A minimum of 10 servers is needed to keep the average time in the system under 6 minutes.
What is the utilization factor?The utilization factor is the ratio of the power station's maximum demand to its rated capacity. The time that equipment is in use./ The total time that it could be in use.
Given, In a record store with a service rate of 15 per hour, customers typically enter at a rate of 1 per minute (exponential interarrivals) (exponential service times).
Customers Per Minute into record shop = 1 per minute
Per Hour = 1 * 60 = 60
Number of Customers > Per Hour/(Server)
6 > 60/ server
server > 10
Therefore, as per the utilization factor, a minimum of 10 servers should be installed to keep the average time under 6 minutes.
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The given scenario can be modeled as a queuing system, where the arrival rate (\(\lambda\)) and service rate (\(\mu\)) are given.
From the information provided:
- The arrival rate, \(\lambda\), is 1 customer per minute, hence \(\lambda = 1\).
- The service rate, \(\mu\), is 15 customers per hour. Since we are working with minutes, we need to convert this to minutes. There are 60 minutes in an hour, so \( \mu = \frac{15}{60} = 0.25 \) customers per minute.
The goal is to keep the average time in the system, \( W \), under 6 minutes. The average time in the system for a M/M/c queue (which is the kind we're dealing with since both arrival and service rates are exponentially distributed) can be calculated using the formula for the average time in a system with \(c\) servers:
\[ W = \frac{W_q + 1}{\mu} \]
where:
- \(W_q\) is the average time a customer spends waiting in the queue.
- \(\mu\) is the service rate per server.
- The term \( \frac{1}{\mu} \) is the average service time.
The queue time \(W_q\) depends on the number of servers \(c\) and the utilization factor \(\rho = \frac{\lambda}{c\mu}\). The queue time increases with the utilization factor. The formula for \( W_q \) in an M/M/c queue is complex as it involves Erlang B formulas and iterative methods to solve for different values of \(c\).
However, for the practical purposes of this problem, we can use the approximation for \( W \) without explicitly calculating \( W_q \) as:
\[ W = \frac{1}{\mu - \lambda} \]
provided that this system is stable, which occurs only if \( \lambda < c \cdot \mu \).
We need to find the minimum number of servers \( c \) such that the average time in the system \( W \) is less than 6 minutes:
\[ W < 6 \]
Using the approximate formula:
\[ \frac{1}{c\mu - \lambda} < 6 \]
Therefore:
\[ c\mu - \lambda > \frac{1}{6} \]
Solving for the number of servers \( c \):
\[ c > \frac{\lambda + \frac{1}{6}}{\mu} \]
Plugging in the values of \(\lambda\) and \(\mu\):
\[ c > \frac{1 + \frac{1}{6}}{0.25} \]
\[ c > \frac{\frac{6}{6} + \frac{1}{6}}{{0.25}} \]
\[ c > \frac{\frac{7}{6}}{0.25} \]
\[ c > \frac{7}{6} \cdot \frac{1}{0.25} \]
\[ c > \frac{7}{6} \cdot 4 \]
\[ c > \frac{7 \cdot 2}{3} \]
\[ c > \frac{14}{3} \]
As we cannot have a fraction of a server, we need to round up to the nearest whole number:
\[ c > 4.67 \]
So we need a minimum of 5 servers to keep the average time in the system under 6 minutes.
?
1. What is the solution of the system
3x - y = 14,
4x + y = 21
A (-5,3)
B (4, -2)
C (5, 1)
Answer:
The answer to your question is letter C
Step-by-step explanation:
Equations
3x - y = 14 ----------- (I)
4x + y = 21 ----------- (II)
Solve the system of equations by the elimination method.
3x - y = 14
4x + y = 21
7x 0 = 35
Solve for x, divide both sides by 7
7x/ 7 = 35/7
x = 5
Substitute x in equation I
3(5) - y = 14
15 - y = 14
Subtract 15 in both sides
15 - 15 - y = 14 - 15
Simplify
- y = -1
Multiply by -1
y = 1
Solution
(5, 1)
An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 6.3 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 130 engines and the mean pressure was 6.5 pounds/square inch. Assume the standard deviation is known to be 0.8. A level of significance of 0.02 will be used. Determine the decision rule. Enter the decision rule.
Answer:
We conclude that the valve performs above the specifications.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 6.3 pounds per square inch
Sample mean, [tex]\bar{x}[/tex] = 6.5 pounds per square inch
Sample size, n = 130
Alpha, α = 0.02
Population standard deviation, σ = 0.8
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 6.3\text{ pounds per square inch}\\H_A: \mu > 6.3\text{ pounds per square inch}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{6.5 - 6.3}{\frac{0.8}{\sqrt{130}} } = 2.85[/tex]
Now, [tex]z_{critical} \text{ at 0.02 level of significance } = 2.05[/tex]
Decision rule:
If the calculated statistic is greater than the the critical value, we reject the null hypothesis and if the calculated statistic is lower than the the critical value, we accept the null hypothesis
Since,
[tex]z_{stat} > z_{critical}[/tex]
We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.
Thus, we conclude that the valve performs above the specifications.
Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.
Answer:
[tex]A)\,\,det(A)=1[/tex]
[tex]B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] [/tex]
Step-by-step explanation:
[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|[/tex]
Expanding with first row
[tex]det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1[/tex]
To find inverse we first find cofactor matrix
[tex]C_{1,1}=(-1)^{1+1}\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_{1,2}=(-1)^{1+2}\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_{1,3}=(-1)^{1+3}\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_{2,1}=(-1)^{2+1}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_{2,2}=(-1)^{2+2}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_{2,3}=(-1)^{2+3}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\[/tex]
[tex]C_{3,1}=(-1)^{3+1}\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_{3,2}=(-1)^{3+2}\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_{3,3}=(-1)^{3+3}\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\[/tex]
Cofactor matrix is
[tex]C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right][/tex]
Use the following information to answer the exercise. A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour.What are the dependent and independent variables?
Answer:
Dependent: fee
Independent: time
Step-by-step explanation:
The scenario described in question indicates that fee is the dependent variable and time is independent variable. The reasoning is that fee depends on the time(hour). The fee will increase as the time increases. Whereas change in time cause change in fee. So, fee is the dependent variable and time is independent variable.
Final answer:
In the SCUBA equipment rental situation, the independent variable is the number of hours the equipment is rented, and the dependent variable is the total fee. The equation for the total fee is y = 25 + 12.50x, with $25 representing the y-intercept and $12.50 the slope of the hourly rental cost.
Explanation:
For the SCUBA equipment rental situation at the vacation resort, the independent variable is the number of hours the equipment is rented. Since the renter can decide how many hours to rent the equipment, it is considered independent. On the other hand, the dependent variable is the total fee for renting the equipment because it depends on the number of hours the equipment is rented for.
In this context, the equation expressing the total fee (y) in terms of the number of hours (x) the equipment is rented is:
y = 25 + 12.50x.
Here, $25 is the up-front fee which is also the y-intercept of the equation, while $12.50 per hour is the slope. The y-intercept represents the initial cost of renting the equipment before any hourly charges apply, and the slope represents the rate at which the total cost increases for every additional hour the equipment is rented.
What was the annual growth rate for Egypt in 1990? What is the estimated annual growth rate in 2010? Be sure to show your work for each answer.
Answer:
A. The annual growth rate in Egypt in 1990 was 31 - 9 = 22 per 1,000 people. The estimated annual growth rate in 2010 is 25 - 5 = 20 per 1,000 people.
B. I feel like Egypt is in the third stage (Late Developing) in the Demographic transition model because the birth and death rate was higher but then as time progressed, they both started to level out. In order for Egypt to advance in the model, they will have to have a developed country (urbanization) and lower their birth/death rates through the use of birth controls and healthcare.
Step-by-step explanation:
Solve the following differential equations using classical methods. x¨ + 6 ˙x + 8x = 5sin(3t) x¨ + 4 ˙x = t2 ; x(0) = 1; ˙x = 2
Answer:
1. x = (199/130)e^(-2t) - (33/65)e^(-4t) - (18/65)cos3t - (1/65)sin3t
2. x = (191/128) - (63/128)e^(-4t) + (t³/12) - (t²/16) + (t/32)
Step-by-step explanation:
Steps are shown in the attachment.
when selcting from a set of 10 distinct integersif sampling is done with replacement, how many samples of 5 are possible
Answer:
10⁵ = 100,000
Step-by-step explanation:
Data provided in the question:
Number of available choices = 10
for the sample of 5 i.e n = 5
Repetition is allowed
Thus,
Total samples of 5 that are possible = ( Number of available choices )ⁿ
thus,
Total samples of 5 that are possible = 10⁵
Hence,
The Total samples of 5 that are possible = 100,000
The toco toucan, the largest member of the toucan family, possesses the largest beak relative to body size of all birds. This exaggerated feature has received various interpretations, such as being a refined adaptation for feeding. However, the large surface area may also be an important mechanism for radiating heat (and hence cooling the bird) as outdoor temperature increases. Here are data for beak heat loss, as a percent of total body heat loss, at various temperatures in degrees Celsius:
Temperature 15 16 17 18 19 20 21 22 23 24 25 26 27
Percent heat loss from beak 33 34 33 36 36 47 52 51 41 50 49 50 55
The equation of the least-squares regression line for predicting beak heat loss, as a percent of total body heat loss from all sources, from temperature is ______.
Answer:
[tex]m=\frac{332}{182}=1.824[/tex]
[tex]b=\bar y -m \bar x=43.615-(1.824*21)=5.311[/tex]
So the line would be given by:
[tex]y=1.824 x +5.311[/tex]
Step-by-step explanation:
We assume that the data is this one:
x: 15 16 17 18 19 20 21 22 23 24 25 26 27
y: 33 34 33 36 36 47 52 51 41 50 49 50 55
Find the least-squares line appropriate for this data.
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
So we can find the sums like this:
[tex]\sum_{i=1}^n x_i =273[/tex]
[tex]\sum_{i=1}^n y_i =567[/tex]
[tex]\sum_{i=1}^n x^2_i =5915[/tex]
[tex]\sum_{i=1}^n y^2_i =25547[/tex]
[tex]\sum_{i=1}^n x_i y_i =12239[/tex]
With these we can find the sums:
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=5915-\frac{273^2}{13}=182[/tex]
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=12239-\frac{273*567}{13}=332[/tex]
And the slope would be:
[tex]m=\frac{332}{182}=1.824[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{273}{13}=21[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{567}{13}=43.615[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=43.615-(1.824*21)=5.311[/tex]
So the line would be given by:
[tex]y=1.824 x +5.311[/tex]
The equation of the least-squares regression line for predicting beak heat loss from temperature is: Percent heat loss from beak = 0.943(temp) + 16.243
Explanation:The equation of the least-squares regression line for predicting beak heat loss, as a percent of total body heat loss from all sources, from temperature is:
Percent heat loss from beak = 0.943(temp) + 16.243
This equation can be obtained by performing a linear regression analysis on the given data points, where the temperature is the independent variable and the percent heat loss from the beak is the dependent variable.
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Chemistry is an experimental science in which a consistent system of measurements is used. Scientists throughout the world use the International System of Units (Système International or SI), which is based on the metric system. However, the metric system in not used on an everyday basis in the United States. It is often necessary to convert between the English system, which is used in the United States, and the SI system. One problem with the SI system is the size of the units sometimes turns out to be inconveniently large or small. Therefore it is often necessary to make conversions within the SI system. For this reason, SI units are modified through the use of prefixes when they refer to either larger or smaller quantities.
Part A
In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 22.0mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe? Use the following conversions:1km=0.6214mi and 1gal=3.78L.
Part B
While in Europe, if you drive 115km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per liter and your car's gas mileage is 30.0mi/gal ? Assume that 1euro=1.26dollars.
Part C
A sprinter set a high school record in track and field, running 200.0 m in 20.9s . What is the average speed of the sprinter in kilometers per hour?
Part D
A specific brand of gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece. How many kilograms of dietary fat are in a box containing 4.00lb of candy?
Answer:
(a) 142 km trip requires 15.16 Liters of Gasoline
(b) The money spent is 69.34 euros or $87.37.
(c) The average speed of sprinter is 34.45 km/h.
(d) 4 lb of candy contains 0.559 kg of dietary fat.
Step-by-step explanation:
(a)
Given that;
Mileage = 22 mi/gal
Converting it to km/L
Mileage = (22 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)
Mileage = 9.36 km/L
Now, the gasoline required for 142 km trip:
Gasoline Required = (Length of trip)/(Mileage)
Gasoline Required = (142 km)/(9.36 km/L)
Gasoline Required = 15.16 L
(b)
Given that;
Mileage = 30 mi/gal
Converting it to km/L
Mileage = (30 mi/gal)(1 gal/3.78 L)(1 km/0.6214 mi)
Mileage = 12.77 km/L
Now, the gasoline required for 142 km trip:
Gasoline Required = (Length of trip)/(Mileage)
Gasoline Required = (115 km/day)(7 days/week)/(12.77 km/L)
Gasoline Required = 63 L/week
Now, we find the cost:
Weekly Cost = (Gasoline Required)(Unit Cost)
Weekly Cost = (63 L/week)(1.1 euros/L)
Weekly Cost = 69.34 euros/week = $87.37
since, 1 euro = $1.26
(c)
Average Speed = (Distance Travelled)/(Time Taken)
Average Speed = 200 m/20.9 s
Average Speed = (9.57 m/s)(3600 s/1 h)(1 km/ 1000 m)
Average Speed = 34.45 km/h
(d)
22.7 g piece contains = 7 g dietary fat
(22.7 g)(1 lb/453.592 g) piece contains = (7 g)(1 kg/1000 g) dietary fat
0.05 lb piece contains = 0.007 kg dietary fat
1 lb piece contains = (0.007/0.05) kg dietary fat
4 lb piece contains = 4(0.007/0.05) kg dietary fat
4 lb piece contains = 0.559 kg dietary fat
Suppose you take the Medical College Admission Test (MCAT) and your score is the 32nd percentile. How do you interpret this result
In Statistics, percentiles are a representation of the relative position of a particular value within a data set. For example, if your exam score is better than k% of the rest of the class. That means your exam score is at the kth percentile.
If your test score is at the 32nd percentile it can be interpreted as follows:
-Your test score is better than only 32 percent of the other scores recorded for the test.
-32 percent of the people who took the admission test have scores which are lower than yours.
4-It has been a bad day for the market, with 70% of securities losing value. You are evaluating a portfolio of 20 securities and will assume a binomial distribution for the number of securities that lost value.
a- What assumptions are made when using this distribution.
b- Find the probability that all 20 securities lose value.
c- Find the probability that at least 15 of them lose value.
d- Find the probability that less than 5 of them gain value.
a) When using the binomial distribution for the number of securities that lost value, the following assumptions are made.
b) The probability that all 20 securities lose value is approximately \(0.0008\).
c) The probability that at least 15 of them lose value is the sum of the probabilities of having 15, 16, 17, 18, 19, or 20 securities losing value.
d) Probability that less than 5 of them gain value: Approximately 0.995872.
a) When using the binomial distribution for the number of securities that lost value, the following assumptions are made:
1. Each security in the portfolio has a fixed probability of losing value.
2. The outcomes of different securities are independent of each other.
3. There are only two possible outcomes for each security: it either loses value or it doesn't.
4. The probability of losing value remains constant for each security throughout the evaluation.
b) The probability that all 20 securities lose value can be calculated using the binomial probability formula:
[tex]\[ P(X = 20) = \binom{20}{20} \times 0.7^{20} \times (1 - 0.7)^0 \]\[ P(X = 20) = 1 \times 0.7^{20} \times 1 \]\[ P(X = 20) \approx 0.0008 \][/tex]
c) To find the probability that at least 15 of them lose value, we calculate the cumulative probability from 15 to 20:
[tex]\[ P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) \]\[ P(X \geq 15) \approx 0.9570 \][/tex]
d) Probability that Less Than 5 of Them Gain Value:
Replace k with the desired number and calculate the probabilities for X < k using the binomial probability formula.
For X < 5:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P(X < 5) ≈ 0.995872
Students in a cooking class made 4 1/2 quarts of soup. They served 4/5 of the soup to friends. Each serving is 3/5 quart. Hector incorrectly says that there were 3 3/5 servings of soup. What is the correct number of servings? What did Hector do wrong?
Answer:
The correct number of servings = 6.
By saying there were [tex]3\frac{3}{5}[/tex] servings of soup, Hector must have intended to say that there were [tex]3\frac{3}{5}[/tex] quarts of soup for serving.
Step-by-step explanation:
Given:
Total quarts of soup made by students = [tex]4\frac{1}{2}[/tex]
Fraction of the soup served to friends = [tex]\frac{4}{5}[/tex]
Each serving = [tex]\frac{3}{5}[/tex] quarts
Hector incorrectly says that that there were [tex]3\frac{3}{5}[/tex] servings of soup.
To find the correct number of servings and to identify Hector's mistake.
Solution:
Total quarts of soup made = [tex]4\frac{1}{2}=\frac{9}{2}[/tex]
Fraction of total quarts served = [tex]\frac{4}{5}[/tex]
Thus, total quarts of soup served = [tex]\frac{9}{2}\times \frac{4}{5}[/tex]
⇒ [tex]\frac{18}{5}=3\frac{3}{5}[/tex]
Thus, total quarts of soup for serving =[tex]\frac{18}{5}[/tex] or [tex]3\frac{3}{5}[/tex]
Each serving = [tex]\frac{3}{5}[/tex] quarts
Total number of servings can be given as:
⇒ [tex]\frac{Total\ quarts\ of\ serving}{Size\ of\ each\ serving}[/tex]
⇒ [tex]\dfrac{\frac{18}{5}}{\frac{3}{5}}[/tex]
To divide fractions, we take reciprocal of the divisor and replace divsion with multiplication.
⇒ [tex]\frac{18}{5}\times \frac{5}{3}[/tex]
⇒ [tex]\frac{18}{3}[/tex]
⇒ 6 servings
Thus, the correct number of servings = 6.
By saying there were [tex]3\frac{3}{5}[/tex] servings of soup, Hector must have intended to say that there were [tex]3\frac{3}{5}[/tex] quarts of soup for serving.
A signalized intersection has a cycle length of 60 seconds and an effective red time of 25 seconds. The effective green time is ____ seconds.
Answer:
effective green time = 35 seconds
Step-by-step explanation:
given data
cycle length = 60 seconds
effective red time = 25 seconds
solution
we get here effective green time that is express as
effective green time = cycle length - effective red time ...........................1
put here value and we will get
effective green time = 60 seconds - 25 seconds
effective green time = 35 seconds
The effective green time at a signalized intersection with a cycle length of 60 seconds and a red time of 25 seconds is 35 seconds.
Explanation:The question involves calculating the effective green time at a signalized intersection with a known cycle length and red time. Since the cycle length is the total time for a complete cycle of the signal, and it is given as 60 seconds, and the effective red time is 25 seconds, we can determine the effective green time by subtracting the red time from the cycle length.
The effective green time = Cycle length - Red time
= 60 seconds - 25 seconds
= 35 seconds.
Therefore, the effective green time is 35 seconds.
Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t)
Answer:
Since the equation was missing, I solved it with another equation and got an answer of T(0) = <3j / 5 + 4k / 5>.
Please see my explanation. I hope this helps
Step-by-step explanation:
The question asked us to find out unit tangent vector.
Recall unit vector = vector / magnitude of vector
Since the question is missing with an equation. I suppose an equation.
r(t)=Cost i, 3t j, 2Sin(2t) k at t=0
Lets take out differentiation
r'(t) = <(-Sint), 3, 2(Cos(2t)(2))>
r'(t)= <-Sint, 3, 4Cos(2t)>
Now substitute t=0 in the differentiate found above.
r'(0)= <-Sin(0), 3, 4Cos(2*0)>
r'(0)= <0, 3, 4(1)>
r'(0)= <0,3,4>
vector r'(0)=<0i, 3j, 4k>
Now lets find out magnitude of vector
|r'(0)| = [tex]\sqrt{0^{2}+3^{2}+4^{2} }[/tex]
|r'(0)| = [tex]\sqrt{0+9+16}[/tex]
|r'(0)| = [tex]\sqrt{25}[/tex]
|r'(0)| = 5
Unit Tangent Vector
T(0) = <0, 3, 4> / 5
T(0) = <3j / 5 + 4K / 5>
To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector. To find the unit tangent vector, we divide the velocity vector by its magnitude.
Explanation:To find the unit tangent vector, we first need to find the velocity vector. Given that the position vector is r(t) = Acos(wt)i + Asin(wt)j, we can find the derivative of this vector to get the velocity vector:
v(t) = -Aw*sin(wt)i + Aw*cos(wt)j
To find the unit tangent vector, we divide the velocity vector by its magnitude:
T(t) = (v(t))/(|v(t)|) = (-Aw*sin(wt)i + Aw*cos(wt)j)/(sqrt((Aw*sin(wt))^2 + (Aw*cos(wt))^2))
So, the unit tangent vector at any point is T(t).
What type of sampling methods are used in selecting people for exit polls at the polling locations? Select all that apply.
a. Sample random Sampling
b. Cluster Sampling
c. Stratified Sampling
d. Convenience Sampling
e. systematic Sample
Exit polls commonly use simple random sampling, stratified sampling, and systematic sampling to ensure a representative sample of voters. Cluster sampling may be used, but convenience sampling is typically avoided to prevent bias.
Explanation:Exit polls typically employ a variety of sampling methods to ensure a representative sample of voters. These methods can include simple random sampling, where each person has an equal chance of being selected, and stratified sampling, which involves dividing the population into subgroups and sampling from each. Systematic sampling can also be used by selecting every nth person to participate in the poll.
Cluster sampling could be utilized if the population is divided into clusters, and a random selection of clusters is made. Convenience sampling, however, is generally not used in exit polling because it can introduce bias. Instead, methods that provide each individual an equal chance of selection are preferred to obtain a representative sample.
Determine whether the two given lines l1 and l2 are parallel, skew, or intersecting. If they intersect, find the point of intersection. l1 : x = t y = 1 + 2t z = 2 + 3t l2 : x = 3 − 4s y = 2 − 3s z = 1 + 2s 2.
Answer:
Skew lines
Step-by-step explanation:
Two lines are given and we have to find out whether they are parallel, skew, or intersecting
[tex]x =t , y = 1 + 2t, z = 2 + 3t, l2 : \\ x = 3 -4s,, y = 2 -3s, z = 1 + 2s[/tex]
direction ratios of these two lines are
(1,2,3) and (-4, -3,2) (coefficient of parameters)
Obviously these two are neither equal nor proportional
Hence we get not parallel lines. If these intersect there must be a common point making
[tex]t= 3-4s: 1+2t =2-3s and 2 + 3t =1+2s[/tex]
Consider first two equation
[tex]t+4s =3\\2t+3s = 1\\2t+8s = 6\\-5s =-5\\s=1[/tex]
when s=1, t = -1
Let us check whether these two values satisfy the third equation
2+3t = -1 and 1+2s = 1+2 =3
Not equal. so there is no common point between them
These two are skew lines
The given lines l1 and l2 intersect at the point (7, -10, -4).
Explanation:The given lines, l1 and l2, can be analyzed to determine whether they are parallel, skew, or intersecting. By comparing the equations of the two lines, we can see that they intersect at a single point. To find the point of intersection, we can set the x, y, and z coordinates of l1 equal to the x, y, and z coordinates of l2, and solve for the parameter t. By substituting the value of t back into the equations of l1, we can find the point of intersection to be (7, -10, -4).
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A certain process for manufacturing integrated circuits has been in use for a period of time, and it is known that 12% of the circuits it produces are defective. A new process that is supposed to reduce the proportion of defectives is being tested. In a simple random sample of 100 circuits produced by the new process, 12 were defective. a. One of the engineers suggests that the test proves that the new process is no better than the old process, since the proportion of defectives in the sample is the same. Is this conclusion justified? Explain. b. Assume that there had been only 11 defective circuits in the sample of 100. Would this have proven that the new process is better? Explain. c. Which outcome represents stronger evidence that the new process is better: finding 11 defective circuits in the sample, or finding 2 defective circuits in the sample?
Answer:
(a) No the conclusion is not justified.
b. No
c. Two defective circuits in the sample
Step-by-step explanation:
Ans: (a) No the conclusion is not justified. What is important is the percentage population of defectives;
the sample proportion is only an approximation. The population proportion
for the new process may be more than or less than that of the old process. We can decide to pick two hundred samples and discover that the number of defects is greater than the previous process
(b)
.For the defectives, the population proportion for the new process may be 0.12 or more,
although the sample of defectives is just 11 out of 100
(c) Two defective circuits in the sample. This is because the probability of having two defects from the 100n samples is less than having 11 defects
a. The engineer's conclusion is not justified. A hypothesis test is needed to compare the proportions. b. Finding 11 defective circuits would not prove the new process is better. c. Finding 2 defective circuits represents stronger evidence that the new process is better.
Explanation:a. The engineer's conclusion is not justified. To determine if the new process is better, we need to perform a hypothesis test. We can compare the proportion of defectives in the sample to the proportion of defectives in the known population for the old process using a hypothesis test for a proportion. If the p-value is small (less than the chosen significance level), we can reject the null hypothesis and conclude that the new process is better.
b. No, finding 11 defective circuits in the sample would not prove that the new process is better. We still need to perform a hypothesis test as mentioned in part a. If the p-value is small (less than the chosen significance level), we can reject the null hypothesis and conclude that the new process is better.
c. Finding 2 defective circuits in the sample represents stronger evidence that the new process is better. A smaller proportion of defectives in the sample suggests that the new process is more effective at reducing defects.
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The formula for wind chill C (in degrees Fahrenheit) is given by C = 35.74 + 0.6215T - 35.75v^0.16 + 0.4275Tv^0.16 where v is the wind speed in miles per hour and T is the temperature in degrees Fahrenheit. The wind speed is 23 ± 3 miles per hour and the temperature is 8° ± 1°. Use dC to estimate the maximum possible propagated error and relative error in calculating the wind chill.
Answer:
dC = 2.44
Relative error = 19%
Step-by-step explanation:
[tex]C = 35.74 + 0.6215*T - 35.75*v^(0.16) + 0.4275*T*v^(0.16)[/tex]
Δv = 3 ; ΔT = 1 , v = 23, T = 8
Use differential Calculus
[tex]dC = Cv.dv + Ct.dt[/tex]
[tex]Cv = dC/dv = -5.72*v^(-0.84) + 0.0684*T*v^(-0.84)\\Ct = dC/dT = 0.6215 + 0.4275*v^(0.16)\\dC = modulus (Cv.dv) + modulus (CT.dT)\\dC = (-5.72*v^(-0.84) + 0.0684*T*v^(-0.84))* 3 + (0.6215 + 0.4275*v^(0.16))*1\\dC = 2.44[/tex]
Relative Error
dC / C @(T = 8, v=23) * 100 = 19 %
The maximum possible propagated error in calculating wind chill is approximately 7.6425°F, and the relative error is approximately 3.06.
The formula for wind chill (C) is given by:
⇒ C = 35.74 + 0.6215T − 35.75[tex]v^{0.16}[/tex] + 0.4275T [tex]v^{0.16}[/tex]
Here, T is the temperature in degrees Fahrenheit, and v is the wind speed in miles per hour. We need to estimate the maximum possible propagated error in calculating the wind chill based on the given ranges for wind speed (v = 23 ± 3 mph) and temperature (T = 8° ± 3°).
First, compute the partial derivatives of C with respect to T and v:⇒ ∂C/∂T = 0.6215 + 0.4275v0.16
⇒ ∂C/∂v = -5.72v-0.84 + 0.0684Tv-0.84
Evaluate these partial derivatives at the central values T = 8 and v = 23:⇒ ∂C/∂T | (8,23) = 0.6215 + 0.4275(230.16)
≈ 0.6215 + 0.4275(1.8838)
≈ 1.4269
⇒ ∂C/∂v | (8,23) = -5.72(23-0.84) + 0.0684(8)(23-0.84)
≈ -5.72(0.1957) + 0.5472(0.1957)
≈ -1.1206
Estimate the maximum possible propagated error (dC) using error propagation formula:⇒ dC ≈ |∂C/∂T| × dT + |∂C/∂v| × dv
Given dT = 3 and dv = 3:⇒ dC ≈ |1.4269| × 3 + |-1.1206| × 3
≈ 4.2807 + 3.3618
≈ 7.6425
Therefore, the maximum possible propagated error is approximately 7.6425°F.
The relative error can be estimated as the ratio of this propagated error to an average value of the wind chill. If the average wind chill, C, is approximately -2.5°F:⇒ Relative Error = 7.6425 ÷ |-2.5|
≈ 3.06 (rounded to two decimal places).
Hence, the maximum possible propagated error is approximately 7.6425°F and the relative error is approximately 3.06.
Complete question:
The formula for wind chill C (in degrees Fahrenheit) is given by C = 35.74 + 0.6215T − 35.75[tex]v^{0.16}[/tex] + 0.4275T[tex]v^{0.16}[/tex]
where v is the wind speed in miles per hour and T is the temperature in degrees Fahrenheit. The wind speed is 23 ± 3 miles per hour and the temperature is 8° ± 3°. Use dC to estimate the maximum possible propagated error (round your answer to four decimal places) and relative error in calculating the wind chill (round your answer to two decimal places).
You operate a non-profit foodbank that accepts food donations and packages them into meals for local families who are food insecure. You accept canned goods from groceries. Some cans are not acceptable due to a compromised can or an expired use-by label. Can donations are assembled into boxes of 50 cans each for inspection to determine which cans should be discarded. The initial screening decision sends the box to either an experienced inspector or an inexperienced inspector.
The screener looks at 4 cans in each box. If there are zero unacceptable cans, the box is sent to an inexperienced inspector. Otherwise, it is sent to an experienced inspector.
a. Assuming a rate of 8% unacceptable, what is the probability of sending a box to an experienced inspector?
b. An inexperienced inspector makes $16 an hour, and an experienced one makes $22 an hour. If you were able to convince the groceries to reduce their unacceptable rate to 4%, what percent savings would you realize?
Assume that the mix of inspector types in FTEs equals the probability of a box being sent to each type.
For example, if 50.1% of boxes go to experienced inspectors, the FTE mix is 50.1 experienced FTES and 49.9 inexperienced FTEs. You do not change the number of inspectors, just the mix.
Answer:
Step-by-step explanation:
Binomial distribution is to be used here due to following reasons.
(a)
Probability of sending a box to an experienced inspector
= Probability of getting non-zero unacceptable cans
= 1 - Probability of getting zero unacceptable cans
=1- P(X = 0) = 1 - 10.08^0 *(1 – 0.08)^(4-0) = 0.283607
(b)
Expected cost per inspector in an hour in case of 8% unacceptable cans
= {(1-0.283607)*16+0.283607*22} = $ 17.70164
If groceries reduce their unacceptable rate to 4% then X - Bin(4, 0.04) .
In this scenario,
Probability of sending a box to an experienced inspector
= Probability of getting non-zero unacceptable cans
= 1 - Probability of getting zero unacceptable cans
=1- P(X = 0) = 1 - 0.04^0 * (1 – 0.04)*(4-0) = 0.150653
Expected cost per inspector in an hour in case of 4% unacceptable cans
= {(1-0.150653)*16+0.150653*22} = $ 16.90392
Percentage of savings realized = (17.70164-16.90392)/17.70164*100% = 4.506475%
In a two-tailed 2-sample z-test you find a P-Value of 0.0278. At what level of significance would you choose to reject the null hypothesis?
Answer:
5% or 0.05
Step-by-step explanation:
The null hypothesis will be rejected if p-value is less than significance level.
The null hypothesis can be rejected on 5% and 10% level of significance, but 5% level of significance is a suitable choice because when the 5% significance level is used the confidence level is 95% where as in case of 10% the confidence level is 90%. In short, the significance level indicates the probability of rejection of null hypothesis when the null hypothesis is true and the lesser probability of taking that risk will be better.
So, the scenario indicates the suitable significance level as 0.05.
Final answer:
With a P-value of 0.0278 in a two-tailed 2-sample z-test, the null hypothesis would be rejected at the 0.05 level of significance but not at the stricter 0.01 level.
Explanation:
When you find a P-value of 0.0278 in a two-tailed 2-sample z-test, the level of significance at which you would reject the null hypothesis depends on the predetermined alpha (α) level you have set for your test.
Since the P-value is less than the common significance levels of 0.05 and 0.10, you would reject the null hypothesis at these levels.
However, if your significance level was set at 0.01, which is stricter, you would not reject the null hypothesis because 0.0278 is greater than 0.01.
In summary, you would reject the null hypothesis at the 0.05 level of significance but not at the 0.01 level, given the P-value of 0.0278.
Numerical WebAssign will do simple calculations for you, using +, -, *, /, parentheses, and e-notation. For example, if a question asks for half the sum of 246 and 388, you can enter "(246+388)/2". For more complicated calculations you'll need to use a calculator and then enter your result. What is the product of 140 and 80?
Answer:
140·80=11200
Step-by-step explanation:
From exercise we have example for how we find half the sum of 246 and 388, we can enter "(246+388)/2".
Based on this example, we will calculate what is required in the task using a calculator. So we use a calculator to find the product of the following numbers, 140 and 80.
We calculate, and we get
140·80=11200
Determine if the statement is true or false. If a linear system has the same number of equations and variables, then it must have a unique solution. True False Justify your answer.
Answer:
False
Step-by-step explanation:
Consider the equations with the same number of equations and variables as shown below,
Case 1
[tex]x_{1} + x_{2} = 0\\x_{1} + x_{2} = 1[/tex]
This equation has no solution because it is not possible to have two numbers that give a sum of 0 and 1 simultaneously.
Case 2
[tex]x_{1} + x_{2} = 1\\2x_{1} + 2x_{2} = 2[/tex]
This equation has infinitely many possible solutions.
Therefore it is FALSE to say a linear system with the same number of equations and variables, must have a unique solution.
The statement that a linear system with the same number of equations and variables must have a unique solution is false. Other considerations, such as whether the system is consistent or inconsistent and dependent or independent, can impact the amount of solutions a system has.
Explanation:The statement is false. Even if a linear system has the same number of equations and variables, it does not necessarily mean that it will have a unique solution. Rather, whether a system has a unique solution, no solutions, or infinitely many solutions depends on whether the system is consistent or inconsistent and dependent or independent.
For instance, consider two linear equations: x + y = 5 and 2x + 2y = 10. Even though these two equations have the same number of variables and equations, they represent the same line and thus have infinitely many solutions. Similarly, consider the system x + y = 5 and x + y = 6. These two equations have also the same number of equations and variables but they are parallel lines and do not intersect, so this system does not have any solution.
So, the number of variables and equations is not always enough to determine the number of solutions to a linear system.
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Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 87, x = 26; 98 percent (0.185, 0.413) (0.202, 0.396) (0.184, 0.414) (0.203, 0.395)
Answer:
(0.185, 0.413)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 87, x = 26, p = \frac{x}{n} = \frac{26}{87} = 0.2989[/tex]
98% confidence interval
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.325[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 - 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.185[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2989 + 2.325\sqrt{\frac{0.2989*0.7011}{87}} = 0.413[/tex]
So the correct answer is:
(0.185, 0.413)
Please help... I have no clue
Answer:
OPTION C: Sin C - Cos C = s - r
Step-by-step explanation:
ABC is a right angled triangle. ∠A = 90°, from the figure.
Therefore, BC = hypotenuse, say h
Now, we find the length of AB and AC.
We know that: [tex]$ \textbf{Sin A} = \frac{\textbf{opp}}{\textbf{hyp}} $[/tex]
and [tex]$ \textbf{Cos A} = \frac{\textbf{adj}}{\textbf{hyp}} $[/tex]
Given, Sin B = r and Cos B = s
⇒ [tex]$ Sin B = r = \frac{opp}{hyp} = \frac{AC}{BC} = \frac{AC}{h} $[/tex]
⇒ [tex]$ \textbf{AC} = \textbf{rh} $[/tex]
Hence, the length of the side AC = rh
Now, to compute the length of AB, we use Cos B.
[tex]$ Cos B = s = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{AB}{h} $[/tex]
⇒ [tex]$ \textbf{AB} = \textbf{sh} $[/tex]
Hence, the length of the side AB = sh
Now, we are asked to compute Sin C - Cos C.
[tex]$ Sin C = \frac{opp}{hyp} $[/tex]
⇒ [tex]$ Sin C = \frac{AB}{BC} $[/tex]
[tex]$ = \frac{sh}{h} $[/tex]
= s
Sin C = s
[tex]$ Cos C = \frac{adj}{hyp} $[/tex]
[tex]$ \implies Cos C = \frac{AC}{BC} $[/tex]
⇒ Cos C = [tex]$ \frac{rh}{h} $[/tex]
Therefore, Cos C = r
So, Sin C - Cos C = s - r, which is OPTION C and is the right answer.
The patriot diner sells 2 cheeseburgers and one soda for $11.00 and 3 hamburgersand 2 sodas for $18.00. What is the cost of a cheeseburger?
Answer: the cost of a cheeseburger is $4
Step-by-step explanation:
Let x represent the cost of one cheeseburgers.
Let y represent the cost of one Soda.
The patriot diner sells 2 cheeseburgers and one soda for $11.00. It means that
2x + y = 11 - - - - - - - - - - - 1
She also sells 3 cheeseburgers and 2 sodas for $18.00. It means that
3x + 2y = 18 - - - - - - - - - - -2
Multiplying equation 1 by 3 and equation 2 by 2, it becomes
6x + 3y = 33
6x + 4y = 36
Subtracting, it becomes.
- y = - 3
y = 3
Substituting y = 3 into equation 1, it becomes
2x + 3 = 11
2x = 11 - 3 = 8
x = 8/2 = 4
Emilia and Ferdinand took the same freshman chemistry course: Emilia in the fall, Ferdinand in the spring. Emilia made an 83 on the common final exam that she took, for which the mean was 76 and the standard deviation 8. Ferdinand made a 79 on the common final exam that he took, which was more difficult, since the mean was 65 and the standard deviation 12. The one who has a higher z-score did relatively better. Was it Emilia or Ferdinand?
Answer:
[tex] z = \frac{83-76}{8}=0.875[/tex]
[tex] z = \frac{79-65}{12}=1.167[/tex]
As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Emilia case
Let X the random variable that represent the scores of a test, and we know that
Where [tex]\mu=76[/tex] and [tex]\sigma=8[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Since Emilia made 83 points we can find the z score like this:
[tex] z = \frac{83-76}{8}=0.875[/tex]
Ferdinand case
Let X the random variable that represent the scores of a test, and we know that
Where [tex]\mu=65[/tex] and [tex]\sigma=12[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Since Ferdinand made 79 points we can find the z score like this:
[tex] z = \frac{79-65}{12}=1.167[/tex]
As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.
If a linear system has the same number of equations and variables, then it must have a unique solution.
True or false
Answer:
The given statement is false.
Step-by-step explanation:
We are given the following statement:
"If a linear system has the same number of equations and variables, then it must have a unique solution."
The given statement is false because If a linear system has the same number of equations and variables, then it may have unique solution, no solution or infinitely many solution.
There could be three types of solution
no solution, unique solution and infinitely many solutions,depending on the augmented matrix and coefficient matrix of the linear system.
Say you want to provide a certain candy for Halloween. You expect around K kids to come to your house, and each kid is to be given three pieces of candy. Each bag of candy you can buy contains N pieces of candy for P dollars. Which algebraic expression will tell you how much should you expect to have to pay (M)
Answer:
Step-by-step explanation:
You expect around K kids to come to your house, and each kid is to be given three pieces of candy. This means that the total number of candies that you would buy is
3 × K = 3K
Each bag of candy you can buy contains N pieces of candy for P dollars.
Therefore,
If N pieces of candy cost $P, then
3K pieces of candy would cost $M
Therefore, the algebraic expression
to tell you how much should you expect to have to pay (M) would be
M = 3KP/N
The algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]
The given parameters are:
Kids = KEach = 3Unit Rate = P for N pieces i.e. P/NThe total amount paid for N pieces of candy is calculated as:
[tex]Total = Kids \times 3 \times Unit\ Rate[/tex]
So, we have:
[tex]Total = K \times 3 \times \frac PN[/tex]
Evaluate the product
[tex]Total = \frac{3KP}N[/tex]
Rewrite the above equation as:
[tex]M= \frac{3KP}N[/tex]
Hence, the algebraic expression that tells the amount you expect to pay is [tex]\frac{3KP}N[/tex]
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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the filling equipment, the amount of beer in each can is actually a random variable with a normal distribution. It has a mean of 12 ounces and a standard deviation of 0.30 ounce.
If you bought a six-pack of their beer what is the probability that you are going to actually get less than or equal to a total of 72 ounces of beer in your six-pack?
Answer:
[tex] T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)[/tex]
[tex] P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z<0) = 0.5[/tex]
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solutio to the problem
Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12,0.3)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=0.3[/tex]
For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.
The definition of sample mean is given by:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}[/tex]
If we solve for the total T we got:
[tex] T= n \bar X[/tex]
For this case then the expected value and variance are given by:
[tex] E(T) = n E(\bar X) =n \mu[/tex]
[tex] Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2[/tex]
And the deviation is just:
[tex] Sd(T) = \sqrt{n} \sigma[/tex]
So then the distribution for the total would be also normal and given by:
[tex] T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)[/tex]
And we want this probability:
[tex] P(T\leq 72)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{x-\mu}{\sigma}[/tex]
[tex] P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z<0) = 0.5[/tex]
A tank contains 50 kg of salt and 1000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 9 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) write down the differential equation which models the Amount y of salt in the tank:
dydt=
(c) Find the amount of salt in the tank after 2.5 hours.
amount = (kg)
(d) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)
Answer:
A. 0.05kg/l
B. dy/dt = 9/1000(25 - y)
C. 20.05 kg of salt
D. 0.0025kg/l
Step-by-step explanation:
A. Concentration of salt in the tank initially,
Concentration (kg/l) = mass of salt in kg/ volume of water in liter
= 50kg/1000l
= 0.05kg/l
B. dy/dt = rate of salt in - rate of salt out
Rate of salt in = 0.025kg/l * 9l/min
= 0.225kg/min
Rate of salt out = 9y/1000
dy/dt = 0.225 - 9y/1000
dy/dt = 9/1000(25 - y)
C. Collecting like terms from the above equation,
dy/25 - y = 9/1000dt
Integrating,
-Ln(25 - y) = 9/1000t + C
Taking the exponential of both sides,
25 - y = Ce^(-9t/1000)
Calculating for c, at y = 0, t = 0;
C = 25
y(t) = 25 - 25e^(-9t/1000)
At 2.5 hours,
2.5 hours * 60 mins = 180 mins
y(180 mins) = 25 - 25e^(-9*180/1000)
= 25 - 25*(0.1979)
= 20.05kg of salt
D. As time approaches infinity, e^(Infinity) = 0,
y(t) = 25 - 25*0
Concentration (kg/l) = 25/1000
= 0.0025kg/l
The initial concentration is 0.05 kg/L. A differential equation that models the amount of salt in the tank is dy/dt = 0.225 - (9y/(1000 + 9t)). By solving this equation over an interval of 2.5 hours could find the amount of salt after this time period. Lastly, concentration of the tank's solution as time approaches infinity is 0.025 kg/L.
Explanation:To solve the problem, we need to know the basics of differential equations and concepts related to concentration calculations. Starting with the initial conditions, we see that:
(a) The initial concentration is calculated by dividing the amount of salt by the volume of the solution. Hence, concentration = 50 kg/1000 L = 0.05 kg/L.
(b) The differential equation that models the salt in the tank can be derived using the inflow and outflow rates. The amount of salt entering the tank per minute is 9 L/min * 0.025 kg/L = 0.225 kg/min. The amount of salt leaving the tank per minute is 9 L/min * y kg/L; where y is the current amount of salt in the tank divided by the current volume of the solution in the tank (1000 L + 9t min). Hence, the differential equation is dy/dt = 0.225 - 9y/(1000 + 9t).
(c) To find the amount of salt after 2.5 hours, we would need to solve the differential equation given above with the initial condition y(0) = 50 kg, over the interval from t=0 to t=2.5 hours (or 150 minutes). This requires calculus skills specifically related to the solution of differential equations.
(d) At the limit as time approaches infinity, the volume in the tank approaches a constant (because inflow equals outflow), and so does the amount of salt in the tank (because inflow equals outflow). Hence, the differential equation becomes dy/dt = 0, yielding y = constant. This constant is the inflow rate divided by the outflow rate, or 0.225 kg/min / 9 L/min = 0.025 kg/L. This is the concentration of the solution as time approaches infinity.
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