Answer:
a) [tex] n =0, \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]
[tex] n =1, \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]
[tex] n =2, \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]
[tex] n =3, \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]
b) n=0
[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]
The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]
And if we replace into the formula of relative error we got:
[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]
n =1
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]
[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]
n =2
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]
[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]
n =3
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]
[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]
[tex] \arctan (\pi/6) = 0.48[/tex]
Step-by-step explanation:
Part a
the general term is given by:
[tex] a_n = \frac{(-1)^n}{2n+1} x^{2n+1}[/tex]
And if we replace n=0,1,2,3 we have the first four terms like this:
[tex] n =0, \frac{(-1)^0}{2*0+1} x^{2*0+1}= x[/tex]
[tex] n =1, \frac{(-1)^1}{2*1+1} x^{2*1+1}= -\frac{x^3}{3}[/tex]
[tex] n =2, \frac{(-1)^2}{2*2+1} x^{2*2+1}= \frac{x^5}{5}[/tex]
[tex] n =3, \frac{(-1)^3}{2*3+1} x^{2*3+1}= -\frac{x^7}{7}[/tex]
Part b
If we use the approximation [tex] arctan x \approx x[/tex] we got:
n=0
[tex] arctan(\pi/6) \approx \pi/6 = 0.523599[/tex]
The real value for the expression is [tex] arctan (\pi/6) = 0.482348[/tex]
And if we replace into the formula of relative error we got:
[tex] \% error= \frac{|0.523599 -0.482348|}{0.482348} * 100= 8.55\%[/tex]
If we add the terms for each value of n and we calculate the error we see this:
n =1
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} = 0.47576[/tex]
[tex] \% error= \frac{|0.47576 -0.482348|}{0.482348} * 100= 1.37\%[/tex]
n =2
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5} = 0.483631[/tex]
[tex] \% error= \frac{|0.483631 -0.482348|}{0.482348} * 100= 0.27\%[/tex]
n =3
[tex] arctan(\pi/6) \approx \pi/6 -\frac{(pi/6)^3}{3} +\frac{(pi/6)^5}{5}-\frac{(pi/6)^7}{7} = 0.48209[/tex]
[tex] \% error= \frac{|0.48209 -0.482348|}{0.482348} * 100= 0.05\%[/tex]
And thn we can conclude that the approximation is given by:
[tex] \arctan (\pi/6) = 0.48[/tex]
Rounded to 2 significant figures
Final answer:
The Maclaurin series for arctan x includes the first four terms: x, -x^3/3, x^5/5, and -x^7/7. To estimate arctan(π/6), we incrementally add these terms, leading to progressively better approximations.
Explanation:
The Maclaurin series for the arctangent function is given by:
arctan x = ∑ n=0 [infinity] (−1)n/(2n +1) x2n+1
(a) Writing out the first 4 terms for n = 0 to 3, we get:
For n = 0: x
For n = 1: −x3/3
For n = 2: x5/5
For n = 3: −x7/7
The series starts with x and alternates between subtracting and adding subsequent odd-powered terms, each divided by the respective odd number.
(b) To estimate arctan(π/6), also known as arctan(1/(√3)), we add terms of the series one by one:
Simplest estimate: arctan(π/6) ≈ π/6
Adding the second term: arctan(π/6) ≈ π/6 - (π/6)3/3
Including the third term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5
Including the fourth term: arctan(π/6) ≈ π/6 - (π/6)3/3 + (π/6)5/5 - (π/6)7/7
By computing these sums, we get increasingly accurate estimates for the value of arctan(π/6).
Scores for a common standardized college aptitude test are normally distributed with a mean of 512 and a standard deviation of 106. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the test has no effect
If 1 of the men is randomly selected, find the probability that his score is at least 559.5.
P(X > 559.5) =
If 18 of the men are randomly selected, find the probability that their mean score is at least 559.5.
P(M > 559.5) =
To find the probability of a man's score being at least 559.5 on the standardized college aptitude test, we can calculate the z-score and find the area under the normal distribution curve. The same process applies to finding the probability of the mean score of a sample of 18 men being at least 559.5.
Explanation:To find the probability that a randomly selected man's score is at least 559.5, we need to calculate the z-score for this value and then find the area under the normal distribution curve to the right of that z-score.
To find the probability that the mean score of 18 randomly selected men is at least 559.5, we first need to find the mean and standard deviation of the sample mean. Then, we can calculate the z-score for the given mean score and find the area under the normal distribution curve to the right of that z-score.
P(X > 559.5) = 1 - P(X ≤ 559.5)
P(M > 559.5) = 1 - P(M ≤ 559.5)
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The probabilities of a score being above 559.5 are as follows: for a single randomly selected individual, the probability is approximately 0.3271; for a group of 18 randomly selected individuals, the probability that their mean score is above 559.5 is approximately 0.0287.
Explanation:This is a problem of statistics, more specifically Normal Distribution and Standard Deviation. In a Normal Distribution, the mean (average) is the center of the distribution and standard deviation measures how spread out the scores are from the mean. The Z-Score gives us a measure of how many standard deviations an element is from the mean.
Firstly, to find the probability that a randomly selected man scores at least 559.5, we find the Z-Score using the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. Thus the Z-Score is Z = (559.5 - 512) / 106 = 0.448. From the Z-table or calculator, we find that P(Z > 0.448) ≈ 0.3271. Therefore, P(X > 559.5) = 0.3271.
Secondly, for a sample of 18 men, we use the formula for the standard deviation of a sample mean, σM = σ / sqrt(n), where σ is the standard deviation, and n is the size of the sample. The new standard deviation becomes σM = 106 / sqrt(18) = 25. This gives Z = (559.5 - 512) / 25 =1.90. From the Z-table or calculator, we find that P(Z > 1.90) ≈ 0.0287. Therefore, P(M > 559.5) = 0.0287.
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I don't know how to simplify this question. Is quotient rule necessary?
Answer:
The answer to your question is [tex]\frac{(x+1)^{2}(x - 23)}{(x-7)^{3}}[/tex], because the directions says to give the answer in factor form.
Step-by-step explanation:
[tex]\frac{3(x+1)^{2}(x-7)^{2}- (x+1)^{3}(2)(x - 7)}{(x - 7)^{4}}[/tex]
Factor like terms (x - 7)(x + 1)²
[tex]\frac{(x - 7)(x + 1)^{2}[3(x - 7) - 2(x + 1)}{(x - 7)^{4}}[/tex]
Simplify
[tex]\frac{(x+1)^{2}[3(x - 7) - 2(x+1)]}{(x-7)^{3}}[/tex]
Expand
[tex]\frac{(x+1)^{2}[3x - 21 -2x - 2]}{(x - 7)^{3}}[/tex]
Simplify
[tex]\frac{(x + 1)^{2}(x - 23)}{(x - 7)^{3}}[/tex]
A manufacturer of skis produces two types: downhill and cross country. The times required for manufacturing and finishing each ski are: manufacturing time per ski, downhill 2.5 hours, cross country 1.5 hours. Finishing time per ski: downhill 0.5 hours, cross country 1.5 hours. The maximum total weekly hours available for manufacturing and finishing the skis are 90 hours and 42 hours. The profit per ski are $50 for downhill and $50 cross country. Determine how many of each kind of ski should be produced to achieve a maximum profit?
Answer:
So to maximize profit 24 downhill and 20 cross country shouldbe produced
Step-by-step explanation:
Let X be the number of downhill skis and Y the number of cross country skis.
Time required for manufacturing and finishing each ski are: manufacturing time per ski, downhill 2.5 hours, cross country 1.5 hours
Finishing time per ski: downhill 0.5 hours, cross country 1.5 hours.
Total manufacturing time taken = (2.5) x+ (1.5+) y = 2.5x+1.5y≤90
total finishing time taken = 0.5x+1.5 y≤42
Profit function
Z = 50x+50y
Objective is to maximize Z
Solving the two equations we get intersecting point is
(x,y) = (24,20)
In the feasible region corner points are (0.28) (36,0)
Profit for these points are
i) 2200 for (24,20)
ii) 1400 for (0,28)
iii) 1800 for (36,0)
So to maximize profit 24 downhill and 20 cross country shouldbe produced.
To determine the optimal production quantity for each type of ski to achieve maximum profit, set up a system of equations using manufacturing time, finishing time, and profit. Graph the equations and find the intersection point.
Explanation:To determine how many of each kind of ski should be produced to achieve a maximum profit, we can set up a system of equations.
Let x be the number of downhill skis and y be the number of cross country skis.
The manufacturing time equation is 2.5x + 1.5y ≤ 90.
The finishing time equation is 0.5x + 1.5y ≤ 42.
The profit equation is 50x + 50y ≤ P, where P is the maximum profit.
We can graph these equations and find the intersection point, which represents the optimal production quantity for each type of ski.
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An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally distributed. What raw score corresponds to the 70th percentile?
Answer:
82.62
Step-by-step explanation:
Mean score (μ) = 80
Standard deviation (σ) = 5
The 70th percentile of a normal distribution has an equivalent z-score of roughly 0.525.
For any given score, X, the z-score can be determined by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For z = 0.525:
[tex]0.525=\frac{X-80}{5}\\ X=82.62[/tex]
A raw score of approximately 82.62 corresponds to the 70th percentile.
Answer: the raw score that corresponds to the 70th percentile is 82.625
Step-by-step explanation:
Since the population of scores in the aptitude test is normally distributed., we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = aptitude test scores.
µ = mean score
σ = standard deviation
From the information given,
µ = 80
σ = 5
We want to find the raw score that corresponds to the 70th percentile.
70th percentile = 70/100 = 0.7
Looking at the normal distribution table, the z score corresponding to 0.7 is 0.525.
Therefore,
0.525 = (x - 80)/5
5 × 0.525 = x - 80
2.625 = x - 80
x = 2.625 + 80
x = 82.625
Quantity A of an ideal gas is at absolute temperature TTT, and a second quantity B of the same gas is at absolute temperature 2T2T. Heat is added to each gas, and both gases are allowed to expand isothermally.
Answer:
The question is incomplete, here is the complete question ; Quantity A of an ideal gas is at absolute temperature T, and a second quantity B of the same gas is at absolute temperature 2T. Heat is added to each gas, and both gases are allowed to expand isothermally. If both gases undergo the same entropy change, is more heat added to gas A or gas B?
a. More heat is added to gas A
b. More heat is added to gas B
c.The same amount of heat is added to each gas
Option B is the correct answer = more heat is added to gas B
Step-by-step explanation:
Considering dQ = dS/T
dQ(A) = dS/T
dQ(B) = dS/2T
From this, it implies that dQ(B) = dQ(A)/2
and as such, more heat is added to gas B or gas B will undergo the greater entropy change
(6, -12). (15. -3)
Find the slope
Answer:
The slope is 1.
Step-by-step explanation:
A first order function has the following format
[tex]y = ax + b[/tex]
In which a is the slope
(6, -12).
This means that when [tex]x = 6, y = -12[/tex]
So
[tex]y = ax + b[/tex]
[tex]-12 = 6a + b[/tex]
(15. -3)
This means that when [tex]x = 15, y = -3[/tex]
So
[tex]y = ax + b[/tex]
[tex]-3 = 15a + b[/tex]
We have to solve the following system of equations:
[tex]-12 = 6a + b[/tex]
[tex]-3 = 15a + b[/tex]
We have to find a
In the second equation i will write as:
[tex]b = -3 - 15a[/tex]
Replacing in the first
[tex]-12 = 6a + b[/tex]
[tex]-12 = 6a - 3 - 15a[/tex]
[tex]-9 = -9a[/tex]
[tex]9a = 9[/tex]
[tex]a = \frac{9}{9}[/tex]
[tex]a = 1[/tex]
The slope is 1.
Answer:
1
Step-by-step explanation:
Slope = (y2-y1)/(x2-x1)
= (-3-(-12))/(15-6)
= (-3+12)/(9)
= 9/9
= 1
3 divided by (4x-15) divided by 5
Answer:
Step-by-step explanation:
3/(4x-15)/5
3÷(4x-15)/5
3 x 5/(4x-15) = 15/(4x-15)
What is the mean? If the answer is a decimal, round it to the nearest tenth.
56 47 48 52 62 59 49 56 43 48
Answer:
The mean is 52.
Step-by-step explanation:
The mean is the sum of all elements divided by the number of elements.
In this problem, we have that:
Elements
56 47 48 52 62 59 49 56 43 48
Sum
[tex]56+47+48+52+62+59+49+56+43+48 = 520[/tex]
Number of Elements
10 elements
The mean is
[tex]M = \frac{520}{10} = 52[/tex]
In a high school graduating class of 128 students, 52 are on the honor roll. Of these, 48 are going on to college; of the other 76 students, 56 are going on to college. What is the probability that a student selected at random from the class is (a) going to college, (b) not going to college, and (c) not going to college and on the honor roll?
Answer:
a. 0.8125
b. 0.1875
c. 0.03125
Step-by-step explanation:
from the information given, we can come up with the following data
Total No of students =128
Total Nos of students going to college = 48+56=104
Total Nos of students not going to college and on the honor roll= 52-48=4
a. To determine the probability of students going to college we have
P(going to college) = (total number of students going to college)/total number of students
P(going to college) = 104/128
P(going to college) = 0.8125
b. To determine the probability of students not going to college, we use the rule that says total probability is 1, hence
Pr(not going to college)=1-Pr(going to college)
Pr(not going to college)=1-0.8125
Pr(not going to college)=0.1875
c. To determine the probability of students NOT going to college and on the pay roll we have
Pr = (Total Number of students not going to college and on the honor roll)/total number of students
Pr=4/128
Pr=0.03125
To find the probabilities, we use conditional probability, complement rule, and joint probability rule.
Explanation:To answer this question, we need to use the concept of conditional probability. Let's break it down:
(a) To find the probability that a student is going to college, we can add the probabilities of two mutually exclusive events: being on the honor roll and going to college, and not being on the honor roll and still going to college. So, P(going to college) = P(on honor roll) * P(going to college | on honor roll) + P(not on honor roll) * P(going to college | not on honor roll).
(b) To find the probability that a student is not going to college, we can use the complement rule: P(not going to college) = 1 - P(going to college).
(c) To find the probability that a student is not going to college and on the honor roll, we can use the joint probability rule: P(not going to college and on honor roll) = P(on honor roll) * P(not going to college | on honor roll).
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Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three halves An implicit solution in the form F(x,y)equalsC is nothingequalsC, where C is an arbitrary constant.
Answer:
Step-by-step explanation:
To solve the differential equation
dy/dx = 5x^4(1 + y²)^(3/2)
First, separate the variables
dy/(1 + y²)^(3/2) = 5x^4 dx
Now, integrate both sides
To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)
dy = (1/cos²u)du
So,
dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du
= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)
Because cos²u + sin²u = 1 (Trigonometric identity),
The equation becomes
[1/(1/cos²u)^(3/2) × 1/cos²u] du
= cos³u/cos²u
= cosu
Integral of cosu = sinu
But y = tanu
Therefore u = arctany
We then have
cos(arctany) = y/√(1 + y²)
Now, the integral of the equation
dy/(1 + y²)^(3/2) = 5x^4 dx
Is
y/√(1 + y²) = x^5 + C
So
y - (x^5 + C)√(1 + y²) = 0
is the required implicit solution
write a function that represents the sequence 7, 14, 21, 28, ...
Answer:
a ₙ = 7n
Step-by-step explanation:
This is an arithmetic sequence, the common difference between each term is 14-7 = 21-14 = 28-21 = 7
to the previous term in the sequence addition of 7 gives the next term.
Arithmetic Sequence:
d = 7
This is the formula of an arithmetic sequence.
a ₙ = a₁ + d(n − 1)
Substitute in the values of
a₁ = 7 and d = 7
a ₙ = 7 + 7(n − 1)
a ₙ = 7 + 7n -7
a ₙ = 7 - 7 +7n = 7n
a ₙ = 7n
Answer:
Step-by-step explanation:
In an arithmetic sequence, consecutive terms differ by a common difference and it is always constant. Looking at the set of numbers,
14 - 7 = 21 - 14 = 28 - 21 = 7
Therefore, it is an arithmetic sequence with a common difference of 7.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = 7
d = 7
The function that represents the sequence would be
Tn = 7 + (n - 1)7
Tn = 7 + 7n - 7
Tn = 7n
Multiple-choice questions each have fourfour possible answers left parenthesis a comma b comma c comma d right parenthesis(a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions. Same question with multiplication rule to find P(WWC) with C as Correct and W as wrong__________.
Answer: 9/64
Step-by-step explanation:
Probability is a chance of prediction. It's a measure of how an event is likely to happen.
P(A) = Number of favorable outcome/Total Number of favorable outcome
Let's make W the correct answer and C the right answer.
The probability of choosing the correct answer from multiple choice question:
P(C) = 1/4
The probability of choosing the wrong answer from multiple choice question:
P(C) =1/4
P(W)= 1 - 1/4
P(W) = 3/4
Therefore, to find P(WWC)
P(WWC) = P(W) × P(W) × P(C)
P(WWC) = 3/4 ×3/4 × 1/4
P(WWC) = 9/64
The probability is 9/64.
All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be proficient in reading, 78% were found to be proficient in mathematics, and 65% were found to be proficient in both reading and mathematics. A student is chosen at random.
What is the probability that the student is proficient in neither reading nor mathematics?
Answer:
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a student is proficient in reading
B is the probability that a student is proficient in mathematics.
C is the probability that a student is proficient in neither reading nor mathematics.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a student is proficient in reading but not mathematics and [tex]A \cap B[/tex] is the probability that a student is proficient in both reading and mathematics.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So
[tex](A \cup B) + C = 1[/tex]
In which
[tex](A \cup B) = a + b + (A \cap B)[/tex]
65% were found to be proficient in both reading and mathematics.
This means that [tex]A \cap B = 0.65[/tex]
78% were found to be proficient in mathematics
This means that [tex]B = 0.78[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.78 = b + 0.65[/tex]
[tex]b = 0.13[/tex]
85% of the students were found to be proficient in reading
This means that [tex]A = 0.85[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]0.85 = a + 0.65[/tex]
[tex]a = 0.20[/tex]
Proficient in at least one:
[tex](A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98[/tex]
What is the probability that the student is proficient in neither reading nor mathematics?
[tex](A \cup B) + C = 1[/tex]
[tex]C = 1 - (A \cup B) = 1 - 0.98 = 0.02[/tex]
There is a 2% probability that the student is proficient in neither reading nor mathematics.
How many quarts of water must be added to 3 gallons of soup that is 60% chicken broth to make the soup 40% chicken broth
Answer:
6 quarts
Step-by-step explanation:
60% 3 gallons = 1.8 gallons of broth
water = 0% broth
1.8=1.2+0.4x
0.6=0.4x
x=1.5
1.5 gallons = 6 quarts
The amount of water must be added to 3 gallons of soup which is 60% chicken broth to make the soup 40% chicken broth is 6 quartz.
What is equation?In other terms, it is a mathematical statement stating that "this is equivalent to that." It appears to be a mathematical expression on the left, an equal sign in the center, and a mathematical expression on the right.
Given:
There are 3 gallons of soup that is 60% chicken broth to make the soup 40% chicken broth,
Assume the number of gallons is x then write the equation as shown below,
60% 3 gallons = 1.8 gallons of broth
water = 0% broth
1.8 = 1.2 + 0.4x
0.6 = 0.4x
x = 1.5
As we know that 1 gallon = 4 quartz,
1.5 gallons = 6 quarts
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Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition, the number of defects per unit is distributed according to a Poisson distribution. What is the probability that there will be a total of 7 defects on four units
Answer:
The probability that there will be a total of 7 defects on four units is 0.14.
Step-by-step explanation:
A Poisson distribution describes the probability distribution of number of success in a specified time interval.
The probability distribution function for a Poisson distribution is:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...[/tex]
Let X = number of defects in a unit produced.
It is provided that there are, on average, 2 defects per unit produced.
Then in 4 units the number of defects is, [tex](2\times4)=8[/tex].
Compute the probability of exactly 7 defects in 4 units as follows:
[tex]P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14[/tex]
Thus, the probability of exactly 7 defects in 4 units is 0.14.
You get 3% commission on all sales. This month, you made a sale of $45,050 and a sale of $6,785.25. What is your commission for the month?
Answer:
Your commission for the month is $1,553.35.
Step-by-step explanation:
You made 2 sales.
In each you got a commission of 3%. Your total commission is the sum of both commisions. So
Sale of $45,050:
You got 3% of the sale. So
0.03*45050 = $1,351.5
Sale of $6,728.25:
You got 3% of this sale. So
0.03*6728.25 = $201.85
Total commision for the month:
$1,351.5 + $201.85 = $1,553.35.
Your commission for the month is $1,553.35.
Answer:
1,553.35
The commission for the month is a total of 1,553.35
This year, a small business had a total revenue of $ 62,100 . If this is 15 % more than their total revenue the previous year, what was their total revenue the previous year?
Answer:
Their total revenue the previous year was $54,000.
Step-by-step explanation:
This question can be solved by a simple rule of three.
This year revenue was $62,100. It was 15% more than last year, so 115% = 1.15 of last year. How much was the revenue last year, that is, 100% = 1?
62,100 - 1.15
x - 1
[tex]1.15x = 62100[/tex]
[tex]x = \frac{62100}{1.15}[/tex]
[tex]x = 54000[/tex]
Their total revenue the previous year was $54,000.
The union of two events A and B is the event that: a) The intersection of A and B does not occur. b) Both A and B occur. c) Either A or B or both occur. d) Either A or B, but not both occur. e) A and B occur at the same time. f) None of the above
Answer:
c) Either A or B or both occur.
Step-by-step explanation:
Suppose that we have two events
Event A
Event B
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a a happens and b does not and [tex]A \cap B[/tex] is the probability that aboth events happen
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The union of events A and B is:
[tex](A \cup B) = a + b + (A \cap B)[/tex]
Which includes either one of them or both.
So the correct answer is:
c) Either A or B or both occur.
Suppose the tank is halfway full of water. The tank has a radius of 2 ft and is 4 ft long. Calculate the force (in lb) on one of the ends due to hydrostatic pressure.
(Assume a density of water rho = 62.4 lb/ft3.)
Answer:
The answer is 332.8 lb
Step-by-step explanation:
See attached picture for the solution
The force on one of the ends due to hydrostatic pressure is 332.8lb
Data;
Density = 62.4 lb/ft^3length = 4ftradius = 2ftForce Due to PressureThe force due to hydrostatic pressure can be calculated as
From the attached diagram;
[tex]F = pressure * area\\density = 62.4 lb/ft^3\\depth of water = 2 - y\\pressure = (2 - y)(62.4)\\pressure = 124.8 - 62.4y\\[/tex]
We can proceed as
[tex]x^2 + (y - 2)^2 = 2^2\\x^2 = 4 - (y - 2)^2\\x = +- \sqrt{4y - y^2}\\[/tex]
this implies that
[tex]2x = 2\sqrt{4y - y^2}[/tex]
The area is given as
[tex]\delta A = (2x)*\delta y\\\delta A = 2\sqrt{4y - y^2 \delta y}[/tex]
The force would be given by
[tex]\delta F = (2-y)(62.4)(2\sqrt{4y - y^2})\delta y[/tex]
The total force is given by
[tex]F = \int\limits^2_0 {(2-y)(62.4)(2\sqrt{4y - y^2}) } \, dy\\F = 124.8\int\limits^2_0 {(2-y)(\sqrt{4y - y^2}) } \, dy\\F = 124.8[-\frac{1}{3}y(y -4)(\sqrt{4y -y^2}]_0^2\\F = 124.8[-\frac{1}{3}(2)(2-4)\sqrt{4(2)-2^2}\\ F = 332.8lb[/tex]
The force on one of the ends due to hydrostatic pressure is 332.8lb
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Suppose that if θ = 1, then y has a normal distribution with mean 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with mean 2 and standard deviation σ. Also, suppose Pr(θ = 1) = 0.5 and Pr(θ = 2) = 0.5.
Step-by-step explanation:
We have two cases for Ф,
1. Ф=1; it implies that Pr(Ф=1)=0.5, while y~N(1,α²)
2. Ф=2; it implies that Pr(Ф=2)=0.5, while y~N(2,α²)
Now,
For 1st case of α=2,
We have marginal probability density formula
p(y)=∑p(yIФ)p(Ф)
=p(yIФ=1)p(Ф=1)+p(yIФ=2)p(Ф=2)
=N(yI1,2²)(1/2)+N(yI2,2²)(1/2)
=(1/2)[N(yI1,2²)+N(yI2,2²)]
Now.
For Pr(Ф=1Iy=1) at α=2
We have,
=p(Ф=1Iy=1)
=[p(y=1,Ф=1)]/[p(y=1)]
=[p(y=1IФ=1)p(Ф=1)]/[p(y=1)]
={(1/[tex]\sqrt{2x-2}[/tex])exp[(-1/(2*2²))(1-1)²(1/2)]}/{(1/[tex]\sqrt{2x-2}[/tex])(1/2)[exp[(-1/(2*2²))(1-1)²]+exp[(-1/(2*2²))(1-2)²]}
=0.53 Answer
Now, to describe the changes in shape of Ф when α is increased and decreased:
The formula for posterioir density is p(ФIy)=p(yIФ)p(Ф)/p(y)
=exp[(-1/(2α²)(y-Ф)²]/{exp[(-1/(2α²)(y-1)²]+exp[(-1/(2α²))(y-2)²]}
Now at Ф=1 and solving the equation, we get
p(Ф=1Iy)=1 / {1+exp[(2y-3)/2α²]}
Similarly at Ф=1 and solving the equation, we get
p(Ф=2Iy)=1 / {1+exp[(2y-3)/2α²]}
Conclusion:
α² → ∞ ⇒p(ФIy) → p(Ф) = 1/2
α² → 0 ⇒ two cases
y > 3/2, α² → 0 ⇒p(Ф=2Iy) → 1
y < 3/2, α² → 0 ⇒p(Ф=1Iy) → 1
The value of θ determines the mean of the normal distribution for y, while σ remains constant. The probabilities of θ being 1 or 2 are both 0.5.
The given information states that if θ = 1, then y has a normal distribution with a mean of 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with a mean of 2 and standard deviation σ.
The probabilities of θ being 1 or 2 are both 0.5.
This means that there is a 50% chance of θ being 1, and a 50% chance of θ being 2.
This information allows us to understand how the value of θ affects the distribution of y. When θ is 1, y follows a normal distribution with mean 1 and standard deviation σ.
When θ is 2, y follows a normal distribution with mean 2 and standard deviation σ. The probabilities of these scenarios happening are equal.
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Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and X2 are independent.
(a) The mgf of binomial distribution Bin(n, p) is (1 − p + pet)n. Use this fact to obtain the distribution of X1 + X2.
(b) Find the probability P(X1 + X2 = 1|X2 = k) for k = 0 and 1. Then use the law of total probability to find P (X1 + X2 = 1)
Answer:
a) X1+X2 have distribution Bi(n1+n2, p)
b)
P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹
P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹
P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Step-by-step explanation:
Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).
b)
Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).
If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹
If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹
Thus,
P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted
Answers:
without replacement: a) 86/969 b)80/323
with replacement: a)853/6859 b)1440/6859
Step-by-step explanation:
a) probability that all three are of same color:if the balls are not replaced
the equal probable results of experiment
the outcome space :
[tex]S=[R_{1} R_{2} R_{3} ,R_{1} R_{2} G_{1} ,R_{1} R_{2} B_{1}..........][/tex], and there are 19.18.17 elements in S.
where
R- event that all three balls are red.
G- events that all three balls are green.
B- events that all three balls are blue.
and the number of possibilities from S that are in Red is 5.4.3 there are 6.5.4 events in blue, and 8.7.6 in green
[tex]P(R)+P(B)+P(G)=(5.4.3/19.18.17)+(6.5.4/19.18.17)+(8.7.6/19.18.17)[/tex]
[tex]=86/969[/tex]
if the balls are placed
the equal probable results of experiment
the outcome space :
[tex]S=[R_{1} R_{2} R_{3}, R_{1} R_{2} G_{1},R_{1} R_{2} B_{1} .........][/tex]
and there are 19.19.19 elements in S.
R- event that all three balls are red.
G- events that all three balls are green.
B- events that all three balls are blue.
and the number of possibilities from S are in red is 5.5.5, there are 6.6.6 events in blue and 8.8.8 in green
thus the result is:
[tex]P(R)+P(B)+P(G)=(5^3/19^3)+(6^3/19^3)+(8^3/19^3)=853/6859[/tex]
b) probability that all three are in different colors:if the balls are not replaced
the equal probable results of experiment
the outcome space :
[tex]S=[R_{1} R_{2} R_{3}, R_{1} R_{2} B_{1} ,R_{1} R_{2} G_{1}..... ][/tex]
and there are 19.18.17 elements in S
and if all balls have to be in different colors.
choose one of the 5 red balls, 1 of the 6 green balls and 1 of the 8 green balls. and for every choice of red, a green and a blue ball they can be permuted 3! ways
[tex]P(R,B,G)=(5.6.8.3!/19.18.17)=80/323[/tex]
if the balls are replaces then
[tex]P(R,B,G)=(5.6.8.3!/19^3)=1440/6859[/tex]
Bespin Car Rental predicts that the annual probability of one of its cars being destroyed in a crash is 1 in 1,000,000. If destroyed, the value of the property damage to the car equals $45,000. Assume that there are no partial losses; the car is either destroyed in a crash or suffers no loss. A) Show the physical damage loss distribution for Bespin Car Rental’s automobiles and calculate the expected value of the physical damage loss. B) Show the calculations for the variance and the standard deviation.
Answer:
(A) The expected loss is $0.045.
(B) The variance and standard deviation of physical damage loss are $2,025 and $45 respectively.
Step-by-step explanation:
The annual probability of Bespin Car Rental's cars being destroyed is 1 in a million, i.e 0.000001.
It is assumed that the car is either destroyed or there was no loss suffered.
The loss amount in case the car is destroyed is, $45,000.
(A)
The distribution for physical damage loss is displayed in the table below.
The Expected value of physical damage loss is:
[tex]E(X)=\sum xP(X)=(45000\times0.000001)+(0\times0.999999)=0.045[/tex]
Thus, the expected loss is $0.045.
(B)
The variance of a random variable X is: Var (X) = E (X²) - [E (X)]².
The variance of physical damage loss is:
Compute the variance as follows:
[tex]Var(X)=E(X^{2})-[E(X)]^{2}\\=\sum x^{2}P(X)-[\sum xP(X)]^{2}\\=[(45000^{2}\times0.000001)+(0^{2}\times0.999999)]-(0.045)^{2}\\=2025-0.002025\\=2024.997975\\\approx2025[/tex]
The standard deviation of physical damage loss is:
[tex]SD=\sqrt{Var(X)}=\sqrt{2025}=45[/tex]
Thus, the variance and standard deviation of physical damage loss are $2,025 and $45 respectively.
In a survey of 859 homeowners with high-speed Internet, the average monthly cost of a high-speed Internet plan was $64.1 with standard deviation $12.62. Assume the plan costs to be approximately bell-shaped. Estimate the number of plans that cost between $51.48 and $76.72. Round to the nearest whole number.
To estimate the number of high-speed Internet plans that cost between $51.48 and $76.72, we can use the standard normal distribution and the z-score formula. The estimated number of plans is 587.
Explanation:To estimate the number of plans that cost between $51.48 and $76.72, we can use the standard normal distribution and the z-score formula. First, we calculate the z-scores for both costs:
z1 = (51.48 - 64.1) / 12.62 = -1.003
z2 = (76.72 - 64.1) / 12.62 = 1.003
Next, we find the area under the standard normal curve between these two z-scores using a z-table or a calculator. Let's assume the area is approximately 0.6827.
Finally, we multiply this area by the total number of homeowners surveyed (859) to estimate the number of plans that fall within this cost range:
Number of plans = 0.6827 * 859 = 586.92
Rounding to the nearest whole number, the estimated number of plans is 587.
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Wes and Tma are a married couple and provide financial assistance to several persons during the current yeaL For the siruations below, determine whether the individuals qualify as Wes and Tina's dependents.
Answer: Hello! Apparently, your question is incomplete since the rest of it is missing and we need the alternatives to work on. Fortunately, we were able to find the whole of it so we can help you! Here it goes:
Wes and Tina are a married couple and provide financial assistance to several persons during the current year. In all of the situations below, assume that any dependency tests not mentioned have been met.
Requirement
For each situation, determine whether the individuals qualify as Wes and Tina's dependents. (Select the best possible answer in each case.)
Requirement a. Brian is age 24 and Wes and Tina's son. He is a full-time student and lives in an apartment near campus. Wes and Tina provide over 50% of his support. Brian works as a waiter and earned $4,200.
Answer: He can't be considered a dependent because his income is higher than $4050. Also, he can't be considered a child since he is older than 23.
Requirement b. Same as Part a except that Brian is a part-time student.
Answer: The status of beig a student doesn't change the fact that he is over 23, so he is still not be claimed a dependent since he can't be considered a child.
Requirement c. Sherry is age 22 and Wes and Tina's daughter. She is a full-time student and lives in the college dormitory. Wes and Tina provide over 50% of her support. Sherry works part-time as a bookkeeper and earned $5,000.
Answer: She may be claimed as a dependent since she meets the four relationship requirements, age, abode and support. Also, being a student is not a strong influence in this case.
Requirement d. Same as Part c except that Sherry is a part-time student.
Answer: Under this condition, she can't be considered a dependent since she becomes an unqualified child. She's not a full-time student and over 18 years old.
Requirement e. Granny, age 82, is Tina's grandmother and lives with Wes and Tina. During the current year, Granny's only sources of income were her Social Security of $4,800 and interest on U.S. bonds of $4,500. Granny uses her income to pay for 40% of her total support, Wes and Tina provide the remainder of Granny's support.
Answer: Granny wouldn't be eligible, since her bonds are higher than $4050 , not considering her Social Security income.
The number of bats in a colony is growing exponentially. After 2 years, there were 180 bats. After 5 years, there were 1440 bats. If the colony continues to grow at the same rate, how many bats are expected to be in the colony after 9 years
Answer:
23040 bats
Step-by-step explanation:
Let N(t) be the number of bats at time t
We know that exponential function
[tex]y=ab^t[/tex]
According to question
[tex]N(t)=ab^t[/tex]
Where t (in years)
Substitute t=2 and N(2)=180
[tex]180=ab^2[/tex]...(1)
Substitute t=5 and N(5)=1440
[tex]1440=ab^5[/tex]...(2)
Equation (1) divided by equation (2)
[tex]\frac{180}{1440}=\frac{ab^2}{ab^5}=\frac{1}{b^{5-2}}[/tex]
By using the property [tex]a^x\div a^y=a^{x-y}[/tex]
[tex]\frac{1}{8}=\frac{1}{b^3}[/tex]
[tex]b^3=8=2\times 2\times 2=2^3[/tex]
[tex]b=2[/tex]
Substitute the values of b in equation (1)
[tex]180=a(2)^2=4a[/tex]
[tex]a=\frac{180}{4}=45[/tex]
Substitute t=9
[tex]N(9)=45(2)^9=23040 bats[/tex]
Hence, after 9 years the expected bats in the colony=23040 bats
To find the number of bats expected to be in the colony after 9 years, we can use the equation for exponential growth. By plugging in the given population values and solving for the growth rate, we can then calculate the population after 9 years.
Explanation:To find the number of bats expected to be in the colony after 9 years, we need to determine the growth rate. Let's use the equation for exponential growth: N = P * e^(kt), where N is the final population, P is the initial population, e is the base of the natural logarithm, k is the growth rate, and t is the time.
We are given the population after 2 years (P = 180) and after 5 years (P = 1440). Plugging these values into the equation, we can solve for k:
180 = P * e^(2k) and 1440 = P * e^(5k).
Dividing the second equation by the first equation, we can eliminate P and solve for e^(3k): 8 = e^(3k).
Taking the natural logarithm of both sides, we get: ln(8) = 3k.
Finally, solving for k, we have: k = ln(8) / 3.
Now, we can use the calculated value of k to find the population after 9 years:
N = P * e^(9k).
Plugging in the value of P and k, we get: N = 180 * e^(9 * ln(8) / 3). Calculating this expression gives us the expected number of bats in the colony after 9 years.
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Kevin Hall is considering an investment that pays 7.70 percent, compounded annually. How much will he have to invest today so that the investment will be worth $30,000 in six years
Answer:
He will have to invest $20,519.84 today.
Step-by-step explanation:
We can solve this question using the simple interest formula:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex].
In this problem, we have that:
[tex]I = 0.077, t = 6, T = 30,000[/tex]
So
[tex]T = E + P[/tex].
[tex]E + P = 30000[/tex]
[tex]E = 30000 - P[/tex]
So
[tex]E = P*I*t[/tex]
[tex]30000 - P= P*0.077*6[/tex]
[tex]30000 - P = 0.462P[/tex]
[tex]1.462P = 30000[/tex]
[tex]P = \frac{30000}{1.462}[/tex]
[tex]P = 20519.84[/tex]
He will have to invest $20,519.84 today.
Kevin Hall needs to invest approximately $19,249.38 today to have $30,000 in six years at an annual interest rate of 7.70%, compounded annually.
To determine how much Kevin Hall should invest today to have $30,000 in six years with a 7.70% annual interest rate, we'll use the formula for present value (PV) of a future amount, which is:
PV = FV / (1 + r)ⁿ
Where:
FV = Future Value = $30,000r = annual interest rate = 7.70% or 0.077n = number of years = 6Plugging in the values:
PV = 30,000 / (1 + 0.077)⁶
Calculating the denominator:
(1 + 0.077)⁶ ≈ 1.5583
Thus,
PV = 30,000 / 1.5583 ≈ $19,249.38
Kevin Hall should invest approximately $19,249.38 today to have $30,000 in six years with a 7.70% annual interest rate, compounded annually.
suppose that you made four measurement of a speed of a rocket: 12.7 km/s, 13.4 km/s, 12.6 km, and 13.3 km/s. compute: the mean, the standard deviations, and the standard deviation of the mean
the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.
To compute the mean, standard deviation, and standard deviation of the mean, we'll follow these steps:
1. Calculate the mean [tex](\( \mu \))[/tex]:
[tex]\[ \mu = \frac{\text{sum of all measurements}}{\text{number of measurements}} \][/tex]
2. Calculate the standard deviation [tex](\( \sigma \))[/tex]:
[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n}} \][/tex]
3. Calculate the standard deviation of the mean [tex](\( \sigma_\bar{x} \))[/tex]:
[tex]\[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \][/tex]
Let's plug in the given measurements:
[tex]\[ x_1 = 12.7 \, \text{km/s} \][/tex]
[tex]\[ x_2 = 13.4 \, \text{km/s} \][/tex]
[tex]\[ x_3 = 12.6 \, \text{km/s} \][/tex]
[tex]\[ x_4 = 13.3 \, \text{km/s} \][/tex]
1. Mean (\( \mu \)):
[tex]\[ \mu = \frac{12.7 + 13.4 + 12.6 + 13.3}{4} \][/tex]
[tex]\[ \mu = \frac{51}{4} \][/tex]
[tex]\[ \mu = 12.75 \, \text{km/s} \][/tex]
2. Standard deviation (\( \sigma \)):
[tex]\[ \sigma = \sqrt{\frac{(12.7 - 12.75)^2 + (13.4 - 12.75)^2 + (12.6 - 12.75)^2 + (13.3 - 12.75)^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.05^2 + 0.65^2 + (-0.15)^2 + 0.55^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.0025 + 0.4225 + 0.0225 + 0.3025}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.75}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{0.1875} \][/tex]
[tex]\[ \sigma \approx 0.433 \, \text{km/s} \][/tex]
3. Standard deviation of the mean (\( \sigma_\bar{x} \)):
[tex]\[ \sigma_\bar{x} = \frac{0.433}{\sqrt{4}} \][/tex]
[tex]\[ \sigma_\bar{x} = \frac{0.433}{2} \][/tex]
[tex]\[ \sigma_\bar{x} \approx 0.217 \, \text{km/s} \][/tex]
So, the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.
Kenneth Brown is the principal owner of Brown Oil, Inc. After quitting his university teaching job, Ken has been able to increase his annual salary by a factor of over 100. At the present time, Ken is forced to consider purchasing some more equipment for Brown Oil because of competition. His alternatives are shown in the following table:
FAVORABLE UNFAVORABLE
MARKET MARKET
EQUIPMENT ($) ($)
Sub 100 300,000
Answer:
Multiple Answers
Step-by-step explanation:
You forgot to put all the question, I attached it to the answer.
The questions we need to response are:
a)What type of decision is Ken facing?
There are three types of decision in probability. This are:
Risky that cover when the event is known and you know the chances of success.
Of uncertainty that cover for a known event but you dont know the possibilities of success.
Of ignorance that cover a unknown event, with unknown possibilities of succes.
So the decision Kenneth is facing is of uncertainty.
(b)What decision criterion should he use?
The criterion decision he should take would be Maximax.
This states that you should select the option that have the maximum gain.
(c)What alternative is best? The best option should be sub 100 because of the decision criterion we decided to use. Sub 100 has the maximum gain.
Final answer:
Kenneth Brown faces a decision on purchasing equipment for his company, Brown Oil, Inc. Business management principles and financial implications play a key role in this decision-making process.
Explanation:
Kenneth Brown, the principal owner of Brown Oil, Inc., faces the decision to purchase equipment for the company due to competition. His alternatives are presented in the form of a table detailing the costs under favorable and unfavorable market conditions.
In this scenario, business management principles come into play as Ken evaluates the potential costs and benefits of investing in new equipment for Brown Oil, Inc. This decision-making process is crucial for the company's competitiveness and growth in the market.
Considering the financial implications and potential outcomes, Ken will need to assess the risks and rewards associated with each equipment option to make an informed decision that aligns with the company's goals and future success.
A disease is infecting a colony of 1000 penguins living on a remote island. Let P(t) be the number of sick penguins t days after the outbreak. Suppose that 50 penguins had the disease initially, and suppose that the disease is spreading at a rate proportional to the product of the time elapsed and the number of penguins who do not have the disease.
(a) Give the mathematical model(differential equation and initial condition) for P.
(b) Find the generalsolution of the differential equation in (a).
(c) Find the particular solution that satisfies the initial condition.
Answer:
a. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. [tex]C = 950[/tex]
c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
Step-by-step explanation:
a. Let the number of penguins who have the disease t days after the outbreak be P
Initial number of penguins = 1000
Therefore, current number of penguins = 1000 - P
And the rate of spread of disease according to the statement is
[tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]
where k is the constant of proportionality
[tex]\frac{dP}{1000-P}=kt.dt[/tex]
Integrating both sides
[tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. Seeing as 50 penguins had the disease initially,
t = 0
P = 50
The general solution of the differential solution becomes
50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)
[tex]C = 1000 - 50 = 950[/tex]
c. Therefore, the solution that satisfies the initial condition is
[tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]