Kristina made the investment of $10,500 at 11% and $18,000 at 13% in each account, after one year if the the total interest was $3,495.00.
Let x be the amount invested at 11% and y be the amount invested at 13%.
The sum of the amounts is the total amount invested, which is $28,500.
Therefore, we have:
x + y = 28,500
We are also given that the total interest earned after one year is $3,495.
We can use the simple interest formula:
I = Prt,
where I is the interest,
P is the principal,
r is the interest rate as a decimal,
and t is the time in years. For the 11% account, we have:
I₁ = 0.11x(1) = 0.11x
For the 13% account, we have:
I₂ = 0.13y(1) = 0.13y
The sum of the interests is equal to $3,495, so we have:
0.11x + 0.13y = 3,495
Multiplying the first equation by 0.11, we get:
0.11x + 0.11y = 3,135
Subtracting this equation from the second equation, we get:
0.02y = 360
Dividing both sides by 0.02, we get:
y = 18,000
Substituting this into the first equation, we get:
x + 18,000 = 28,500x = 10,500
Therefore, Kristina invested $10,500 at 11% and $18,000 at 13%.
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The mean age of the employees at a company is 40. The standard deviation of the ages is 3. Suppose the same people were working for the company 5 years ago. What were the mean and the standard deviation of their ages then?
The mean and standard deviation of the employees' ages five years ago were 35 and 3, respectively.
Given that the mean age of the employees in a company is 40 and the standard deviation of their ages is 3. We need to find the mean and standard deviation of their ages five years ago. We know that the mean age of the same group of people five years ago would be 40 - 5 = 35.
Also, the standard deviation of a group remains the same, so the standard deviation of their ages five years ago would be the same, i.e., 3.
Therefore, the mean and standard deviation of the employees' ages five years ago were 35 and 3, respectively.
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Suppose we fit the model Y₁ = B+; to the data (x1, Y1),..., (xn, Yn) using least squares. (Note that there is no intercept.) Suppose the data were actually generated from the model Y;= x² + €is where i~ N(0, 1). Find the mean and variance of B (conditional on x1,...,xn).
To find the mean and variance of B (conditional on x₁, ..., xₙ), we need to consider the least squares estimation process and the properties of the error term €.
In the given model, Yᵢ = xᵢ² + €ᵢ, we have a quadratic relationship between the response variable Y and the predictor variable x, and the error term € follows a normal distribution with mean 0 and variance 1.
The least squares estimation aims to minimize the sum of squared residuals, which can be represented as:
∑(Yᵢ - Bxᵢ²)²
Taking the derivative with respect to B and setting it to zero, we can solve for the value of B that minimizes the sum of squared residuals. However, in this case, since there is no intercept term, the derivative simplifies to:
∑xᵢ²(Yᵢ - Bxᵢ²) = 0
Expanding this equation, we get:
∑xᵢ⁴B = ∑xᵢ²Yᵢ
Solving for B, we have:
B = (∑xᵢ²Yᵢ) / (∑xᵢ⁴)
The mean of B, denoted as E(B), can be calculated by taking the expected value of B given x₁, ..., xₙ:
E(B | x₁, ..., xₙ) = E(∑xᵢ²Yᵢ) / E(∑xᵢ⁴)
Since x₁, ..., xₙ are assumed to be fixed (non-random), we can treat them as constants. Therefore, we can take them out of the expectation:
E(B | x₁, ..., xₙ) = (∑xᵢ²E(Yᵢ)) / (∑xᵢ⁴)
Now, since E(Yᵢ) = E(xᵢ² + €ᵢ) = xᵢ² + E(€ᵢ) = xᵢ², we can simplify the expression further:
E(B | x₁, ..., xₙ) = (∑xᵢ⁴) / (∑xᵢ⁴) = 1
Therefore, the mean of B (conditional on x₁, ..., xₙ) is 1.
To calculate the variance of B (conditional on x₁, ..., xₙ), we need to consider the properties of the error term €. Since € follows a normal distribution with mean 0 and variance 1, it is independent of x₁, ..., xₙ.
The variance of B, denoted as Var(B | x₁, ..., xₙ), can be calculated as follows:
Var(B | x₁, ..., xₙ) = Var(∑xᵢ²Yᵢ) / Var(∑xᵢ⁴)
Again, since x₁, ..., xₙ are constants, we can take them out of the variance:
Var(B | x₁, ..., xₙ) = (∑xᵢ⁴Var(Yᵢ)) / (∑xᵢ⁴)
Since Var(Yᵢ) = Var(xᵢ² + €ᵢ) = Var(€ᵢ) = 1, the expression simplifies to:
Var(B | x₁, ..., xₙ) = (∑xᵢ⁴) / (∑xᵢ⁴) = 1
Therefore, the variance of B (conditional on x₁
, ..., xₙ) is 1.
In summary, the mean of B (conditional on x₁, ..., xₙ) is 1, and the variance of B (conditional on x₁, ..., xₙ) is 1.To find the mean and variance of B (conditional on x₁, ..., xₙ), we need to consider the least squares estimation process and the properties of the error term €.
In the given model, Yᵢ = xᵢ² + €ᵢ, we have a quadratic relationship between the response variable Y and the predictor variable x, and the error term € follows a normal distribution with mean 0 and variance 1.
The least squares estimation aims to minimize the sum of squared residuals, which can be represented as:
∑(Yᵢ - Bxᵢ²)²
Taking the derivative with respect to B and setting it to zero, we can solve for the value of B that minimizes the sum of squared residuals. However, in this case, since there is no intercept term, the derivative simplifies to:
∑xᵢ²(Yᵢ - Bxᵢ²) = 0
Expanding this equation, we get:
∑xᵢ⁴B = ∑xᵢ²Yᵢ
Solving for B, we have:
B = (∑xᵢ²Yᵢ) / (∑xᵢ⁴)
The mean of B, denoted as E(B), can be calculated by taking the expected value of B given x₁, ..., xₙ:
E(B | x₁, ..., xₙ) = E(∑xᵢ²Yᵢ) / E(∑xᵢ⁴)
Since x₁, ..., xₙ are assumed to be fixed (non-random), we can treat them as constants. Therefore, we can take them out of the expectation:
E(B | x₁, ..., xₙ) = (∑xᵢ²E(Yᵢ)) / (∑xᵢ⁴)
Now, since E(Yᵢ) = E(xᵢ² + €ᵢ) = xᵢ² + E(€ᵢ) = xᵢ², we can simplify the expression further:
E(B | x₁, ..., xₙ) = (∑xᵢ⁴) / (∑xᵢ⁴) = 1
Therefore, the mean of B (conditional on x₁, ..., xₙ) is 1.
To calculate the variance of B (conditional on x₁, ..., xₙ), we need to consider the properties of the error term €. Since € follows a normal distribution with mean 0 and variance 1, it is independent of x₁, ..., xₙ.
The variance of B, denoted as Var(B | x₁, ..., xₙ), can be calculated as follows:
Var(B | x₁, ..., xₙ) = Var(∑xᵢ²Yᵢ) / Var(∑xᵢ⁴)
Again, since x₁, ..., xₙ are constants, we can take them out of the variance:
Var(B | x₁, ..., xₙ) = (∑xᵢ⁴Var(Yᵢ)) / (∑xᵢ⁴)
Since Var(Yᵢ) = Var(xᵢ² + €ᵢ) = Var(€ᵢ) = 1, the expression simplifies to:
Var(B | x₁, ..., xₙ) = (∑xᵢ⁴) / (∑xᵢ⁴) = 1
Therefore, the variance of B (conditional on x₁
, ..., xₙ) is 1.
In summary, the mean of B (conditional on x₁, ..., xₙ) is 1, and the variance of B (conditional on x₁, ..., xₙ) is 1.To find the mean and variance of B (conditional on x₁, ..., xₙ), we need to consider the least squares estimation process and the properties of the error term €.
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Scholars are interested in whether women and men have a difference in the amount of time they spend on sports video games (1 point each, 4 points in total) 4A. What is the independent variable? 4B. What is the dependent variable? 4C. Is the independent variable measurement data or categorical data? 4D. Is the dependent variable discrete or continuous?
Answer:4A. The independent variable in this study is gender (male/female).4B. The dependent variable in this study is the amount of time spent on sports video games.4C. The independent variable is categorical data.4D. The dependent variable is continuous.
An independent variable is a variable that is manipulated or changed to determine the effect it has on the dependent variable. In this study, the independent variable is gender because it is the variable that the researchers are interested in testing to see if it has an impact on the amount of time spent playing sports video games.
The dependent variable is the variable that is measured to see how it is affected by the independent variable. In this study, the dependent variable is the amount of time spent playing sports video games because it is the variable that is being tested to see if it is affected by gender.
Categorical data is data that can be put into categories such as gender, race, and ethnicity. In this study, the independent variable is categorical data because it involves the two categories of male and female.
Continuous data is data that can be measured and can take on any value within a certain range such as height or weight. In this study, the dependent variable is continuous data because it involves the amount of time spent playing sports video games, which can take on any value within a certain range.
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The position function s(t)=t 2
−6t−40 represents the position of the back of a car backing out of a driveway and then driving in a straight line, where s is in feet and t is in seconds. In this case, s(t)=0 represents the time at which the back of the car is at the garage door, so s(0)=−40 is the starting position of the car, 40 feet inside the garage. Part 1 - 1 point Part 2 - 1 point Determine the velocity of the car when s(t)=14.
Part 1: Finding the derivative of the position function to get the velocity function, the position function is given by: 's(t) = t^2 - 6t - 40' To find the velocity function, we need to take the derivative of the position function with respect to time: 'v(t) = s'(t) = 2t - 6' Therefore, the velocity function is given by: 'v(t) = 2t - 6'
Part 2: Determining the velocity of the car when s(t) = 14, We are given that 's(t) = 14', and we need to find the velocity of the car at this point. To do this, we can substitute 's(t) = 14' into the velocity function: 'v(t) = 2t - 6', We get: 'v(t) = 2t - 6 = 2(2.8284...) - 6 ≈ -1.34', Therefore, the velocity of the car when 's(t) = 14' is approximately '-1.34' feet per second.
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x1 x2 x3 x4 x5
5 numbers ranging from 1 to 15, and x1 < x2 < x3 < x4 < x5
how many combinations that x1 + x2 + x3 +x4 + x5 = 30
The total number of combinations that x1 + x2 + x3 + x4 + x5 = 30 is:
C(16, 4) + C(15, 4) + C(14, 4) + C(13, 4) + C(12, 4)= 1820 + 1365 + 1001 + 715 + 495
= 5396.
Given that there are 5 numbers ranging from 1 to 15 and x1 < x2 < x3 < x4
< x5. We are to find how many combinations that x1 + x2 + x3 + x4 + x5 =
30.
We are given the following:
5 numbers ranging from 1 to 15.x1 < x2 < x3 < x4 < x5
We are to find how many combinations that x1 + x2 + x3 + x4 + x5 = 30.
Now, if x1 = 1, then we need to find 4 numbers from 2 to 15 which add up to 29.
x1 can be any one of the five numbers:
1, 2, 3, 4, 5.
Therefore, let's consider each of the 5 cases:
Case 1: x1 = 1
If x1 = 1, then we need to find 4 numbers from 2 to 15 which add up to
29 - 1 = 28.
There are 13 numbers from 2 to 15.
So, using the formula of choosing k elements out of n (with the order not mattering), we can find the number of ways to do this as:
C(4 + 13 - 1, 4) = C(16, 4)
Case 2: x1 = 2
If x1 = 2, then we need to find 4 numbers from 3 to 15 which add up to 29 - 2 = 27.
There are 12 numbers from 3 to 15.
So, the number of ways to do this as:
C(4 + 12 - 1, 4) = C(15, 4)
Case 3: x1 = 3
If x1 = 3, then we need to find 4 numbers from 4 to 15 which add up to
29 - 3 = 26.
There are 11 numbers from 4 to 15.
So, the number of ways to do this as:
C(4 + 11 - 1, 4) = C(14, 4)
Case 4: x1 = 4
If x1 = 4, then we need to find 4 numbers from 5 to 15 which add up to
29 - 4 = 25.
There are 10 numbers from 5 to 15.
So, the number of ways to do this as:
C(4 + 10 - 1, 4) = C(13, 4)
Case 5: x1 = 5
If x1 = 5, then we need to find 4 numbers from 6 to 15 which add up to
29 - 5 = 24.
There are 9 numbers from 6 to 15.
So, the number of ways to do this as:
C(4 + 9 - 1, 4) = C(12, 4)
Hence, the total number of combinations that x1 + x2 + x3 + x4 + x5 = 30 is:
C(16, 4) + C(15, 4) + C(14, 4) + C(13, 4) + C(12, 4)= 1820 + 1365 + 1001 + 715 + 495
= 5396.
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(Theoretical Probability MC)
A bucket contains three slips of paper. One of the following colors is written on each slip of paper: Red, Blue, and Yellow.
List 1 List 2 List 3 List 4
Red Red, Blue Red, Red Red, Red, Red
Blue Red, Yellow Red, Blue Red, Blue, Yellow
Yellow Blue, Red Red, Yellow Red, Yellow, Blue
Red Blue, Yellow Blue, Blue Blue, Blue, Blue
Blue Yellow, Red Blue, Red Blue, Red, Yellow
Yellow Yellow, Blue Blue, Yellow Blue, Yellow, Red
Red Red Yellow, Yellow Yellow, Yellow, Yellow
Blue Blue Yellow, Red Yellow, Red, Blue
Yellow Yellow Yellow, Blue Yellow, Blue, Red
Which list gives the sample space for pulling two slips of paper out of the bucket with replacement?
List 1
List 2
List 3
List 4
The list of the sample space for two slips of paper is
Red Red, Blue Red, Yellow RedRed Blue, Blue Blue, Yellow BlueRed Yellow, Blue Yellow, Yellow YellowHow to determine the list of the sample space for two slips of paperFrom the question, we have the following parameters that can be used in our computation:
Slips of paper = 3
Also, we have
Colours = Red, Blue, and Yellow.
When two colors are selected out of the bucket with replacement, we have the following list
Red Red, Blue Red, Yellow Red
Red Blue, Blue Blue, Yellow Blue
Red Yellow, Blue Yellow, Yellow Yellow
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Dell Eatery employs one worker whose job it is to load apple pies on outgoing company cars. Cars arrive at the loading gate at an average of 48 per day, or 6 per hour, according to a Poisson distribution. The worker loads them at a rate of 8 per hour, following approximately the exponential distribution in service times. a. Determine the operating characteristics of this loading gate problem. [6 Marks] b. What is the probability that there will be more than six cars either being loaded or waiting? [2 Marks] Formulae L= μ−λ
λ
W= μ−λ
1
L q
W q
rho
P 0
= μ(μ−λ)
λ 2
= μ(μ−λ)
λ
= μ
λ
=1− μ
λ
P n>k
=( μ
λ
) k+1
The required probability is 0.4408.
The operating characteristics of the loading gate problem are:
L = λ/ (μ - λ)
W = 1/ (μ - λ)
Lq = λ^2 / μ (μ - λ)
Wq = λ / μ (μ - λ)
ρ = λ / μ
P0 = 1 - λ / μ
Where, L represents the average number of cars either being loaded or waiting.
W represents the average time a car spends either being loaded or waiting.
Lq represents the average number of cars waiting.
Wq represents the average waiting time of a car.
ρ represents the utilization factor.
ρ = λ / μ represents the ratio of time the worker spends loading cars to the total time the system is busy.
P0 represents the probability that the system is empty.
The probability that there will be more than six cars either being loaded or waiting is to be determined. That is,
P (n > 6) = 1 - P (n ≤ 6)
Now, the probability of having less than or equal to six cars in the system at a given time,
P (n ≤ 6) = Σn = 0^6 [λ^n / n! * (μ - λ)^n]
Putting the values of λ and μ, we get,
P (n ≤ 6) = Σn = 0^6 [(6/ 48)^n / n! * (8/ 48)^n]
P (n ≤ 6) = [(6/ 48)^0 / 0! * (8/ 48)^0] + [(6/ 48)^1 / 1! * (8/ 48)^1] + [(6/ 48)^2 / 2! * (8/ 48)^2] + [(6/ 48)^3 / 3! * (8/ 48)^3] + [(6/ 48)^4 / 4! * (8/ 48)^4] + [(6/ 48)^5 / 5! * (8/ 48)^5] + [(6/ 48)^6 / 6! * (8/ 48)^6]P (n ≤ 6) = 0.5592
Now, P (n > 6) = 1 - P (n ≤ 6) = 1 - 0.5592 = 0.4408
Therefore, the required probability is 0.4408.
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The height in meters of a projectile at t seconds can be found by the function h(t)=-4.9t^(2)+60t+1.2. Find the height of the projectile 4 seconds after it is launched..
The height of the projectile 4 seconds after it is launched is 164 meters.
The height of a projectile at any given time can be determined using the function h(t) = -4.9t^2 + 60t + 1.2, where h(t) represents the height in meters and t represents time in seconds.
To find the height of the projectile 4 seconds after it is launched, we substitute t = 4 into the function and evaluate it.
Substituting t = 4 into the function, we have:
h(4) = -4.9(4)^2 + 60(4) + 1.2
Simplifying the equation, we get:
h(4) = -4.9(16) + 240 + 1.2
= -78.4 + 240 + 1.2
= 162.8 + 1.2
= 164
This means that after 4 seconds, the projectile reaches a height of 164 meters above the ground. The height can be interpreted as the vertical distance from the ground level.
Therefore, the value obtained is 164 which is the height of the projectile 4 seconds after it is launched.
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At a running race, the ratio of female runners to male runners is 3 to 2. there are 75 more female runners than male runners. determine which of the equations could be used to solve for the amount of male runners (m) in the race and which could not. select true or false for each statement.
The equations that could be used to solve for the number of male runners (m) in the race are (m+75)/m = 3 / 2 and 150 + 2m = 3m. The correct options are A and B.
Given that at a running race, the ratio of female runners to male runners is 3 to 2.
There are 75 more female runners than male runners.
The ratio is written as,
f/ m = 3 / 2
There are 75 more female runners than male runners.
f = m + 75
The equation can be written as,
f / m = 3 / 2
( m + 75 ) / m = 3 / 2
Or
150 + 2m = 3m
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Place the following numbers on the number line in ASCENDING order. − 4 3 ,−3,0.75,−1.8,−3.5 Anu has 9 problems to solve for his math homework. He has solved 2/3 of the problems already. How many of his problems has he solved?
The numbers can be arranged in ascending order as: −4.3,−3.5,−3,−1.8,0.75. Anu has solved 6 out of the 9 math problems assigned to him.
To determine the number of problems Anu has solved, we first calculate 2/3 of the total number of problems. If Anu has 9 problems in total, 2/3 of 9 can be found by multiplying 9 by 2/3. Using the formula for finding a fraction of a number, we have (9 * 2) / 3 = 18 / 3 = 6.
Therefore, Anu has solved 6 problems out of the 9 in his math homework.
In conclusion, the numbers −4.3,−3.5,−3,−1.8,0.75 can be arranged in ascending order, and Anu has solved 6 out of the 9 problems for his math homework.
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Suppose a certain item increased in price by 18% a total of 5 times and then decreased in price by 5% a total of 2 times. By what overall percent did the price increase?
Round your answer to the nearest percent.
In the United States, the annual salary of someone without a college degree is (on average) $31,377, whereas the annual salary of someone with a college degree is (on average) $48,598. If the cost of a four-year public university is (on average) $16,891 per year, how many months would it take for the investment in a college degree to be paid for by the extra money that will be earned with this degree?
Round your answer to the nearest month.
Note: You should not assume anything that is not in the problem. The calculations start as both enter the job market at the same time.
The price increased by approximately 86% overall.
The item's price increased by 18% five times, resulting in a cumulative increase of (1+0.18)^5 = 1.961, or 96.1%. Then, the price decreased by 5% twice, resulting in a cumulative decrease of (1-0.05)^2 = 0.9025, or 9.75%. To calculate the overall percent increase, we subtract the decrease from the increase: 96.1% - 9.75% = 86.35%. Therefore, the price increased by approximately 86% overall.
To determine how many months it would take for the investment in a college degree to be paid for, we calculate the salary difference: $48,598 - $31,377 = $17,221. Dividing the cost of education ($16,891) by the salary difference gives us the number of years required to cover the cost: $16,891 / $17,221 = 0.98 years. Multiplying this by 12 months gives us the result of approximately 11.8 months, which rounds to 12 months.
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1. For the equation x^2/x+3=1/2
do the following:
2 a) Use the Intermediate Value Theorem to prove that the given equation has at least one solution in the interval 0 < x < 2.
b) Find all solutions to the given equation that are in the interval 0 < x < 2.
Given equation is `x^2 / x + 3 = 1 / 2` To use the Intermediate Value Theorem (IVT), we must show that
`f(x) = x^2 / x + 3 - 1/2` is continuous in the given interval 0 < x < 2.
To demonstrate that f(x) is continuous in this interval, we must first check that f(x) is defined for all x in 0 < x < 2.
x + 3 ≠ 0
x ≠ -3
As a result, f(x) is defined for all x ≠ -3, which is also in the given interval. Since f(x) is a polynomial, it is continuous in all x in the domain, including the given interval 0 < x < 2. This implies that f(x) is defined for all x in the interval `(0, 2)`. Let's evaluate f(0) and f(2):f(0) = 0^2 / 0 + 3 - 1/2
= 0 - 1/2 = -1/2f(2)
= 2^2 / 2 + 3 - 1/2
= 4 / 5 - 1/2
= 3/10 Since f(0) and f(2) have opposite signs, we may use the IVT to conclude that there exists at least one real solution for the given equation in the interval `(0, 2)`.
Let us now proceed to find all solutions to the given equation that are in the interval `(0, 2)`.
`x^2 / x + 3 = 1 / 2``x^2 = x / 2 + 3 / 2``x^2 - x / 2 - 3 / 2 = 0`
We must first solve the quadratic equation `x^2 - x / 2 - 3 / 2 = 0` in order to find the solutions to the given equation. Using the quadratic formula, we get:`x = [-(-1/2) ± √((-1/2)^2 - 4(1)(-3/2))]/(2(1))`
`x = [1/2 ± √(1/4 + 6)]/2`
`x = [1/2 ± √25/4]/2`
`x = [1/2 ± 5/2]/2`
Thus, the two solutions to the given equation in the interval `(0, 2)` are:`x = (1 + 5) / 4 = 3/2`
`x = (1 - 5) / 4 = -1/2`
The solution x = -1/2 is not in the interval `(0, 2)`, but it satisfies the given equation. As a result, the two solutions to the given equation are:`x = 3/2` and `x = -1/2`.
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suppose trains arrive at a busy train station at a rate of 1 every 4.64 minutes. what is the probability that the next train arrives 4.92 minutes or more from now? round your answer to 4 decimal places.
We can round the complementary probability to 4 decimal places. Since trains arrive at a rate of 1 every 4.64 minutes, the average time between two consecutive trains is 4.64 minutes.
The rate at which trains arrive at the busy train station is 1 train every 4.64 minutes.
To find the probability that the next train arrives 4.92 minutes or more from now, we need to calculate the complementary probability, which is the probability that the next train arrives within 4.92 minutes from now.
To find this probability, we can subtract the probability of the next train arriving within 4.92 minutes from 1.
Let's calculate the probability of the next train arriving within 4.92 minutes.
Since trains arrive at a rate of 1 every 4.64 minutes, the average time between two consecutive trains is 4.64 minutes.
The probability of the next train arriving within 4.92 minutes is equal to the ratio of 4.92 minutes to the average time between two consecutive trains.
Probability = 4.92 / 4.64
Now, let's calculate the complementary probability:
Complementary probability = 1 - Probability
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Suppose we are preparing a lovely Canard `a l’Orange (roast duck with orange sauce). We first take our duck out of a 36◦F refrigerator and place it in a 350◦F oven to roast. After 10 minutes the internal temperature is 53◦F. If we want to roast the duck until just under well-done (about 170◦F internally), when will it be ready
The duck will be ready in approximately 78.82 minutes when roasted at 350°F to reach an internal temperature of just under 170°F.
To determine when the duck will be ready, we can use the concept of thermal equilibrium and the principle of heat transfer.
Let's assume that the rate of temperature increase follows a linear relationship with time. This allows us to set up a proportion between the temperature change and the time taken.
The initial temperature of the duck is 36°F, and after 10 minutes of roasting, the temperature reaches 53°F. This means the temperature has increased by 53°F - 36°F = 17°F in 10 minutes.
Now, let's calculate the rate of temperature increase:
Rate of temperature increase = (Change in temperature) / (Time taken)
= 17°F / 10 minutes
= 1.7°F per minute
To find out when the duck will reach an internal temperature of 170°F, we can set up the following equation:
Change in temperature = Rate of temperature increase * Time taken
Let's solve for the time taken:
170°F - 36°F = 1.7°F per minute * Time taken
134°F = 1.7°F per minute * Time taken
Time taken = 134°F / (1.7°F per minute)
Time taken ≈ 78.82 minutes
Therefore, when roasted at 350°F for 78.82 minutes, the duck will be done when the internal temperature reaches slightly about 170°F.
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Suppose we are given a list of floating-point values x 1
,x 2
,…,x n
. The following quantity, known as their "log-sum-exp", appears in many machine learning problems: l(x 1
,…,x n
)=ln(∑ k=1
n
e x k
). 1. The value p k
=e x k
often represents a probability p k
∈(0,1]. In this case, what is the range of possible x k
's? 2. Suppose many of the x k
's are very negative (x k
≪0). Explain why evaluating the log-sum-exp formula as written above may cause numerical error in this case. 3. Show that for any a∈R, l(x 1
,…,x n
)=a+ln(∑ k=1
n
e x k
−a
) To avoid the issues you explained in question 2, suggest a value a that may improve computing l(x 1
,…,x n
)
To improve computing l (x1, x n) any value of a can be used. However, to avoid underflow, choosing the maximum value of x k, say a=max {x1, x n}, is a good choice. The value of pk is within the range of (0,1]. In this case, the range of possible x k values will be from infinity to infinity.
When the values of x k are very negative, evaluating the log-sum-exp formula may cause numerical errors. Due to the exponential values, a floating-point underflow will occur when attempting to compute e-x for very small x, resulting in a rounded answer of zero or a float representation of zero.
Let's start with the right side of the equation:
ln (∑ k=1ne x k -a) = ln (e-a∑ k=1ne x k )= a+ ln (∑ k=1ne x k -a)
If we substitute l (x 1, x n) into the equation,
we obtain the following:
l (x1, x n) = ln (∑ k=1 ne x k) =a+ ln (∑ k=1ne x k-a)
Based on this, we can deduce that any value of a would work for computing However, choosing the maximum value would be a good choice. Therefore, by substituting a with max {x1, x n}, we can compute l (x1, x n) more accurately.
When pk∈ (0,1], the range of x k is.
When the x k values are very negative, numerical errors may occur when evaluating the log-sum-exp formula.
a + ln (∑ k=1ne x k-a) is equivalent to l (x1, x n), and choosing
a=max {x1, x n} as a value may improve computing l (x1, x n).
Given a list of floating-point values x1, x n, the log-sum-exp is the quantity given by:
l (x1, x n) = ln (∑ k= 1ne x k).
When pk∈ (0,1], the range of x k is from. This is because the value of pk=e x k often represents a probability pk∈ (0,1], so the range of x k values should be from. When x k is negative, the log-sum-exp formula given above will cause numerical errors when evaluated. Due to the exponential values, a floating-point underflow will occur when attempting to compute e-x for very small x, resulting in a rounded answer of zero or a float representation of zero.
a+ ln (∑ k=1ne x k-a) is equivalent to l (x1, x n).
To improve computing l (x1, x n) any value of a can be used. However, to avoid underflow, choosing the maximum value of x k, say a=max {x1, x n}, is a good choice.
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a band od dwarves is looking for a new mountain to claim and start mining it. It turns out the mountain Is full of gold, then they recieve 100 gold pieces, if IT's full Of silver they get 30 gold pieces, and If there's a dragon there, they get no gold or silver but instead have To pay 80 gold pieces to keep from eating them.
they've identified mr.bottle snaps a potential candidate to claim and start mining. the probability Of funding gold at mt.bottlesnaap is 20%, silver is 50%, and a dragon is 30% what therefore to the nearest gold piece Is the expected value for the dwarves in mining mt. bottlesnap
The expected value for the dwarves in mining Mt. Bottlesnap is 11 gold pieces (rounded to the nearest gold piece).
Let G be the amount of gold pieces the dwarves receive from mining Mt. Bottlesnaap, S be the amount of gold pieces they receive if it's full of silver, and D be the amount of gold pieces they lose if there's a dragon.
We are given:
P(G) = 0.2, with G = 100
P(S) = 0.5, with S = 30
P(D) = 0.3, with D = -80
The expected value of mining Mt. Bottlesnaap can be calculated as:
E(X) = P(G) * G + P(S) * S + P(D) * D
Substituting the given values, we get:
E(X) = 0.2 * 100 + 0.5 * 30 + 0.3 * (-80)
= 20 + 15 - 24
= 11
Therefore, the expected value for the dwarves in mining Mt. Bottlesnap is 11 gold pieces (rounded to the nearest gold piece).
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For the following functions, please list them again but in the order of their asymptotic growth rates, from the least to the greatest. For those functions with the same asymptotic growth rate, please underline them together to indicate that. n!,log 2
(n!),3 n
,(log 2
n) n
,log 2
n n
,(log 10
n) 2
,log 10
n 10
,n 1/2
,5 n/2
The functions can be ordered as follows: 1/2, log₂(n), log₂(n) * n, log₁₀(n), 2, n, 3ⁿ, 5n/2, 10, n!, where the underlined functions have the same asymptotic growth rate.
To order the functions based on their asymptotic growth rates:
1. 1/2: This is a constant value, which does not change as the input size increases.
2. log₂(n): The logarithm grows at a slower rate than any polynomial function.
3. log₂(n) * n: The product of logarithmic and linear terms exhibits a higher growth rate than log₂(n) alone, but still slower than polynomial functions.
4. log₁₀(n) and 2: Both log₁₀(n) and 2 have the same asymptotic growth rate, as logarithmic functions with different bases have equivalent growth rates.
5. n: Linear growth indicates that the function increases linearly with the input size.
6. 3ⁿ: Exponential growth indicates that the function grows at a much faster rate compared to polynomial or logarithmic functions.
7. 5n/2: This is a linear function with a constant factor, which grows at a slightly slower rate than n.
8. 10: This is a constant value, similar to 1/2, indicating no growth with the input size.
9. n!: Factorial growth represents the fastest-growing function among the listed functions.
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Q1 On average, the number of road accidents per year in a specific town is 325 . a) Calculate the mean and standard deviation of the number of accidents per week. b) What is the probability of there being less than 5 accidents in a given week? Q2 A factory requires at least three electrical generators to power it's machines during a power outage. If each generator independently has a 18% probability of failing during a power outage, how many generators should the factory purchase if the factory manager wants at least a 95% probability of successfully powering the machines during a power outage? Hint: the factory will only need a few generators, try an increasing number of generators until the required probability is acheived.
The average number of road accidents per year in a specific town is 325. We will assume that the distribution of the number of accidents per week follows a Poisson distribution with a rate of 325/52 = 6.25 accidents per week.
To find the mean and standard deviation of the number of accidents per week: Mean = rate = 6.25 accidents per week Standard deviation = sqrt(rate) = sqrt(6.25) = 2.5 accidents per week.
Therefore, the mean and standard deviation of the number of accidents per week are 6.25 and 2.5, respectively)To find the probability of there being less than 5 accidents in a given week:
P(X < 5) = P(X ≤ 4)
(since the number of accidents is a discrete random variable)Using the Poisson distribution with a rate of 6.25 accidents per week, we getups(X ≤ 4) = 0.2656 (to 4 decimal places)
Taking the natural logarithm of both Sides[tex]'ll(0.82)X ≥ ln(0.95)X ≥ ln(0.95)/ln(0.82)X ≥ 8.47[/tex] (rounded up)
Therefore, the factory should purchase at least 9 generators to achieve at least a 95% probability of successfully powering the machines during a power outage.
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Which of these functions has;
(i) the smallest growth rate?
(ii) which has the largest growth rate?, as N tends to infinity.
f1(N) = 10 N
f2(N) = N log(N)
f3(N) = 2N
f4(N) = 10000 log(N)
f5(N) = N2
(i) The function with the smallest growth rate as N tends to infinity is f3(N) = 2N. (ii) The function with the largest growth rate as N tends to infinity is f5(N) = N^2.
(i) The function with the smallest growth rate as N tends to infinity is f1(N) = 10N.
To compare the growth rates, we can consider the dominant term in each function. In f1(N) = 10N, the dominant term is N. Since the coefficient 10 is a constant, it does not affect the growth rate significantly. Therefore, the growth rate of f1(N) is the smallest among the given functions.
(ii) The function with the largest growth rate as N tends to infinity is f5(N) = N^2.
Again, considering the dominant term in each function, we can see that f5(N) = N^2 has the highest exponent, indicating the largest growth rate. As N increases, the quadratic term N^2 will dominate the other functions, such as N, log(N), or 2N. The growth rate of f5(N) increases much faster compared to the other functions, making it have the largest growth rate as N tends to infinity.
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Let X={X 1
,X 2
,X 3
} be a set of i.i.d. random variables with joint pdf Y 1
= smallest random variable of {X 1
,X 2
,X 3
} Y 2
= middle random variable of {X 1
,X 2
,X 3
} Y 3
= largest random variable of {X 1
,X 2
,X 3
} Hint: List all permutations of {X 1
,X 2
,X 3
}, then use the transformation of Theorem 9, adding over the permutations. Theorem 9. If a random variable X with pdf p X
(x) is mapped by a monotone function g(x) to a random variable Y, then pdf of Y is p Y
(y)={ ∑ i=1
l
p X
(g −1
(Y)) ∣
∣
∂Y
∂X
∣
∣
,
0,
X=g −1
(Y)
elsewhere
The joint PDF of {Y1, Y2, Y3} is as follows. The joint PDF is calculated from the transformation formula. For the joint PDF of Y, we'll start by finding the distribution functions F(y1), F(y2), and F(y3).
There are 6 potential ways in which X1, X2, and X3 can be sorted, based on the theorem statement. As a result, each summand in the equation below is replicated six times. We have:
∑pY(y1, y2, y3)=∑6(i=1)pX(xi)∣∣∣∂xi∂yi∣∣∣
where {x1, x2, x3} is any permutation of {y1, y2, y3}.Let's take into account the particular permutation {y1 < y2 < y3}. In this case, we must find the range of X1, X2, and X3 values that correspond to a specific set of Y1, Y2, and Y3 values. We have the following inequalities:
x1 < y1; x1 < y2; x1 < y3; x2 > y1; x2 < y2; x2 < y3; x3 > y1; x3 > y2; x3 < y3.
If we subtract the first inequality from the second, we get x2 - x1 > 0. Similarly, the inequality x3 - x2 > 0 and the inequality x3 - x1 > 0 can be obtained from the last two inequalities. This implies that 0 < x2 - x1 < x3 - x2 < x3 - x1. Let d1 = x2 - x1 and d2 = x3 - x2. We have d1 + d2 = x3 - x1 < ∞, so both d1 and d2 are bounded. The bounds of d1 and d2 are 0 and ∞, respectively, because they are both positive and d1 < d2. We have the following conditional probabilities of interest:
P(Y2 > y2 | Y1 = y1, Y3 = y3) = P(X2 > y2 | X1 < y1, X2 > y1, X3 > y3) = 1 - Fx2(y2 | x1 < y1, x2 > y1, x3 > y3).P(Y1 < y1, Y2 > y2 | Y3 = y3) = P(X1 < y1, X2 > y2 | X1 < y1, X2 > y1, X3 > y3) = P(X2 > y2 | X1 < y1, X2 > y1, X3 > y3) = 1 - Fx2(y2 | x1 < y1, x2 > y1, x3 > y3).P(Y1 < y1, Y2 < y2 | Y3 = y3) = P(X1 < y1, X2 < y2 | X1 < y1, X2 < y2, X3 > y3) = P(X2 < y2 | X1 < y1, X2 < y2, X3 > y3) = Fx2(y2 | x1 < y1, x2 < y2, x3 > y3)
We thus have:
pY(y1, y2, y3) = 6pX(x1)pX(x2)pX(x3) ∣∣∣∂x1∂y1∣∣∣∣∣∂x2∂y2∣∣∣∣∣∂x3∂y3∣∣∣ 1{y1 y1; x2 < y2; x2 < y3; x3 > y1; x3 > y2; x3 < y3.
Subtracting the first inequality from the second, we get x2 - x1 > 0. Similarly, the inequality x3 - x2 > 0, and the inequality x3 - x1 > 0 can be obtained from the last two inequalities. This implies that 0 < x2 - x1 < x3 - x2 < x3 - x1. Let d1 = x2 - x1 and d2 = x3 - x2. We have d1 + d2 = x3 - x1 < ∞, so both d1 and d2 are bounded. The bounds of d1 and d2 are 0 and ∞, respectively, because they are both positive and d1 < d2. We need to find the conditional probabilities of interest.
Finally, we can find the joint PDF of {Y1, Y2, Y3} using the transformation formula.The joint PDF is given as follows: pY(y1, y2, y3) = 6pX(x1)pX(x2)pX(x3) ∣∣∣∂x1/∂y1∣∣∣∣∣∂x2/∂y2∣∣∣∣∣∂x3/∂y3∣∣∣ 1{y1}
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If f(x) is a linear function, f(−4)=1, and f(5)=−1, find an equation for f(x) f(x)=
Therefore, the equation for the linear function f(x) is f(x) = (-2/9)x + 1/9.
To find an equation for the linear function f(x), we can use the point-slope form of a linear equation, which is:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line, and m is the slope of the line.
Given the points (-4, 1) and (5, -1), we can calculate the slope (m) using the formula:
m = (y₂ - y₁) / (x₂ - x₁)
m = (-1 - 1) / (5 - (-4))
= -2 / 9
Now, we can select one of the points and substitute the values into the point-slope form to find the equation. Let's choose the point (-4, 1):
y - 1 = (-2/9)(x - (-4))
y - 1 = (-2/9)(x + 4)
y - 1 = (-2/9)x - 8/9
Adding 1 to both sides:
y = (-2/9)x - 8/9 + 1
y = (-2/9)x - 8/9 + 9/9
y = (-2/9)x + 1/9
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The purchase price for a used car, including finance charges is $7242. A down payment of $450 was made. The remainder was paid in 24 equal monthly payments. Find the monthly payment.
If the purchase price for a used car, including finance charges is $7242, a down payment of $450 was made and the remainder was paid in 24 equal monthly payments, then the monthly payment is $283.
To calculate the monthly payment, follow these steps:
The formula to find the purchase price of the car is as follows: Purchase price of the car = Down payment + Remaining amount. ⇒Remaining amount = Purchase price of the car - Down payment. = 7242- 450= $6792.The monthly amount can be calculated by dividing the remaining amount by the number of monthly payments. So, the formula to calculate the monthly amount will be as follows: Monthly amount= Remaining amount/ Number of monthly payments= 6792/24= $283Therefore, the monthly payment would be $283.
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Ivan has 18 bulbs to make a string of holiday lights. How many distinct arrangements can he make if he has 6 red bulbs, 6 green bulbs, 4 blue bulbs, and 2 yellow bulbs?
Ivan can make 133,056,000 distinct arrangements with the 18 bulbs.
To determine the number of distinct arrangements Ivan can make with the given bulbs, we can use the concept of permutations.
The total number of bulbs Ivan has is 18, consisting of 6 red bulbs, 6 green bulbs, 4 blue bulbs, and 2 yellow bulbs.
We can calculate the distinct arrangements using the formula for permutations with repetition. The formula is given by:
P = n! / (n1! * n2! * n3! * ... * nk!)
Where P represents the total number of distinct arrangements, n is the total number of objects (bulbs), and ni represents the number of objects of each type.
Substituting these values into the formula, we get:
P = 18! / (6! * 6! * 4! * 2!)
Calculating this expression gives us:
P = (18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6!) / (6! * 6! * 4! * 2!)
Simplifying the equation, the factorials in the numerator and denominator cancel out:
P = 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7
Evaluating this expression, we find:
P = 133,056,000
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Question Simplify: ((4)/(2n))^(3). You may assume that any variables are nonzero.
The simplified expression is 8/n^(3).
To simplify the expression ((4)/(2n))^(3), we can first simplify the fraction inside the parentheses by dividing both the numerator and denominator by 2. This gives us (2/n) raised to the third power:
((4)/(2n))^(3) = (2/n)^(3)
Next, we can use the exponent rule which states that when a power is raised to another power, we can multiply the exponents. In this case, the exponent on (2/n) is raised to the third power, so we can multiply it by 3:
(2/n)^(3) = 2^(3)/n^(3) = 8/n^(3)
Therefore, the simplified expression is 8/n^(3).
This expression represents a cube of a fraction with numerator 8 and denominator n^3. This expression is useful in various applications such as calculating the volume of a cube whose edges are defined by (4/2n), which is equivalent to half of the edge of a cube of side length n. The expression 8/n^3 can also be used to evaluate certain integrals and solve equations involving powers of fractions.
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15, 6, 14, 7, 14, 5, 15, 14, 14, 12, 11, 10, 8, 13, 13, 14, 4, 13, 3, 11, 14, 14, 12
compute the standard deviation for both sample and population
The sample standard deviation of the given data is approximately 4.0 while the population standard deviation is approximately 3.94.
The formula for computing standard deviation is as follows:
[tex]\[\large\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i-\mu)^2}{n-1}}\][/tex]
where:x is the individual value.μ is the mean (average).n is the number of values.[tex]\(\sigma\)[/tex] is the standard deviation.
A standard deviation is the difference between the average and the square root of the variance of a set of data. Standard deviation measures the amount of variability or dispersion for a subject set of data. We will compute both the sample standard deviation and the population standard deviation.
To calculate the sample standard deviation, we can use the same formula as we did in the population standard deviation, but we must divide by n - 1 instead of n. Thus:
[tex]\[\large s = \sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}\][/tex]
where:[tex]\(\sigma\)[/tex] is the standard deviation.x is the individual value.μ is the mean (average).n is the number of values. [tex]\(\sigma\)[/tex] is the standard deviation.
For the given data 15, 6, 14, 7, 14, 5, 15, 14, 14, 12, 11, 10, 8, 13, 13, 14, 4, 13, 3, 11, 14, 14, 12
we first calculate the mean.
µ = (15+6+14+7+14+5+15+14+14+12+11+10+8+13+13+14+4+13+3+11+14+14+12) / 23=10.6
After that, we compute the standard deviation (sample).
s = √ [ (15-10.6)² + (6-10.6)² + (14-10.6)² + (7-10.6)² + (14-10.6)² + (5-10.6)² + (15-10.6)² + (14-10.6)² + (14-10.6)² + (12-10.6)² + (11-10.6)² + (10-10.6)² + (8-10.6)² + (13-10.6)² + (13-10.6)² + (14-10.6)² + (4-10.6)² + (13-10.6)² + (3-10.6)² + (11-10.6)² + (14-10.6)² + (14-10.6)² + (12-10.6)² ] / 22
s = 4.0
The sample standard deviation is approximately 4.0.
For the population standard deviation, we should replace n-1 by n in the above formula. Thus:
σ = √ [ (15-10.6)² + (6-10.6)² + (14-10.6)² + (7-10.6)² + (14-10.6)² + (5-10.6)² + (15-10.6)² + (14-10.6)² + (14-10.6)² + (12-10.6)² + (11-10.6)² + (10-10.6)² + (8-10.6)² + (13-10.6)² + (13-10.6)² + (14-10.6)² + (4-10.6)² + (13-10.6)² + (3-10.6)² + (11-10.6)² + (14-10.6)² + (14-10.6)² + (12-10.6)² ] / 23
σ = 3.94 (approximately)
Therefore, the population standard deviation is approximately 3.94.
The sample standard deviation of the given data is approximately 4.0 while the population standard deviation is approximately 3.94.
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Given and integer A and B, find the number X such that X*(X+1) falls between [A,B] both inclusive. Assume: 1 <=A<=B<=1e9 and X is non-negative Give an algo. that solves this problem.
This algorithm has a time complexity of O(log B), where B is the given upper bound. It efficiently finds the maximum X that satisfies the given condition within the given range [A, B].
To find the number X such that X*(X+1) falls between [A, B] inclusively, you can use a binary search algorithm. Here's an algorithm that solves the problem:
Set the initial range for X as [0, B].
While the range is valid (lower bound <= upper bound):
a. Calculate the middle value of the range: mid = (lower bound + upper bound) / 2.
b. Calculate the value of mid*(mid+1).
c. If the calculated value is less than A, update the lower bound to mid + 1.
d. If the calculated value is greater than B, update the upper bound to mid - 1.
e. If the calculated value is within the range [A, B], return mid as the answer.
If the loop finishes without finding a solution, return -1 to indicate that no such X exists.
The binary search algorithm works by repeatedly dividing the search range in half until the desired value is found or the range becomes invalid.
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Given the demand equation x=10+20/p , where p represents the price in dollars and x the number of units, determine the elasticity of demand when the price p is equal to $5.
Elasticity of Demand = Therefore, demand is elastic unitary inelastic when price is equal to $5 and a small increase in price will result in an increase in total revenue. little to no change in total revenue.
a decrease in total revenue.
This value is negative, which means that the demand is elastic when p = 5. An elastic demand means that a small increase in price will result in a decrease in total revenue.
Given the demand equation x = 10 + 20/p, where p represents the price in dollars and x the number of units, the elasticity of demand when the price p is equal to $5 is 1.5 (elastic).
To calculate the elasticity of demand, we use the formula:
E = (p/q)(dq/dp)
Where:
p is the price q is the quantity demanded
dq/dp is the derivative of q with respect to p
The first thing we must do is find dq/dp by differentiating the demand equation with respect to p.
dq/dp = -20/p²
Since we want to find the elasticity when p = 5, we substitute this value into the derivative:
dq/dp = -20/5²
dq/dp = -20/25
dq/dp = -0.8
Now we substitute the values we have found into the formula for elasticity:
E = (p/q)(dq/dp)
E = (5/x)(-0.8)
E = (-4/x)
Now we find the value of x when p = 5:
x = 10 + 20/p
= 10 + 20/5
= 14
Therefore, the elasticity of demand when the price p is equal to $5 is:
E = (-4/x)
= (-4/14)
≈ -0.286
This value is negative, which means that the demand is elastic when p = 5.
An elastic demand means that a small increase in price will result in a decrease in total revenue.
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A family wants to have a $160,000 college fund for their children at the end of 18 years. What contribution must be made at the end of each quarter if their investment pays 7.7%, compounded quarterly? (Round your answer to the nearest cent.) $
The contribution to be made at the end of each quarter is $54,547.22.
Given: $160,000, r = 7.7%, n = 4, t = 18 years
To calculate: the contribution to be made at the end of each quarter
We know that;
A = P(1 + r/n)^(nt)
where, A = Amount after time t
P = Principal (initial amount)
r = Annual interest rate
n = Number of times the interest is compounded per year
t = Time the money is invested
The formula can be rearranged as;P = A / (1 + r/n)^(nt)
Using the values given above;
P = $160,000 / (1 + 7.7%/4)^(4*18)
P = $160,000 / (1 + 0.01925)^(72)
P = $160,000 / (1.01925)^(72)
P = $160,000 / 2.9357
P = $54,547.22
Therefore, the contribution to be made at the end of each quarter is $54,547.22.
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In this problem, you will show that equality can be considered as a special case of congruence. Using our definition of congruence, what does a≡b(mod0) mean? Show your work.
"a ≡ b(mod0) means that a and b are equal."
Given, a≡b(mod0)To find what a≡b(mod0) means, we need to understand the definition of congruence.
Two integers are said to be congruent modulo n if their difference is divisible by n.
That is, a ≡ b(mod n) if n divides a-b where n is a positive integer.
Now, substituting 0 in place of n, we get, a ≡ b(mod 0) if 0 divides a-b or in other words a-b = 0. Hence, a ≡ b(mod 0) if a = b.
Since the difference between a and b must be divisible by n, and since 0 is divisible by every integer, the only way for a ≡ b(mod 0) is when a = b.
So, a ≡ b(mod0) means that a and b are equal.
Hence, the answer is "a ≡ b(mod0) means that a and b are equal."
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Solve the following equation by using the Quadratic Formula. When necessary, give answers in simplest radical form. 3x^(2)+4x+1=5
Given equation is 3x²+4x+1 = 5We need to solve the above equation using the quadratic formula.
[tex]x = (-b±sqrt(b²-4ac))/2a[/tex]
[tex]x = (-4±sqrt(4²-4(3)(1)))/2(3)x = (-4±sqrt(16-12))/6x = (-4±sqrt(4))/6[/tex]
Where a, b and c are the coefficients of quadratic On comparing the given equation with the quadratic equation.
[tex]ax²+bx+c=0[/tex]
We get a=3, b=4 and c=1 Substitute the values of a, b and c in the quadratic formula to get the roots of the equation. Solving the equation we get,
[tex]x = (-4±sqrt(4²-4(3)(1)))/2(3)x = (-4±sqrt(16-12))/6x = (-4±sqrt(4))/6[/tex]
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