40. Events A2 and A3 are not independent as P(A2 and A3) does not equal P(A2)P(A3). If the individual did not like vehicle #1, the probability of liking at least one of the other two vehicles is 0.73.
Explanation:
Probability of Likes and Independency of Events
To find the probability that the individual likes both vehicle #1 and vehicle #2, which is P(A1 ∩ A2), we use the formula: P(A1 ∩ A2) = P(A1) + P(A2) - P(A1 ∪ A2). Given that P(A1) = 0.55, P(A2) = 0.65, and P(A1 ∪ A2) = 0.80, we can calculate P(A1 ∩ A2):
P(A1 ∩ A2) = 0.55 + 0.65 - 0.80 = 0.40.
The conditional probability P(A2|A3) represents the probability of liking vehicle #2 given that the individual already likes vehicle #3. Using the definition of conditional probability:
P(A2|A3) = P(A2 ∩ A3) / P(A3) = 0.50 / 0.70 = approximately 0.714.
To determine if A2 and A3 are independent events, we can check if P(A2 ∩ A3) equals P(A2)P(A3) which would be 0.65 * 0.70 = 0.455. Since P(A2 ∩ A3) = 0.50, which is not equal to 0.455, A2 and A3 are not independent.
Moreover, if A2 and A3 were independent, then knowing that A3 occurred would not change the probability of A2 happening. However, P(A2|A3) is not equal to P(A2), indicating their dependence. Lastly, when the individual did not like vehicle #1, the probability that they liked at least one of the other two vehicles can be found using the probability of the union of events, subtracting the probability of liking vehicle #1:
P(A2 ∪ A3 | not A1) = P(A2 ∪ A3) - P(A1) + P(A1 ∩ A2 ∩ A3) = 0.88 - 0.55 + 0.40 = 0.73.
A tank contains 70 kg of salt and 1000 L of water. A solution of a concentration 0.035 kg of salt per liter enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank initially? concentration = (kg/L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. dy/dt = (kg/min) y(0) = (kg) (c) Solve the initial value problem in part (b). y(t) = (d) Find the amount of salt in the tank after 5 hours. amount = (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = (kg/L)
Answer:
(a) Initial concentration of salt = 0.07 kg/L
(b) [tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c) [tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)Therefore the amount of salt after 5 hours is =38.18 kg
(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L
Step-by-step explanation:
Given that,
A tank contains 70 kg of salt and 1000 L of water.
(a)
[tex]\textrm{Concentration of salt }=\frac{\textrm{mass of salt}}{\textrm{volume of water}}[/tex]
[tex]=\frac{70}{1000} kg/L[/tex]
=0.07 kg/L
(b)
Let Y(t) be the amount of salt at any instant time t.
Therefore
[tex]Y'(t)= Y_{in}-Y_{out}[/tex]
[tex]Y_{in}=\textrm{concentration of salt} \times \textrm{rate of enter}[/tex]
=(8×0.035) kg/min
=0.28 kg/min
Since the rate of water in and out are same , the amount of solution remain constant.
[tex]Y_{out}= (\frac{y}{1000}\times 8) kg/min[/tex]
[tex]=\frac{y}{125} kg/L[/tex]
[tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c)
The above equation can be rewrite as
[tex]Y'(t) +\frac{y}{125} =0.28[/tex]
The coefficient of y is p(t) [tex]=\frac{1}{125}[/tex]
The integrating factor of the D.E is[tex]=e^{\int p(t) dt=[/tex] [tex]e^{\int \frac{1}{125} dt[/tex] [tex]=e^{\frac{1}{125} t[/tex]
Multiplying the integrating factor of both sides of D.E
[tex]e^{\frac{1}{125} t} \ \frac{dY}{dt} +e^{\frac{1}{125} t} .\frac{1}{125} Y=0.28 \ e^{\frac{1}{125} t}[/tex]
Integrating both sides
[tex]\int e^{\frac{1}{125} t} \ \ dY+\int e^{\frac{1}{125} t} .\frac{1}{125} Y \ dt=\int0.28 \ e^{\frac{1}{125} t}\ dt[/tex]
[tex]\Rightarrow e^{\frac{1}{125} t} \ Y= \frac{0.28e^{\frac{1}{125}t} }{\frac{1}{125}} +C[/tex]
[tex]\Rightarrow Y=35+Ce^{-\frac{1}{125}t}[/tex]
At initial when t= 0, y =70
Therefore
[tex]70=35+Ce^0[/tex]
⇒C= 70-35
⇒C=35
Therefore
[tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)
When t= 5 hour = 300 min
To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E
Therefore the amount of salt after 5 hours is = [tex]Y(300)=35+35e^{-\frac{300}{125} }[/tex]
= 38.18 Kg
(e)
When t=∞
[tex]Y(\infty )= 35 +e^{-\infty}[/tex]
= 35 Kg [tex][e^{-\infty}=0][/tex]
Since the amount of water is remain same i.e 1000 L
Therefore the concentration of salt is [tex]=\frac{35}{1000}kg/L[/tex]
=0.035 kg/L
This problem involves using differential equations to model the concentration of salt in a tank over time. The initial concentration is determined, an initial value problem is set up, the initial value problem is solved, the amount of salt after a specific time is found, and the approaching concentration as time goes to infinity is identified.
Explanation:
(a) The initial concentration of the solution can be found by dividing the amount of salt by the volume of water. So, 70 kg / 1000 L = 0.07 kg/L.
(b) Let y be the amount of salt in the tank at time t. Then, dy/dt = 0.035 kg/L * 8 L/min - 0.07 kg/L * 8 L/min since salt is being added at a rate of 0.035 kg/L *8 L/min and leaving at a concentration of 0.07 kg/L * 8 L/min. Thus, dy/dt = 0.28 kg/min - 0.56 kg/min = -0.28 kg/min. The initial condition is y(0) = 70 kg.
(c) To solve this differential equation, we use the method of separation of variables. We get y(t) = 70 - 0.28t in kg.
(d) After 5 hours, or 300 minutes, the amount of salt in the tank is y(300) = 70 - 0.28*300 = 16 kg.
(e) As time approaches infinity, the concentration of salt will approach the incoming salt concentration, which is 0.035 kg/L since the solution is mixed and leaves at that rate.
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A certain kind of sheet metal has, on average, 9 defects per 11 square feet. Assuming a Poisson distribution, find the probability that a 14 square foot metal sheet has at least 10 defects. Round your answer to four decimals.
To find the probability that a 14 square foot metal sheet has at least 10 defects, we can use the Poisson distribution formula.
Explanation:To find the probability that a 14 square foot metal sheet has at least 10 defects, we can use the Poisson distribution formula.
The Poisson distribution formula is given by: P(x) = (e^-λ * λ^x) / x!
Where:
x is the number of defects (in this case, at least 10)λ is the average number of defects per unit area (in this case, 9 defects per 11 square feet)e is Euler's number, approximately 2.71828To find P(x ≥ 10), we need to calculate the probabilities for x = 10, 11, 12, 13, 14 and sum them up.
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The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to determine whether the variance of the population is significantly more than 0.003.
At 95% confidence, the null hypothesis:
a. should be rejected
b. should not be rejected
c. should be revised
d. None of these alternatives is correct
Answer:
Option b. should not be rejected
Step-by-step explanation:
We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.
We have to test whether the variance of the population is significantly more than 0.003, i.e.;
Null Hypothesis, [tex]H_0[/tex] : [tex]\sigma[/tex] = [tex]\sqrt{0.003}[/tex]
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\sigma > \sqrt{0.003}[/tex]
The test statistics used here for testing variance is;
T.S. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]
where, s = sample standard deviation = 0.06
n = sample size = 26 cans
So, Test statistics = [tex]\frac{(26-1)0.06^{2} }{0.003 }[/tex] ~ [tex]\chi^{2}__2_5[/tex]
= 30
So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.
We examined the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The scatterplot showed a linear form with strong positive association. Here is are the StatCrunch linear regression results. The r-sq value is about 0.75. What does this tell us? Simple linear regression results: Dependent Variable: Metascore Independent Variable: Rotten Tomatoes Metascore 21.605526+0.62434658 Rotten Tomatoes Sample size: 72 R (correlation coefficient) 0.86590 102 R-sq 0.74978458 Estimate of error standard deviation: 7.7978789a. 75 % of the Metascore ratings are accurately predicted by the regression line. b. Rotten Tomato ratings explain about 75 % of the variation in Metascore ratings.c. For each one point increase in Rotten Tomato ratings, we predict a 0.75 point increase in Metascore ratings.
Answer:
Therefore Rotten Tomato ratings explain about 75% of the variation in Metascore ratings is the correct answer here.
Step-by-step explanation:
The R2 value for regression or the coefficient of determination represents the proportion of variation in dependent variable that is explained by regression ( the rest of it is residual variation )
The given question is about the analysis of a linear regression model to examine the relationship between Rotten Tomatoes ratings and Metascore ratings for popular movies.
Explanation:The given question pertains to linear regression analysis in statistics. It involves examining the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The r-squared value of 0.75 indicates that approximately 75% of the variation in Metascore ratings can be explained by the linear relationship with Rotten Tomatoes ratings. Additionally, for every one-point increase in Rotten Tomatoes ratings, the regression model predicts a 0.75-point increase in Metascore ratings.
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On April 1, the unpaid balance in an account was $218. A payment of $30 was made on April 11. On April 21, a purchase was made. The finance charge rate was per month of the average daily balance. Find the new balance at the end of April. Group of answer choices
The question is not complete and the complete question is;
On April 1, the unpaid balance in an account was $218. A payment of $30 was made on April 11. On April 21, a $40 purchase was made. The finance charge rate was 15% per month of the average daily balance. Find the new balance at the end of April.
Answer:
Average daily balance = $211.33
Step-by-step explanation:
From the question, we know the following;
On April 1, the unpaid balance was $218.
On April 11, a payment of $30 was made.
On April 21, a $40 purchase was made.
Now, from april 1st till 10th, the unpaid balance was $218.
Since a $30 deposit was paid on the 11th,it means that from april 11th till 20th, the unpaid balance was $218 - $30 = $188
Also since a $40 purchase was made on April 21st,it means that from april 21st till 30th, the unpaid balance was $188 + $40 = 228
Now, since we want to find the average daily balance for the month, let's find the total daily balance and divide by the number of days in the month.
Total balance was:
10 days from april 1st till 10th = 10 x $218 = $2180.
10 days from april 11th till 20th = 10 x $188 = $1880.
10 days from april 21st till 30th = 10 x 228 = $2280.
Adding up, the total =$2180 + $1880+ $2280 = $6340
Now the average daily balance for the month which has 30 days = $6340/30 = $211.33
Matrix multiplication was used to encode a message using the given encoding matrix: [i 47 A=1 |-1 -3] The original message was converted to row matrices of size: 1x 2 and each was multiplied by A. sp 0 A 1 B 2 C 3 D 4 E F G H I J K L M N O P Q R 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 14 | 15 | 16 | 17 | 18 S T U V W X Y Z 19 20 21 22 23 24 25 26 Coded message: 11 52 -8 -9 -13 -39 5 20 12 56 5 20 -2 7 9 41 25 100 Find A and use it to decode the message
Answer:
[tex]A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right][/tex] message SHOW_ME_THE_MONEY_
Step-by-step explanation:
The matrix
[tex]A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\[/tex]
We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.
Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table
Encoded message Decoded message message in letters by association
11 52 19 8 S H
-8 -9 15 23 O W
-13 -39 0 13 _ M
5 20 5 0 E _
12 56 20 8 T H
5 20 5 0 E _
-2 7 13 15 M O
9 41 14 5 N E
25 100 25 0 Y _
Then the message decoded is SHOW_ME_THE_MONEY_
Final answer:
To decode the message, first find the inverse of the encoding matrix A, then multiply the encoded message matrices by this inverse. Match the resulting numbers to letters using the given table to reveal the original message.
Explanation:
The question involves the process of decoding a coded message that was encrypted using matrix multiplication with a given encoding matrix. Firstly, we need to decode the message by multiplying the encoded row matrices by the inverse of the encoding matrix A, which is provided as A = [[1 47] [-1 -3]]. Then, we will use the inverse of matrix A to transform the coded row matrices back to their original message form.
Once we have the row matrices that represent our numbers, we match these numbers to the corresponding letters using the given table, ultimately revealing the decoded message.
For example, if we have a decoded row matrix of [3 1], it would correspond to the letters "C" and "A" according to the provided letter to number mapping.
One model of a forklift truck can raise a maximum of 1750 kilograms. Write an in equality to describe the maximum number of 40 kilogram boxes that this forklift truck can raise
Answer:
The forklift truck can raise maximum of 43 boxes.
Step-by-step explanation:
We are given the following in the question:
Maximum limit of forklift truck =
1750 kilograms
Let x be the number of 40 kilogram boxes that this forklift truck can raise.
Thus, we can write the following inequality that describes the maximum number of boxes raised.
[tex]40x \leq 1750\\\\x \leq \dfrac{1750}{40}\\\\x \leq 43.75\\Max(x) = 43[/tex]
Thus, the forklift truck can raise maximum of 43 boxes.
A distribution of grades in an introductory statistics course (where A = 4, B = 3, etc) is: X 0 1 2 3 4 P(X) 0.1 0.17 0.21 0.32 0.2 Part a: Find the probability that a student has passed this class with at least a C (the student's grade is at least a 2).
Answer:
The probability that a student has passed this class with at least a C is 0.73
Step-by-step explanation:
A = 4 B = 3 C = 2 D = 1 E = 0
P(0) = 0.1
P(1) = 0.17
P(2) = 0.21
P(3) = 0.32
P(4) = 0.2
Probability that student passed this class with at least a C mean the score must be more than C
P(X\geq 2) = P(2) + P(3) + P(4)\\
P(X\geq 2) = 0.21 + 0.32 + 0.2\\
P(X\geq 2) = 0.73
Final answer:
The probability that a student has passed an introductory statistics class with at least a C grade is found by summing the probabilities for grades C and above, resulting in a probability of 0.73 or 73%.
Explanation:
The question asks for the probability that a student has passed an introductory statistics class with at least a C grade, which is represented as a grade of at least 2 when A = 4, B = 3, and so forth. The distribution of grades and their probabilities are given as follows: X represents the grade, and P(X) represents the probability of obtaining that grade. Therefore, to find the probability of passing the class with at least a C, we need to sum the probabilities of getting a grade of 2 or higher.
P(X=2) = 0.21
P(X=3) = 0.32
P(X=4) = 0.2
Adding these probabilities together:
P(X≥2) = 0.21 + 0.32 + 0.2 = 0.73
Therefore, the probability that a student has passed this class with at least a C is 0.73 or 73%.
When five basketball players are about to have a free-throw competition, they often draw names out of a hat to randomly select the order in which they shoot. What is the probability that they shoot free throws in alphabetical order? Assume each player has a different name. P(shoot free throws in alphabetical order)equals nothing (Type an integer or a simplified fraction.)
Answer:
1/120
Step-by-step explanation:
Since the players have different names, there is only one possible arrangement in which they are in alphabetical order. The total number of ways to order 5 basketball players (n) is:
[tex]n = 5!=5*4*3*2*1\\n=120[/tex]
Therefore, there is a 1/120 probability that they shoot free throws in alphabetical order.
To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. If the population standard deviation is 8.2 years, computer the standard error of the mean. (Round to one decimal place) What is the probability that the sample mean age of the employees will be within 2 years of the population mean age
Answer:
The standard error of the mean is 1.3.
87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
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, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\sigma = 8.2, n = 40[/tex]
Computer the standard error of the mean
[tex]s = \frac{8.2}{\sqrt{40}} = 1.3[/tex]
The standard error of the mean is 1.3.
What is the probability that the sample mean age of the employees will be within 2 years of the population mean age
This is the pvalue of Z when [tex]X = \mu + 2[/tex] subtracted by the pvalue of Z when [tex]X = \mu - 2[/tex]. So
[tex]X = \mu + 2[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{\mu + 2 - \mu}{1.3}[/tex]
[tex]Z = 1.54[/tex]
[tex]Z = 1.54[/tex] has a pvalue of 0.9382
-----
[tex]X = \mu - 2[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{\mu - 2 - \mu}{1.3}[/tex]
[tex]Z = -1.54[/tex]
[tex]Z = -1.54[/tex] has a pvalue of 0.0618
0.9382 - 0.0618 = 0.8764
87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age
The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a mean of 7 people per hour. How many people do you expect to arrive during a 55-minute period?
Answer:
I Expect 6.4167 persons arriving during a 55.minute period.
Step-by-step explanation:
If the mean in one hour (60 minutes) is 7, then in a 55 minute period you can expect, by using a rule of three, 7*55/60 = 6.4167 persons. This is because the amount of expected people is always proportional to the amount of time spent waiting for poisson distributions.
An experiment to investigate the survival time in hours of an electronic component consists of placing the parts in a test cell and running them under elevated temperature conditions. Six samples were tested with the following resulting failure times (in hours): 34, 40, 46, 49, 61, 64. (a) Calculate the sample mean and sample standard deviation of the failure time. (b) Determine the range of the true mean at 95% confidence level. (c) If a seventh sample is tested, what is the prediction interval (95% confidence level) of its failure time
Answer:
Step-by-step explanation:
The detailed steps and appropriate formular is as shown in the attached file.
The answers are :
(a) [tex]\text{Sample Mean is}[/tex] [tex]49[/tex] [tex]\text{and Sample Standard Deviation is}[/tex] [tex]11.7[/tex].
(b) [tex]95\%[/tex][tex]\text{Confidence Interval for True Mean} (36.72, 61.28).[/tex]
(c) [tex]95\% \text{Prediction Interval for a New Sample is} (16.54, 81.46)[/tex]
(a) Calculate the Sample Mean and Sample Standard Deviation
First, let's calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).
Given data: [tex]34, 40, 46, 49, 61, 64[/tex]
[tex]\text{Sample Mean} (\(\bar{x}\))[/tex]
[tex]\[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1}{6} (34 + 40 + 46 + 49 + 61 + 64) = \frac{294}{6} = 49\][/tex]
Sample Standard Deviation (s)
[tex]\[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}][/tex]
[tex]\[\begin{aligned}s & = \sqrt{\frac{1}{5} ((34-49)^2 + (40-49)^2 + (46-49)^2 + (49-49)^2 + (61-49)^2 + (64-49)^2)} \\& = \sqrt{\frac{1}{5} (225 + 81 + 9 + 0 + 144 + 225)} \\& = \sqrt{\frac{1}{5} \times 684} \\& = \sqrt{136.8} \\& = 1.7\end{aligned}\][/tex]
(b) Determine the Range of the True Mean at [tex]95\%[/tex] Confidence Level
To find the 95% confidence interval for the mean, we use the formula:
[tex]\[\bar{x} \pm t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}\][/tex]
For [tex]\(n = 6\)[/tex], degrees of freedom [tex]= \(n-1 = 5\)[/tex]. Using a t-table, the critical value for [tex]95\%[/tex] confidence and [tex]5[/tex] degrees of freedom is approximately [tex]2.571[/tex]
[tex]\[\begin{aligned}\text{Margin of Error} & = t_{\alpha/2, n-1} \frac{s}{\sqrt{n}} \\& = 2.571 \times \frac{11.7}{\sqrt{6}} \\& = 2.571 \times 4.776 \\& = 12.28\end{aligned}\][/tex]
So, the [tex]95\%[/tex] confidence interval is:
[tex]\[49 \pm 12.28 \Rightarrow (36.72, 61.28)\][/tex]
(c) Prediction Interval ([tex]95\%[/tex] Confidence Level) for a Seventh Sample
The prediction interval for a new observation is calculated using:
[tex]\[\bar{x} \pm t_{\alpha/2, n-1} s \sqrt{1 + \frac{1}{n}}\][/tex]
[tex]\[\begin{aligned}\text{Prediction Interval} & = 49 \pm 2.571 \times 11.7 \times \sqrt{1 + \frac{1}{6}} \\& = 49 \pm 2.571 \times 11.7 \times \sqrt{1.1667} \\& = 49 \pm 2.571 \times 11.7 \times 1.080 \\& = 49 \pm 32.46\end{aligned}\][/tex]
So, the prediction interval is:
[tex]\[(16.54, 81.46)\][/tex]
"6. A brand name has a 40% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 6 randomly selected consumers. a. What is the probability that exactly 5 of the 6 consumers recognize the brand name? b. What is the probability that all of the selected consumers recognize the brand name? c. What is the probability that at least 5 of the selected consumers recognize the brand name? d. If 6 consumers are randomly selected, is 5 and unusually high number of consumers that recognize the brand name?"
Answer:
(a) The probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.
(b) The probability that all of the selected consumers recognize the brand name is 0.0041.
(c) The probability that at least 5 of the selected consumers recognize the brand name is 0.041.
(d) The events of 5 customers recognizing the brand name is unusual.
Step-by-step explanation:
Let X = number of consumer's who recognize the brand.
The probability of the random variable X is, P (X) = p = 0.40.
A random sample of size, n = 6 consumers are selected.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X = 5) as follows:
[tex]P(X=5)={6\choose 5}0.40^{5}(1-0.40)^{6-5}\\=6\times0.01024\times0.60\\=0.036864\\\approx0.0369[/tex]
Thus, the probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.
(b)
Compute the value of P (X = 6) as follows:
[tex]P(X=6)={6\choose 6}0.40^{6}(1-0.40)^{6-6}\\=1\times 0.004096\times1\\=0.004096\\\approx0.0041[/tex]
Thus, the probability that all of the selected consumers recognize the brand name is 0.0041.
(c)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = P (X = 5) + P (X = 6)
= 0.0369 + 0.0041
= 0.041
Thus, the probability that at least 5 of the selected consumers recognize the brand name is 0.041.
(d)
An event is considered unusual if the probability of its occurrence is less than 0.05.
The probability of 5 customers recognizing the brand name is 0.0369.
This probability value is less than 0.05.
Thus, the events of 5 customers recognizing the brand name is unusual.
A magazine article conducted a survey of 525 people in New York City and found that 30% of the population believe that the Yankees will miss the playoffs this year. In the accompanying dialogue, the article states, we are 92% confident that the true proportion of people in New York City who believe that the Yankees will miss the playoffs this year lies between 25% and 35% . What does 30% represent in the article?
Answer:
In this case, the 30% represents the proportion of the sample. It is a statistic that can be used to estimate a parameter of the population.
Step-by-step explanation:
In this case, the 30% represents the proportion of this specific sample (survey taken by the magazine).
It is a statistic that can be used to estimate a parameter of the population. In this case, it may be used to estimate the true proportion of "people in New York who believe that the Yankees will miss the playoffs this year".
If a new sample is taken, a new statistic will be calculated that may or may not be equal to 30%.
The 30% in the article represents the proportion of people in New York City who believe that the Yankees will miss the playoffs this year. The article provides a confidence interval of 25% to 35% for this proportion, indicating a 92% confidence level.
Explanation:The 30% in the article represents the proportion of people in New York City who believe that the Yankees will miss the playoffs this year. The article states that the survey found that 30% of the population had this belief. Additionally, the article provides a confidence interval of 25% to 35% for this proportion, implying that there is a 92% confidence that the true proportion lies within this range.
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Clayton and Timothy took different sections of Introduction to Economics. Each section had a different final exam. Timothy scored 83 out of 100 and had a percentile rank in his class of 72. Clayton scored 85 out of 100 but his percentile rank in his class was 70. Who performed better with respect to the rest of the students in the class, Clayton or Timothy? Explain your answer.
Answer:
Since Timothy scored on the higher percentile, he performed better with respect to the rest of the students in the class.
Step-by-step explanation:
When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.
So, when comparing scores, whoever scored on the higher percentile scored more than a higher percentage of students, so had the better relative score.
In this problem, we have that
Timothy scored on the 72nd percentile.
Clayton scored on the 70th percentile.
Since Timothy scored on the higher percentile, he performed better with respect to the rest of the students in the class.
determine the solution on the interval [0, 2pi] for the equation
[tex]sec^{2} x-2=0[/tex]
Step-by-step explanation:
sec² x − 2 = 0
sec² x = 2
cos² x = ½
cos x = ±√½
x = π/4, 3π/4, 5π/4, 7π/4
Assumptions underlying the independent-measures t-test According to the theory of stereotype threat, situational pressures can lead to decreased performance on tests of cognitive abilities. Joshua Aronson tested how Caucasian engineering students performed on a math test when placed under a form of situational pressure by randomly assigning 100 of these students to either the control or experimental groups. The control group was told that they were taking a test of general math ability. Members of the experimental group were presented with several news articles discussing the increasing difference in math scores between Asian and Caucasian students and were told that the purpose of the test was to explore these differences.
Would it be valid for Dr. Aronson to use the independent-measures t-test to test whether drawing attention to stereotypes about racial groups and math ability affects math scores?
A. No, because the two populations from which the samples are selected are not normally distributed.
B. Yes, because the sample size is large and there is no reason to believe the assumptions of the independent-measures t-test are violated.
C. No, because the two groups studied are not independent.
D. No, because the variances within the two samples are different.
The answer is B. It is valid for Dr. Aronson to use the independent-measures t-test to investigate the effect of stereotype threat on math test performance among Caucasian engineering students. The t-test, which compares the means of two groups, is appropriate as long as the assumptions of independent measures, normal distribution, and homogeneity of variances are met.
Explanation:The answer is B. The independent-measures t-test is a statistical test used to compare the means of two independent groups to determine if the differences between them are statistically significant. It is valid for Dr. Aronson to use this test because he has two groups of participants (control and experimental) who are separate and independently assigned. The large sample size also makes it more likely that the distribution of scores follows a normal pattern, even if this is not guaranteed.
The assumptions underlying the independent-measures t-test include the independence of observations, normal distribution of the population, and homogeneity of variance (equal variances). In this scenario, nothing suggests these assumptions would be violated. The students are independently assigned to the groups and their scores would not affect one another, satisfying the independence of observations.
Although we do not know for sure if the populations from which the samples come are normally distributed or if the variances within both samples are equal, these are assumptions of the test and not data dependent, therefore we can't definitively argue they are reasons not to use the test.
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The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given the half-life of radium is 1690 years: (a) Write a differential equation that models the amount A of radium present at time t (b) Find the general solution of the differential equation. (c) Find the particular solution of the differential equation with the initial condition A(0)=10 g. (d) How much radium will be present in the sample at t=300 years?
Answer:
a) [tex]\frac{dm}{dt} = -k\cdot m[/tex], b) [tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex], c) [tex]m(t) = 10\cdot e^{-\frac{t}{2438.155} }[/tex], d) [tex]m(300) \approx 8.842\,g[/tex]
Step-by-step explanation:
a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:
[tex]\frac{dm}{dt} = -k\cdot m[/tex]
b) The general solution is found after separating variables and integrating each sides:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]\tau[/tex] is the time constant and [tex]k = \frac{1}{\tau}[/tex]
c) The time constant is:
[tex]\tau = \frac{1690\,yr}{\ln 2}[/tex]
[tex]\tau = 2438.155\,yr[/tex]
The particular solution of the differential equation is:
[tex]m(t) = 10\cdot e^{-\frac{t}{2438.155} }[/tex]
d) The amount of radium after 300 years is:
[tex]m(300) \approx 8.842\,g[/tex]
Answer:
a) -dm/m = k*dt
b) m=m₀*e^(-k*t)
c) m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)
d) m= 8.842 gr
( same results obtained previously by Xero099)
Step-by-step explanation:
Since the amount of radium is proportional to its quantity . Denoting as m the mass of Radium , and t as time. we have:
a) -dm/dt = k*m , where k is the proportionality constant
b) solving the differential equation;
-dm/m = k*dt
∫dm/m = -∫k*dt
ln m = -k*t + C
for t=0 , we have m=m₀ (initial mass)
then
ln m₀ = 0 + C → C=ln m₀
therefore
ln (m/m₀) = -k*t
m=m₀*e^(-k*t)
c) for t= 1690 years , the quantity of radium is half , then m=m₀/2 , thus
ln (m₀/2/m₀) = -k*t
ln (1/2) = -k*t
(ln 2 )/t = k
k = ln 2 / 1690 years = 4.1* 10⁻⁴ years⁻¹
then for m₀= 10 g
m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)
d) at t= 300 years
m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*300 years) = 8.842 gr
then the mass is m= 8.842 gr
Check our blood pressure: In a recent study, the Centers for Disease Control and prevention reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.7 and a standard deviation of 10.1. (a) What proportion of women have blood pressures lower than 68?
Answer:
The proportion of the women that have blood pressures lower than 68 is 0.1038 or 10.38%.
Step-by-step explanation:
Given:
Blood pressure required (x) = 68
Mean blood pressure (μ) = 80.7
Standard deviation (σ) = 10.1
The distribution is normally distributed.
So, first, we will find the z-score of the distribution using the formula:
[tex]z=\frac{x-\mu}{\sigma} [/tex]
Plug in the values and solve for 'z'. This gives,
[tex]z=\frac{68-80.7}{10.1}=-1.26[/tex]
So, the z-score of the distribution is -1.26.
Now, we need the probability [tex]P(x\leq 68 )=P(z\leq -1.26)[/tex].
From the normal distribution table for z-score equal to -1.26, the value of the probability is 0.1038. This is the area to the left of the curve or less than z-score value which is what we need.
Therefore, the proportion of the women that have blood pressures lower than 68 is 0.1038 or 10.38%.
A study of stress on the campus of your university reported a mean stress level of 76 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 4 for 95% confidence. The study was based on a random sample of 49 undergraduates.Give the 95% confidence interval.
Answer:
The 95% confidence interval for population mean is (72, 80).
Step-by-step explanation:
The (1 - α) % confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]
The Margin of error for this confidence interval is:
[tex]MOE=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]
The confidence interval for μ can also be written as:
[tex]CI=\bar x\pm MOE[/tex]
Given:
[tex]\bar x=76\\MOE=4[/tex]
Compute the 95% confidence interval for population mean as follows:
[tex]CI=\bar x\pm MOE\\=76\pm4\\=(72, 80)[/tex]
Thus, the 95% confidence interval for population mean is (72, 80).
Answer:
[tex] \bar X \pm ME[/tex]
And if we find the limits we got:
[tex] 76-4 = 72[/tex]
[tex] 76 + 4 = 80[/tex]
So then the 95% confidence interval would be (72.0,80.0)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X = 76[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n=49 represent the original sample size
Confidence =95% or 0.95
ME=4 represent the margin of error.
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The margin of error is defined as:
[tex] ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The formula for the confidence interval is equivalent to:
[tex] \bar X \pm ME[/tex]
And if we find the limits we got:
[tex] 76-4 = 72[/tex]
[tex] 76 + 4 = 80[/tex]
So then the 95% confidence interval would be (72.0,80.0)
Taylor took 6 hours to drive home from college for Thanksgiving break, a total distance of 290 miles.
She was able to average 50 miles per hour for part of the trip but had to slow down to 45 miles per hour
for the rest of the time due to poor weather. How many hours did she drive at each speed?
Answer: at 50 mph, she drove for 4 hours.
at 45 mph, she drove for 2 hours.
Step-by-step explanation:
Let t represent the time that she spent driving at 50 miles per hour.
Taylor took 6 hours to drive home from college for Thanksgiving break. This means that the time that she spent driving at 45 miles per hour is (6 - t) hours.
Distance = speed × time
Distance covered while driving 50 miles per hour is
50t
Distance covered while driving 45 miles per hour is
45(6 - t)
Since the total distance that she drove is 290 miles, it means that
50t + 45(6 - t) = 290
50t + 270 - 45t = 290
50t - 45t = 290 - 270
5t = 20
t = 20/5 = 4
At 45 miles per hour, she drove at
6 - 4 = 2 hours
we have 350 m^2 of material to build a box whose base width is four times its base length. determine the dimensions of the box that will maximize the enclosed volume
A box with a base width four times its length, built from 350 m^2 material, will have maximum volume when its dimensions are: length = sqrt(35) m, width = 4sqrt(35) m, and height = 350/(12*sqrt(35)) m.
Explanation:We know that the box's area, which is the sum of all the faces, is given by 2lw+2lh+2wh = 350 square meters, where l is the length, w is the width, and h is the height. Since the base width is four times the base length, we can say w=4l. Substituting this into the equation, we solve for h, h = (350 - 2l^2) / (6l).
The volume of the box is given by V = lwh. Substituting the value of h from the previous equation, we get V = l * 4l * (350 - 2l^2) / (6l) = 4/3 *l^2 *(175 - l^2). To maximize the volume, we need to find the value of l when the derivative of the volume equation is 0. Differentiating, setting the derivative equal to 0, and solving for l, we find l = sqrt(35), and substituting this into the earlier equation, we find h = 350/(12*sqrt(35)). The dimensions of the box that will maximize volume are, length = sqrt(35) m, width = 4sqrt(35) m, and height = 350/(12*sqrt(35)) m.
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Choose the correct meaning of a double-blind, placebo-controlled experiment. In a double-blind, placebo-controlled experiment, a subject does not know whether he or she received a treatment or an inactive substance. subjects with similar characteristics are assigned to the same group. subjects are assigned to different treatment groups, one of which receives an inactive substance that appears to be a treatment, through random selection. a subject who received an inactive substance reports an improvement in health or behavior. neither the subjects nor the people administering the treatments know who received a treatment and who received an inactive substance.
Answer:
Step-by-step explanation:
Hello!
A double-blind experiment is a type of experiment where both the experimental units and the researchers that analyze the data don't know what kind of treatment was applied to each subject.
This means, that if it is a placebo-controlled experiment. The subjects will be randomly assigned either the medicament to test ("treatment" group) or the placebo ( "control" group) but they will not know which one they are taking. This way the placebo effect is eliminated.
On the other hand, in other to eliminate the observer bias, the researchers will also not know which patient belongs to the "control" group and wich patient belongs to the "treatment" group.
I hope it helps!
The base of a rectangular tank measures 10 ft by 20 ft. The tank is 16 ft tall, and its top is 10 ft below ground level. The tank is full of water weighing 62.4 lb/ft3. How much work does it take to empty the tank by pumping the water to ground level? Give your answer to the nearest ft ∙ lb.
Answer:
Step-by-step explanation:
Given a rectangular tank with dimension (10ft by 20ft by 16ft)
Then the volume of the tank is
Volume =length × breadth ×height
Volume=10×20×16=3200ft³
V=3200ft³
Then,
∆F= weight density × volume
∆F= 62.4×3200
∆F= 199,680 lb
Then,
Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft
Then ∆x=(26-x)ft
Work is given as
W= -∫F∆xdx. From 0 to 16
W= -∫199,680(26-x)dx From 0 to 16
W=-199,680∫26-x dx From 0 to 16
W=-199,680 (26x-x²/2). From 0 to 16
W=-199,680(26×16-0.5×16² -0-0)
W=-199,680(288)
W=-57,507,840J
The work done to empty the tank is 57,507,840J
Answer:
2 X 10∧6 ft.lb
Step-by-step explanation:
Volume of rectangular tank = Base area X Height = (10 X 20 X 16)ft³ = 3200ft³
Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg
Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)
∴ Δg = 90706.37 * 9.8 * 3.05 = 2.711 X10∧6 J or 2 X 10∧6 ft.lb
According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction. That is, these are tweets that are not replied to or retweeted (Sysomos website, January 5, 2015).
Suppose we randomly select 100 tweets.
(a) What is the expected number of these tweets with no reaction?
(b) What are the variance and standard deviation for the number of these tweets with no reaction?
Answer:
Expected Number=71
Variance =4959
Standard Deviation= 70.42
Step-by-step explanation:
Expected Number = E (x) = np
Here p = 0.71 and q= 1- 0.71= 0.29 n= 100
Expected Number = E (x) = np = 0.71*100= 71
b) The Variance and Standard Deviation for the number of these tweets with no reaction
Suppose the number of tweets are hundred then the number of tweets with no reaction would be 71 .
Variance = E(x²) - [E(x)]² = (100)²- (71)²= 10000- 5041= 4959
Standard Deviation= √4959= 70.42
For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that 52% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 1,000 companies. Compute the margin of error and a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage. If required, round your answer to four decimal places. Round intermediate calculations to four decimal places.
Final answer:
To find the margin of error and 95% confidence interval for the proportion of companies likely to require higher health care contributions, we use a sample proportion of 0.52 and a sample size of 1000 to calculate a standard error of 0.0158. Multiplying this by the z-value of 1.96 gives a margin of error of 0.0310. The 95% confidence interval is thus (0.4890, 0.5510).
Explanation:
The question asks us to calculate the margin of error and a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage, based on a sample of 1,000 companies where 52% are likely to require higher contributions.
We start by identifying the sample proportion (p-hat) and the sample size (n):
p-hat = 0.52
n = 1000
The formula to calculate the standard error (SE) of the proportion is:
[tex]SE = \sqrt{((p-hat * (1 - p-hat)) / n)[/tex]
Using this formula, we calculate the SE:
[tex]SE = \sqrt{((0.52 * (1 - 0.52)) / 1000)[/tex]
[tex]SE = \sqrt{((0.52 * 0.48) / 1000)[/tex]
[tex]SE = \sqrt{(0.2496 / 1000)[/tex]
[tex]SE = \sqrt{(0.0002496)[/tex]
SE = 0.0158
Now we determine the z-value for a 95% confidence interval, which is commonly 1.96.
The margin of error (ME) is calculated by multiplying the z-value by the SE:
ME = z * SE
ME = 1.96 * 0.0158
ME = 0.0310
Finally, we calculate the 95% confidence interval with the formula:
(p-hat - ME, p-hat + ME)
(0.52 - 0.0310, 0.52 + 0.0310)
(0.4890, 0.5510)
Therefore, the margin of error is 0.0310 and the 95% confidence interval for the proportion of companies is (0.4890, 0.5510).
What is the probability that a randomly selected individual is more likely to buy a product emphasized as "Made in our country," given the individual is at least 55 years of age?
Answer:
The probability that a person with at least 55 years is more likely to buy the product will be 31.12%
Step-by-step explanation:
The exercise gives a condition which decreases the table to particular rows. The condition is "more likely" to buy a product, we will only work with the row with the label is "more likely". Therefore, the denominator of our equation will be 1301.
The exercise asks for the probability that a person with at least 55 years given the condition. So, we have:
Probability = (405/1301) x100 = 31.12%
Answer:
0.727
Step-by-step explanation:
The probability of an event can be estimated by taking the ratio of the number of times the events occur and the total number of possible outcomes. In the problem, the condition given is that the person is at least 55 years of age. Considering the column for 55+, the total number of people in the age bracket is 557 and 405 people are more likely to purchase the product. Thus, the probability is: 405/557 = 0.727.
A rectangle is growing such that the length of a rectangle is 5t+4 and its height is t4, where t is time in seconds and the dimensions are in inches. Find the rate of change of area, A, with respect to time. g
Answer:
Therefore the rate of change of area [tex]25t^4+16t^3[/tex] square inches/ s
Step-by-step explanation:
Given that,
A rectangle is growing such that the length of rectangle is(5t+4) and its height is t⁴.
Where t is in second and dimensions are in inches.
The area of a rectangle is = length× height
Therefore the area of the rectangle is
A(t) = (5t+4) t⁴
⇒A(t) = t⁴(5t+4)
To find the rate change of area we need to find out the first order derivative of the area.
Rules:
[tex](1)\frac{dx^n}{dx} = nx^{n-1}[/tex]
[tex](2) \frac{d}{dx}(f(x) .g(x))= f'(x)g(x)+f(x)g'(x)[/tex]
A(t) = t⁴(5t+4)
Differentiate with respect to t
[tex]\frac{d}{dt} A(t)=\frac{d}{dt} [t^4(5t+4][/tex]
[tex]\Rightarrow \frac{dA(t)}{dt} =(5t+4)\frac{dt^4}{dt} +t^4\frac{d}{dt} (5t+4)[/tex]
[tex]\Rightarrow \frac{dA(t)}{dt} = (5t+4)4t^{4-1}+t^4.5[/tex]
[tex]\Rightarrow \frac{dA(t)}{dt} =4t^3(5t+4)+5t^4[/tex]
[tex]\Rightarrow \frac{dA(t)}{dt} = 20t^4+16t^3+5t^4[/tex]
[tex]\Rightarrow \frac{dA(t)}{dt} = 25t^4+16t^3[/tex]
Therefore the rate of change of area [tex]25t^4+16t^3[/tex] square inches/ s
The rate of change of the rectangle's area, with respect to time, is obtained by differentiating the expression for the area with respect to time. Using the product rule, the derivative is found to be (5t4 + 4t3)(5+4t) inches per second.
Explanation:The subject of this question is primarily calculus, focusing on the concept of derivatives and rates of change. The area, A, of the rectangle can be found by multiplying the length by the height, hence A = (5t+4)t4. To find the rate of change of area with respect to time, we differentiate A with respect to time, t. This gives us the derivative dA/dt = d/dt [(5t+4)t4]. Using the product rule for differentiation (uv)' = u'v + uv', we find dA/dt = (5t4 + 4t3)(5+4t). Hence, the rate of change of the area with respect to time is (5t4 + 4t3)(5+4t) inches per second.
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PLEASE HELPPPPPP
The perimeter is 50 ft and the length is 15 ft.
What's the width?
Answer:
10
Step-by-step explanation:
I assume it's a rectangle, therefore:
P = 50
a = 15 ft
b = ?
P = 2a + 2b
50 = 2 * 15 + 2b
50 = 30 + 2b
2b = 50 - 30
2b = 20
b = 10
he labor force participation rate is the number of people in the labor force divided by the number of people in the country who are of working age and not institutionalized. The BLS reported in February 2012 that the labor force participation rate in the United States was 63.7% (Calculatedrisk). A marketing company asks 120 working-age people if they either have a job or are looking for a job, or, in other words, whether they are in the labor force. What is the probability that fewer than 60% of those surveyed are members of the labor force?
Answer:
[tex] z= \frac{0.6-0.637}{0.0439} =-0.843[/tex]
So then we can find the probability like this:
[tex]P(p<0.6) = P(Z<-0.843)[/tex]
And using the normal standard table or excel we got:
[tex]P(p<0.6) = P(Z<-0.843)=0.1996[/tex]
Step-by-step explanation:
For this case we can check if we can use the normal approximation for the proportion and we have this:
[tex] np = 120*0.637 =76.44 >10[/tex]
[tex] n(1-p) = 120*(1-0.637) = 43.56>10[/tex]
Then we can conclude that we can use the normal approximation. And we have this:
[tex] p\sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]
So the mean is given by:
[tex]\mu_p = 0.637[/tex]
And the deviation is given by:
[tex]\sigma_p = \sqrt{\frac{0.637*(1-0.637)}{120}}= 0.0439[/tex]
And for this case we want to find this probability:
[tex] P( p<0.6)[/tex]
And we can use the z score given by:
[tex] z = \frac{p -\mu}{\sigma_p}[/tex]
And for this case the z score is:
[tex] z= \frac{0.6-0.637}{0.0439} =-0.843[/tex]
So then we can find the probability like this:
[tex]P(p<0.6) = P(Z<-0.843)[/tex]
And using the normal standard table or excel we got:
[tex]P(p<0.6) = P(Z<-0.843)=0.1996[/tex]