Answer:
a) [tex]X \sim Binom(n=16, p=0.2)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The expected value is given by this formula:
[tex]E(X) = np=16*0.2=3.2[/tex]
b) [tex]P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Part a
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=16, p=0.2)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The expected value is given by this formula:
[tex]E(X) = np=16*0.2=3.2[/tex]
Part b
For this case we want this probability:
[tex] P(X=0)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And using this function we got:
[tex]P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815[/tex]
Final answer:
The expected number of phishing emails to be caught by the filter is 3.2, and the probability of the filter detecting none of them is approximately 0.02867.
Explanation:
To solve the student's question regarding phishing email filters, we'll apply basic probability concepts.
Part (a): Expected number of phishing emails caught by the filter
Given that software filters catch 20 percent of phishing URLs, and Jason receives 16 phishing e-mails, the expected number of e-mails caught can be calculated as:
Expected number = Total emails × Catch rate
= 16 × 0.20 = 3.2
So, the filter is expected to catch 3.2 phishing e-mails.
Part (b): Probability of the filter detecting none of them
If the filter's catch rate is 20%, then the chance of not catching a single phishing email is the complement, which is 80% or 0.8 for each email. For none of the 16 emails to be caught, we raise this probability to the power of 16:
Probability = (0.8)^16
≈ 0.02867 (rounded to 5 decimal places)
Thus, the chance of the filter not detecting any of the phishing emails is approximately 0.02867.
Prove the superposition principle for nonhomogeneous equations. Suppose that y1 is a solution to Ly1 = f(x) and y2 is a solution to Ly2 = g(x) (same linear operator L). Show that y = y1 + y2 solves Ly = f(x) + g(x). Differential Equation.
Answer:
Step-by-step explanation:
Given Data
Suppose That [tex]y_{1}[/tex] is a solution of L[tex]y_{1}[/tex] = F(x)
and [tex]y_{2}[/tex] is a solution of [tex]Ly_{2}[/tex] = g(x)
L is liner operator
∴ [tex]L(y_{1}+y_{2} )[/tex] = L[tex]y_{1}[/tex] +[tex]Ly_{2}[/tex]
L(y) = F(x) + g(x)
[tex]y_{1}+y_{2}[/tex] is the solution to L(y) = F(x) + g(x)
E.g the liner operator be L = [tex]\frac{d}{dx}[/tex]
[tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex] = f(x)
[tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = g(x)
[tex]\frac{d}{dx}[/tex] [tex](y_{1}+y_{2} )[/tex] = [tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex] + [tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = f(x) + g(x)
Final Answer:
y = y1 + y2 solves the equation Ly = f(x) + g(x).
Explanation:
Certainly! To prove the superposition principle for nonhomogeneous linear differential equations, we will utilize the properties of linear operators and the given solutions for the differential equations.
Let's define L as a linear differential operator. Being linear implies that for any functions u(x) and v(x), and any constants a and b, the operator satisfies the following properties:
1. L(u + v) = L(u) + L(v) (additivity)
2. L(au) = aL(u) (homogeneity)
Given there are two functions y1(x) and y2(x) that are solutions to the nonhomogeneous linear differential equations:
Ly1 = f(x)
Ly2 = g(x)
We need to prove that if you take a linear combination of y1 and y2, the result y = y1 + y2 will also be a solution to the combined nonhomogeneous equation Ly = f(x) + g(x).
Here's how we do it:
Consider a linear combination of y1 and y2, denoted as y = y1 + y2. Apply the linear operator L to both sides of this equation:
L(y) = L(y1 + y2)
Since L is a linear operator, we can apply the additivity property:
L(y) = L(y1) + L(y2)
Now, we know that y1 and y2 are solutions to their respective nonhomogeneous equations, so we can substitute f(x) for L(y1) and g(x) for L(y2):
L(y) = f(x) + g(x)
Thus, we have shown that y = y1 + y2 solves the equation Ly = f(x) + g(x).
This is the proof of the superposition principle for nonhomogeneous linear differential equations. It shows that solutions to such equations can be added together to obtain a new solution corresponding to the sum of the nonhomogeneous parts.
A number between 1 and 15
is chosen at random. What
is the probability that the
number chosen will be a
multiple of 5? please help just learning this will mark brainest got people answering it maki g up junk to get the answer then needing help
The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].
Solution:
Given data:
Number between 1 and 15 is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14.
Total numbers between 1 and 15 = 13
N(S) = 13
Multiple of 5 between 1 and 15 = 5, 10
Number of multiples of 5 between 1 and 15 = 2
N(A) = 2
Probability of multiple of 5 between 1 and 15:
[tex]$P(A)=\frac{N(A)}{N(S)}[/tex]
[tex]$P(A)=\frac{2}{13}[/tex]
The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].
Final answer:
To determine the probability of selecting a multiple of 5 between 1 and 15, count the multiples (5 and 10) and divide by the total number (15), resulting in a probability of 2/15.
Explanation:
The question is related to the concept of probability in mathematics. To find the probability that a randomly chosen number between 1 and 15 is a multiple of 5, you simply count how many multiples of 5 are in that range and divide by the total number of possibilities (15). The multiples of 5 between 1 and 15 are 5 and 10, so there are 2 favorable outcomes. Hence, the probability is 2 divided by 15, which equals 2/15.
Based on outcomes since 1967, the probability a team from the National League wins the baseball World Series in any given year is 0.44. Based on outcomes since 1967, the probability a team from the National Football conference wins the Superbowl in any given year is 0.53. Assume whether a team from the National League wins the World Series in a year is independent of whether a team from the National Football Conference wins the Superbowl in the same year. What is the probability a team from the National League wins the World Series in a year if a team from the National Football Conference wins the Superbowl in the same year
Answer:
Probability a team from national football conference wins the same year will be equal to 0.53
Step-by-step explanation:
Probability that a National League team wins baseball world series = 0.44
Probability that a National Football conference team wins the Superbowl = 0.53
The two events are independent, since the sports played are different (baseball and football). This is a logical assumption to make.
Since these events are independent, the occurrence of one event will not change the probability of the other event. This means that it does not matter whether the national league team won or lost the world series. The probability of the football team winning the Superbowl will remain the same, which is 0.53.
Dandelions are studied for their effects on crop production and lawn growth. In one region, the mean number of dandelions per square meter was found to be 2. We are interested in the number of dandelions in this region. (a) Find the probability of no dandelions in a randomly selected area of 1 square meter in this region. (Round your answer to four decimal places.) (b) Find the probability of at least one dandelion in a randomly selected area of 1 square meter in this region. (Round your answer to four decimal places.)
Answer:
(a) The probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.
(b) The probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.
Step-by-step explanation:
Let X = number of dandelions per square meter.
The average number of dandelions per square meter is, λ = 2.
The random variable X follows a Poisson distribution with parameter λ = 2.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X = 0) as follows:
[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.13534\times 1}{1}=0.1353[/tex]
Thus, the probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.
(b)
Compute the value of P (X ≥ 1) as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
[tex]=1-0.1353\\=0.8647[/tex]
Thus, the probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.
To find the probability of no dandelions, use the Poisson distribution formula with the mean value of 2. To find the probability of at least one dandelion, use the complement rule.
Explanation:To find the probability of no dandelions in a randomly selected area of 1 square meter in this region, we need to use the Poisson distribution. The mean number of dandelions per square meter is 2. The probability of no dandelions is given by the formula: P(X=0) = e^(-2) * (2^0 / 0!). Calculating this gives us approximately 0.1353.
To find the probability of at least one dandelion, we can use the complement rule. The probability of at least one dandelion is equal to 1 minus the probability of no dandelions. So, P(X >= 1) = 1 - P(X=0). Calculating this gives us approximately 0.8647.
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The function f(x,y)=2x + 2y has an absolute maximum value and absolute minimum value subject to the constraint 9x^2 - 9xy + 9y^2 =25. Use Lagrange multipliers to find these values.
Answer:
the minimum is located in x = -5/3 , y= -5/3
Step-by-step explanation:
for the function
f(x,y)=2x + 2y
we define the function g(x)=9x² - 9xy + 9y² - 25 ( for g(x)=0 we get the constrain)
then using Lagrange multipliers f(x) is maximum when
fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →
fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0
g(x) =0 → 9x² - 9xy + 9y² - 25 = 0
subtracting the second equation to the first we get:
2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0
- 18*y + 9*x + 18*x - 9*y = 0
27*y = 27 x → x=y
thus
9x² - 9xy + 9y² - 25 = 0
9x² - 9x² + 9x² - 25 = 0
9x² = 25
x = ±5/3
thus
y = ±5/3
for x=5/3 and y=5/3 → f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 → f(x)= -20/3 (minimum)
finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y
thus the minimum is located in x = -5/3 , y= -5/3
A fishing company operates a search plane to find schools of fish in the ocean. Schools of fish are randomly located in the ocean. On average, there is one school of fish per 100,000 square miles of ocean. During any one day, the plane can search 10,000 square miles.
COMPLETE QUESTION:
What is the expected number of schools of fish found in one day of searching?
Answer: 0.1
Step-by-step explanation:
One school of fish per 100,000 square miles of ocean
Plane can search 10,000 square miles
Therefore,
Expected number of schools of fish found in one day of searching
= 10,000 / 100,000
= 0.1
plsss help i don't knwo thai
Question 4: The value of x is 4.5.
Question 5: The value of x is 15.6.
Solution:
Question 4:
The given triangle is a right triangle.
Using trigonometric formulas for a right triangle.
[tex]$\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}[/tex]
[tex]$\sin C=\frac{AB}{BC}[/tex]
[tex]$\sin 45^\circ=\frac{x}{6.4}[/tex]
[tex]$\frac{1}{\sqrt{2} } =\frac{x}{6.4}[/tex]
Multiply 6.4 on both sides, we get
[tex]$\frac{6.4}{\sqrt{2} }=x[/tex]
x = 4.5
The value of x is 4.5.
Question 5:
The given triangle is a right triangle.
Using trigonometric formulas for a right triangle.
[tex]$\sin \theta=\frac{\text { opposite }}{\text { hypotenuse }}[/tex]
[tex]$\sin E=\frac{EF}{DF}[/tex]
[tex]$\sin 60^\circ=\frac{x}{18}[/tex]
[tex]$\frac{\sqrt3}{2} =\frac{x}{18}[/tex]
Multiply 18 on both sides, we get
x = 15.6
The value of x is 15.6.
A magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 respondents, 13% chose chocolate pie, and the margin of error was given as plus or minus 4 percentage points. What values do p, q,n,E and p represents? If the confidence level is 90%, what is the value of α.
Answer:
The answer to the given problem is given below.
Step-by-step explanation:
What values do p, q,n,E and p represents?
The value of p is the sample proportion.
The value of q is found from evaluating 1− p.
The value of n is the sample size.
The value of E is the margin of error.
The value of p is the population proportion.
If the confidence level is 90%, what is the value of α?
α = 1- 0.90
α = 0.10
Consider the population of four juvenile condors. Their weights in pounds are : 4, 5, 7, 12 (a) Let x be the weight of a juvenile condor. Write the possible unique values for x: (NOTE: Separate each value in the list with a comma.) . (b) Find the mean of the population: (c) Let x¯ be the average weight from a sample of two juvenile condors. List all possible outcomes for x¯. (If a value occurs twice, make sure to list it twice.) This is the sampling distribution for samples of size 2: (NOTE: Separate each value in the list with a comma.) . (d) Find the mean of the sampling distribution: Note: You can earn partial credit on this problem.
Answer:
a) 4, 5, 7, 12
b) 7
c) 4.5, 6.5, 8, 6, 8.5, 9.5
d) 7.167
Step-by-step explanation:
We are given the following in the question:
4, 5, 7, 12
a) unique values for x
4, 5, 7, 12
b) mean of the population
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\mu =\displaystyle\frac{28}{4} = 7[/tex]
c) sampling distribution for samples of size 2
Sample size, n = 2
Possible samples of size 2 are (4,5),(4,7),(4,12),(5,7),(5,12),(7,12)
Sample means are:
[tex]\bar{x_1} = \dfrac{4+5}{2} = 4.5\\\\\bar{x_2} = \dfrac{4+7}{2} = 6.5\\\\\bar{x_3} = \dfrac{4+12}{2} = 8\\\\\bar{x_4} = \dfrac{5+7}{2} = 6\\\\\bar{x_5} = \dfrac{5+12}{2} = 8.5\\\\\bar{x_6} = \dfrac{7+12}{2} = 9.5[/tex]
Thus, the list is 4.5, 6.5, 8, 6, 8.5, 9.5
d) mean of the sampling distribution
[tex]\bar{x} = \dfrac{4.5 + 6.5 + 8+ 6 + 8.5+ 9.5}{6} = \dfrac{43}{6} = 7.167[/tex]
The unique values for x are 4, 5, 7, and 12 pounds. The population mean is 7 pounds, and the sampling distribution for the average weight from a sample of two juvenile condors has possible outcomes of 4.5, 5.5, 6, 8, 8.5, and 9.5 pounds, with a mean of 7 pounds.
Explanation:To answer the student's questions regarding juvenile condor weights:
Possible unique values for x: The unique weights of the juvenile condors are 4, 5, 7, 12 pounds.Mean of the population: The mean (μ) is the average of all the values. To calculate the mean, add all the weights and divide by the number of condors: (4 + 5 + 7 + 12) / 4 = 28 / 4 = 7 pounds.Possible outcomes for ¯x (the average weight from a sample of two juvenile condors): To list all possible outcomes for ¯x, calculate the mean of all possible pairs of condor weights:Since some pairs can be chosen in two ways (e.g., condor 1 and 2, or condor 2 and 1), record the mean for both occurrences. However, in this example, no average is repeated because all condor weights are unique.Mean of the sampling distribution: We can find this by averaging the possible outcomes for ¯x: (4.5 + 5.5 + 8 + 6 + 8.5 + 9.5) / 6 = 42 / 6 = 7 pounds, which is identical to the population mean.
On average, it takes Han Solo 45 seconds to check the coordinates and make the jump into hyperspace. The standard deviation on this important task is 5 seconds. When Han and Chewbacca and their passengers are leaving for Alderaan they make the jump in 33 seconds or less. What is the probability of such an accomplishment?
Answer:
0.0082 or 0.82%
Step-by-step explanation:
Given:
Mean of jump time (μ) = 45 s
Standard deviation (σ) = 5 s
Time for jump required for accomplishment (x) = 33 s
The distribution is normal distribution.
So, first, we will find the z-score of the distribution using the formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Plug in the values and solve for 'z'. This gives,
[tex]z=\frac{33-45}{5}=-2.4[/tex]
So, the z-score of the distribution is -2.4.
Now, we need the probability [tex]P(x\leq 33)=P(z\leq -2.4)[/tex].
From the normal distribution table for z-score equal to -2.4, the value of the probability is 0.0082 or 0.82%.
Therefore, the probability of making a jump in 33 seconds or less is 0.0082 or 0.82%.
Using the normal distribution, it is found that there is a 0.0082 = 0.82% probability of such an accomplishment.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.In this problem:
Mean of 45 seconds, hence [tex]\mu = 45[/tex].Standard deviation of 5 seconds, hence [tex]\sigma = 5[/tex].The probability is the p-value of Z when X = 33, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{33 - 45}{5}[/tex]
[tex]Z = -2.4[/tex]
[tex]Z = -2.4[/tex] has a p-value of 0.0082.
0.0082 = 0.82% probability of such an accomplishment.
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The length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes. Find the probability that it will take between 31 and 40 minutes to complete the test.
Answer:
Probability = 0.23572 .
Step-by-step explanation:
We are given that the length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes.
Let X = length of time needed to complete a certain test
Since, X ~ N([tex]\mu,\sigma^{2}[/tex])
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1) where, [tex]\mu[/tex] = 35 and [tex]\sigma[/tex] = 15
So, P(31 < X < 40) = P(X < 40) - P(X <= 31)
P(X < 40) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{40-35}{15}[/tex] ) = P(Z < 0.33) = 0.62930
P(X <= 31) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{31-35}{15}[/tex] ) = P(Z < -0.27) = 1 - P(Z <= 0.27)
= 1 - 0.60642 = 0.39358
Therefore, P(31 < X < 40) = 0.62930 - 0.39358 = 0.23572 .
To find the probability that it takes between 31 and 40 minutes to complete the test when the test completion times are normally distributed with a mean (μ) of 35 minutes and a standard deviation (σ) of 15 minutes, we can use the properties of the normal distribution.
Firstly, we need to standardize our times (31 and 40 minutes) to find the corresponding z-scores. The z-score is a measure of how many standard deviations an element is from the mean. We use the following formula to calculate the z-score:
\[ z = \fraction{(X - \mu)}{\sigma} \]
where:
- \( X \) is the value for which we are finding the z-score
- \( \mu \) is the mean
- \( σ \) is the standard deviation
We will calculate the z-scores for both 31 minutes and 40 minutes.
For \( X = 31 \):
\[ z_{31} = \fraction{(31 - 35)}{15} = \fraction{-4}{15} \approx -0.267 \]
For \( X = 40 \):
\[ z_{40} = \fraction{(40 - 35)}{15} = \fraction{5}{15} \approx 0.333 \]
Next, we look up these z-scores in the standard normal distribution (z-distribution) table, which will give us the area to the left of each z-score.
If we don't have the z-distribution table available, we might use statistical software or a calculator with a normal distribution function. But let's proceed as if we are using the z-table.
Let's say our z-table gives us the following areas:
Area to the left of \( z_{31} \approx -0.267 \): Approximately 0.3944 (please note that actual values will depend on the specific z-table or calculator you are using).
Area to the left of \( z_{40} \approx 0.333 \): Approximately 0.6304.
Now we want the area between these two z-scores, which represents the probability that the test completion time is between 31 and 40 minutes. To find this, we can subtract the area for \( z_{31} \) from the area for \( z_{40} \):
Probability \( P(31 < X < 40) = P(z_{40}) - P(z_{31}) \)
Substituting the values,
\[ P(31 < X < 40) = 0.6304 - 0.3944 = 0.2360 \]
So the probability that it will take between 31 and 40 minutes to complete the test is approximately 0.2360, or 23.60%.
The number of cars sold weekly by a new automobile dealership grows according to a linear growth model. The first week the dealership sold six cars ( P 0 = 6 ). The second week the dealership sold eight cars ( P 1 = 8 ). Write the recursive formula for the number of cars sold, P n , in the ( n + 1 )th week. P n = P n − 1 + Write the explicit formula for the number of cars sold, P n , in the ( n + 1 )th week. P n = If this trend continues, how many cars will be sold in the fourth week?
(a) Recursive formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week: [tex]P_n = P_{n-1} + 2[/tex]
(b) Explicit formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week:
[tex]P_n = 6 + 2n[/tex]
(c) In the fourth week, the dealership will sell 12 cars.
To find the recursive formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week, we can observe the pattern from the given information.
Given data:
[tex]P_0 = 6[/tex] (number of cars sold in the first week)
[tex]P_1 = 8[/tex] (number of cars sold in the second week)
We can see that each week, the number of cars sold increases by 2. So, the recursive formula can be expressed as:
[tex]P_n = P_{n-1} + 2[/tex]
This formula states that the number of cars sold in the nth week [tex](P_n)[/tex] is equal to the number of cars sold in the previous week [tex](P_{n-1})[/tex] plus 2.
Next, let's find the explicit formula for the number of cars sold, [tex]P_n[/tex], in the (n + 1)th week.
To do this, we need to identify the initial value ([tex]P_0[/tex]) and the common difference (d) in the arithmetic sequence. In this case, the initial value is 6 (P₀ = 6), and the common difference is 2 (the number of cars sold increases by 2 each week).
The explicit formula for an arithmetic sequence is given by:
[tex]P_n = P_0 + n * d[/tex]
Substitute the given values:
[tex]P_n = 6 + n * 2[/tex]
Therefore, the explicit formula for the number of cars sold, P_n, in the (n + 1)th week is [tex]P_n = 6 + 2n[/tex].
Now, let's find how many cars will be sold in the fourth week (n = 3):
[tex]P_3 = 6 + 2 * 3[/tex]
[tex]P_3 = 12[/tex]
In the fourth week, the dealership will sell 12 cars.
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The recursive formula for the number of cars sold weekly is Pn = Pn - 1 + 2, where Pn represents the number of cars in the nth week. The explicit formula is Pn = 6 + 2n. Based on this model, the dealership would sell 14 cars in the fourth week.
Explanation:In your problem, the number of cars sold each week is increasing by a constant amount, following a linear growth model. The first week, 6 cars were sold and the second week, 8 cars were sold. This indicates an increase of 2 cars from week 1 to week 2.
To write the recursive formula for the number of cars sold, Pn, in the (n + 1)th week, we determine the growth by subtraction: P1 - P0 = 8 - 6 = 2. So, every week, the number of cars sold increases by 2. The recursive formula would therefore be Pn = Pn - 1 + 2.
For the explicit formula which gives us the number of cars sold in the n-th week directly, we observe that it started with 6 cars (base) and increments by 2 each week. Therefore, the explicit formula would be Pn = 6 + 2n.
By the fourth week, the number of cars sold would be P4 = 6 + 2 * 4 = 6 + 8 = 14 cars sold.
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Multiply 2 1/2 x 3 2/3
Answer: 9.16666666667
Step-by-step explanation:
(a) Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y' = 4x − 4xy, y(1) = 0. (Round your answer to four decimal places.) y(1.4) =
Answer:
[tex]y\left(1.4\right)=0.992[/tex].
Step-by-step explanation:
The Euler's method states that [tex]y_{n+1}=y_n+h \cdot f \left(x_n, y_n \right)[/tex], where [tex]x_{n+1}=x_n + h[/tex].
To find [tex]y\left(1.4 \right)[/tex] for [tex]y'=- 4 x y + 4 x[/tex] when [tex]y\left(1 \right)=0[/tex], with step size [tex]h=0.2[/tex] using the Euler's method you must:
We have that [tex]h=0.2=\frac{1}{5}[/tex], [tex]x_0=1[/tex], [tex]y_0=0[/tex], [tex]f(x,y)=- 4 x y + 4 x[/tex].
Step 1.
[tex]x_{1}=x_{0}+h=1+\frac{1}{5}=\frac{6}{5}[/tex]
[tex]y\left(x_{1}\right)=y\left( \frac{6}{5} \right)=y_{1}=y_{0}+h \cdot f \left(x_{0}, y_{0} \right)=0+h \cdot f \left(1, 0 \right)=0 + \frac{1}{5} \cdot \left(4.0 \right)=0.8[/tex]
Step 2.
[tex]x_{2}=x_{1}+h=\frac{6}{5}+\frac{1}{5}=\frac{7}{5}=1.4[/tex]
[tex]y\left(x_{2}\right)=y\left( \frac{7}{5} \right)=y_{2}=y_{1}+h \cdot f \left(x_{1}, y_{1} \right)=0.8+h \cdot f \left(\frac{6}{5}, 0.8 \right)=0.8 + \frac{1}{5} \cdot \left(0.96 \right)=0.992[/tex]
The answer is [tex]y\left(1.4\right)=0.992[/tex]
Suppose Gabe, an elementary school student, has just finished dinner with his mother, Judy. Eyeing the nearby cookie jar, Gabe asks his mother if he can have a cookie for dessert. She tells Gabe that she needs to check his backpack to make sure that he finished his homework. Gabe cannot remember where he left his backpack, but he knows for sure that he did not complete his homework and will not be allowed to eat a cookie. Gabe believes his only option is to quickly steal a cookie while his mother is out of the room. Judy then leaves the room to look for Gabe's backpack. Assume that Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. After waiting and not seeing his mother, Gabe decides that he needs a cookie and begins to take one from the jar. Assuming it takes Gabe 15 seconds to grab a cookie from the jar and devour it without a trace, what is the probability that his mother returns in time to catch Gabe stealing a cookie? Please round your answer to the nearest two decimal places.
Answer:
the probability that his mother would return on time to catch Gabe stealing is given by:[tex]\frac{15}{60}=0.25[/tex].
Step-by-step explanation:
probability is given by; the required outcome over the number of possible outcome.
Step1; we have to subtract the first initial 30 secs that Gabe spent wondering if he will be caught. i.e 90 sec-30 secs= 60 Sec.
Step2: we divide the number of seconds that he can finish a cookie without being caught over the number of seconds left before his mother returns. that is why we have [tex]\frac{15}{60} = 0.25[/tex]
The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.
Calculation of the probability:Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. And, it takes Gabe 15 seconds to grab a cookie from the jar.
Now the probability is
[tex]= 15 \div (90 - 30)\\\\= 15 \div 60[/tex]
= 0.25
Hence, we can conclude that The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.
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Consider a sample with a mean of 500 and a standard deviation of 100. What are the z-scores for the following data values: 560, 650, 500, 450, and 300? z-score for 560 z-score for 650 z-score for 500 z-score for 450 z-score for 300
The z-scores for the given data values are approximately:
560: 0.6
650: 1.5
500: 0
450: -0.5
300: -2
Given the mean (μ) of 500 and the standard deviation (σ) of 100, we can calculate the z-scores for the provided data values:
The z-score (also known as the standard score) measures how many standard deviations a data point is away from the mean. It is calculated using the formula:
z = (x - μ) / σ
Where:
x is the data value
μ is the mean of the sample
σ is the standard deviation of the sample
For x = 560:
z = (560 - 500) / 100 = 0.6
For x = 650:
z = (650 - 500) / 100 = 1.5
For x = 500:
z = (500 - 500) / 100 = 0
For x = 450:
z = (450 - 500) / 100 = -0.5
For x = 300:
z = (300 - 500) / 100 = -2
So, the z-scores for the given data values are approximately:
560: 0.6
650: 1.5
500: 0
450: -0.5
300: -2
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The z-scores for the data values 560, 650, 500, 450, and 300 in a sample with a mean of 500 and a standard deviation of 100 are 0.6, 1.5, 0, -0.5, and -2 respectively. They are computed using the formula z = (X - μ) / σ.
Explanation:This question refers to the concept of z-scores in statistics, which is a part of Mathematics. A z-score indicates how many standard deviations a given data point is from the mean. The formula to calculate the z-score is: z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.
The given sample has a mean (μ) of 500 and a standard deviation (σ) of 100. Let's calculate the z-scores:
For X = 560: z = (560 - 500) / 100 = 0.6 For X = 650: z = (650 - 500) / 100 = 1.5 For X = 500: z = (500 - 500) / 100 = 0 For X = 450: z = (450 - 500) / 100 = -0.5 For X = 300: z = (300 - 500) / 100 = -2
So the z-scores for the data values 560, 650, 500, 450, and 300 are 0.6, 1.5, 0, -0.5, and -2 respectively.
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a medical director found mean blood pressure x = 126.07 for an SRS of 72 executives. The standard deviation of the blood pressures of all executives is σ = 15. Give a 90% confidence interval for the mean blood pressure μ of all executives.
Answer:
[tex]126.07-1.64\frac{15}{\sqrt{72}}=123.17[/tex]
[tex]126.07+1.64\frac{15}{\sqrt{72}}=128.97[/tex]
So on this case the 90% confidence interval would be given by (123.17;128.97)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]126.07-1.64\frac{15}{\sqrt{72}}=123.17[/tex]
[tex]126.07+1.64\frac{15}{\sqrt{72}}=128.97[/tex]
So on this case the 90% confidence interval would be given by (123.17;128.97)
The Brutus Gourmet Company produces delicious organic dog treats for canines with discriminating tastes. Management wants the box-filling line to be set so that the process average weight per packet is 127 grams. To make sure that the process is in control, an inspector at the end of the filling line periodically selects a random box of 8 packets and weighs each packet. When the process is in control, the range in the weight of each sample has averaged 10 gramsThe results from the last 5 samples are shown below:Sample Sample Average Sample Range1 124 92 134 83 126 14 127 85 135 7The LCLR = ______________ .
The LCLR is approximately 14.5148.
We have,
To calculate the lower control limit (LCLR) for the range in a control chart, we use the formula:
LCLR = D3 * Average Range
Given the sample ranges of the last 5 samples:
9, 14, 8, 7, and 5, we can calculate the average range:
Average Range = (9 + 14 + 8 + 7 + 5) / 5 = 8.6
The D3 value for a sample size of 8 is typically 1.693.
The lower control limit (LCLR) for the range.
LCLR = D3 * Average Range
= 1.693 * 8.6 = 14.5148
(rounded to four decimal places)
Thus,
The LCLR is approximately 14.5148.
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To find LCLR, calculate the average range of the samples and subtract a constant (d2) from it. In this case, LCLR = 53.977 grams.
Explanation:In this question, we are given the results from the last 5 samples of the weight of packets from the box-filling line. The process average weight per packet is 127 grams. We are asked to find the Lower Control Limit for Range (LCLR) to check if the process is in control. To find LCLR, we need to calculate the average range of the samples and subtract a constant (d2) from it.
First, we calculate the average range of the samples by summing the ranges and dividing by the number of samples. In this case, the sum of the ranges is 92 + 83 + 14 + 85 + 7 = 281. So, the average range is 281 / 5 = 56.2.
Next, we need to find the value of d2, which depends on the sample size. In this case, the sample size is 8. Looking up the value of d2 for a sample sisizeze of 8 in the control chart constants table, we find that it is 2.223. Finally, we can calculate LCLR by subtracting 2.223 from the average range: LCLR = 56.2 - 2.223 = 53.977 grams.
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The time it takes to manufacture a product is modeled by a continuous distribution. The time to manufacture one unit can take anywhere from 5 to 6 minutes with equal probability. What distribution can be used to model the random variable, production time
Answer:
The distribution of the time it takes to manufacture the products can be explained by the continuous Uniform distribution.
Step-by-step explanation:
An Uniform distribution is the probability distribution of outcomes that are equally likely, i.e. all the outcomes has the same probability of occurrence.
Uniform distribution are discrete and continuous.
A discrete uniform distribution describes the probability distribution of discrete random variable that assumes discrete values. For example, roll of a die.
A continuous uniform distribution describes probability distribution of continuous random variable that assumes values in a specified interval. For example, time it takes to reach school from home.
In this case let the random variable X be defined as the time it takes to manufacture a product.
To manufacture 1 unit the time taken is between 5 to 6 minutes.
Every value in the interval 5 - 6 has equal probability.
The distribution of the time it takes to manufacture the products can be explained by the continuous Uniform distribution.
The probability density function of a continuous Uniform distribution is:
[tex]f(x)=\left \{ {{\frac{1}{b-a};\ x\ \epsilon\ [a, b]} \atop {0;\ otherwise}} \right.[/tex]
An oil exploration company currently has two active proj- ects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with and . a. If the Asian project is not successful, what is the proba- bility that the European project is also not successful? Explain your reasoning. b. What is the probability that at least one of the two proj- ects will be successful? c. Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful? P(A) 5 .4 P(B) 5 .7
Answer:
a) P(B'|A') = P(B') = 1 - 0.7 = 0.3.
The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).
b) P(A u B) = 0.82
c) P[(A n B')|(A u B)] = 0.146
Step-by-step explanation:
P(A) = 0.4
P(B) = 0.7
A and B are independent events.
P(A') = 1 - 0.4 = 0.6
P(B') = 1 - 0.7 = 0.3
a) If the Asian project is not successful, what is the proba- bility that the European project is also not successful?
P(B'|A') = P(B') = 1 - 0.7 = 0.3
The reasoning is that: since, events A and B are independent, then, events A' and B' are also independent and for independent events, P(A|B) = P(A).
b) The probability that at least one of the two projects will be successful = P(A u B)
P(A u B) = P(A) + P(B) - P(A n B) = 0.4 + 0.7 - (0.4)(0.7) = 0.82
OR
P(A u B) = P(A n B') + P(A' n B) + P(A n B) = (0.4)(0.3) + (0.6)(0.7) + (0.4)(0.7) = 0.82
c) Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?
P[(A n B')|(A u B)] = }P(A n B') n P(A u B)]/P(A u B) = P(A n B')/P(A u B) = (0.4)(0.3)/(0.82)
P[(A n B')|(A u B)] = 0.146
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A manufacturing process outputs parts having a normal distribution with a mean of 30 cm and standard deviation of 2 cm. From a production sample of 80 parts, what proportion of the sample can be expected to fall between 28 and 32 cm
Answer:
68% of the sample can be expected to fall between 28 and 32 cm
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 30
Standard deviation = 2
What proportion of the sample can be expected to fall between 28 and 32 cm
28 = 30 - 2
28 is one standard deviation below the mean
32 = 30 + 2
32 is one standard deviation above the mean.
By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm
Gabe Amodeo, a nuclear physicist, needs 60 liters of a 60% acid solution. He currently has a 40% solution and a 70% solution. How many liters of each does he need
to make the needed 60 liters of 60% acid solution?
Answer:
20 liters of the 40% solution
60-20=40 liters of the 70% solution
Step-by-step explanation:
40%x+70%(60-x)=60%(60)
0.4x+0.7(60-x)=0.6(60)
0.4x+42-0.7x=36
-0.3x+42=36
-0.3x=-42+36
-0.3x=-6
0.3x=6
3x=60
x=60/3
x=20 liters of the 40% solution
60-20=40 liters of the 70% solution
check: 0.4*20=8
0.7*40=28
8+28=36 and 0.6*60=36 also
Solve for x: 2 over x - 2 + 7 over x^2 - 4 = 5 over x
A. x = -4/3 and x = −5
B. x = -4/3 and x = 5
C. x = 4/3 and x = −5
D. x = 4/3 and x = 5
Answer:
B. x = -4/3 and x = 5
Step-by-step explanation:
2 / (x − 2) + 7 / (x² − 4) = 5 / x
2 / (x − 2) + 7 / ((x + 2) (x − 2)) = 5 / x
Multiply both sides by x − 2:
2 + 7 / (x + 2) = 5 (x − 2) / x
Multiply both sides by x + 2:
2 (x + 2) + 7 = 5 (x − 2) (x + 2) / x
Multiply both sides by x:
2x (x + 2) + 7x = 5 (x − 2) (x + 2)
Simplify:
2x² + 4x + 7x = 5 (x² − 4)
2x² + 4x + 7x = 5x² − 20
0 = 3x² − 11x − 20
Factor:
0 = (3x + 4) (x − 5)
x = -4/3 or x = 5
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000. Find the mean and standard deviation of X - for samples of size 100. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
Answer:
(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411
(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621
Step-by-step explanation:
We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800 and [tex]\sigma[/tex] = 750.
Let X = randomly selected element of the population
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)
P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642
P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .
(b) Now, we are given sample of size, n = 100
So, Mean of X, X bar = 57,800 same as before
But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75
The z probability is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)
P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)
P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621
P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)
This probability is that much small that it is very close to 0
Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .
Using the normal distribution and the central limit theorem, it is found that:
There is a 0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.For samples of size 100, the mean is of 57800 and the standard deviation is 75.There is a 0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.By the Central Limit Theorem, for sampling distributions of samples of size n, the mean is [tex]\mu[/tex] and the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].In this problem:
Mean of 57800, thus [tex]\mu = 57800[/tex].Standard deviation of 750, thus [tex]\sigma = 750[/tex].The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{58000 - 57800}{750}[/tex]
[tex]Z = 0.27[/tex]
[tex]Z = 0.27[/tex] has a p-value of 0.6064.
X = 57000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57000 - 57800}{750}[/tex]
[tex]Z = -1.07[/tex]
[tex]Z = -1.07[/tex] has a p-value of 0.1423.
0.6064 - 0.1423 = 0.4641.
0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.
For samples of size 100, [tex]n = 100[/tex], and then:
[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]
For samples of size 100, the mean is of 57800 and the standard deviation is 75.
Then, the probability is:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{58000 - 57800}{75}[/tex]
[tex]Z = 2.7[/tex]
[tex]Z = 2.7[/tex] has a p-value of 0.9965.
X = 57000:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{57000 - 57800}{75}[/tex]
[tex]Z = -10.7[/tex]
[tex]Z = -10.7[/tex] has a p-value of 0.
0.9965 - 0 = 0.9965.
0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
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A rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches. Find the rate of change of area, A, with respect to time.
Answer:
[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.
Step-by-step explanation:
given that a rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches
Area of rectangle = length * width
A = [tex]\sqrt{t} (5t+4)\\= 5t\sqrt{t} +4\sqrt{t}[/tex]
To find rate of change of A with respect to t, we can find derivative of A with respect to t.
WE get
[tex]\frac{dA}{dt} =5(\frac{3}{2} )\sqrt{t} +\frac{4}{2\sqrt{t} } \\= 3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex]
Hence rate of change of area with respect to time is
[tex]3\sqrt{t} +\frac{2}{\sqrt{t}}[/tex] square inches per seconds.
Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order. y dA, D is bounded by y = x − 42; x = y2 D
To set up the iterated integrals for a given region, we need to find the limits of integration for both x and y. In this case, the region D is bounded by y = x - 42 and x = y^2. By visualizing the region, we can determine the limits of integration for both orders of integration. We can then choose the easier order to evaluate the double integral.
Explanation:To set up the iterated integrals, we need to find the limits of integration for both x and y. First, let's visualize the region D bounded by y = x - 42 and x = y^2. The curve y = x - 42 intersects the parabola x = y^2 at two points: (7, -35) and (-7, -49).
To set up the iterated integral with dx dy order, the outer integral will have limits of integration for y that go from -49 to -35. The inner integral will have limits of integration for x that go from the parabola x = y^2 to the line x = y + 42.
The iterated integral with dy dx order can be set up by reversing the order of integration. The outer integral will have limits of integration for x that go from -7 to 7, and the inner integral will have limits of integration for y that go from the line y = x - 42 to the parabola y = sqrt(x).
To evaluate the double integral using the easier order, we can choose either order and find the corresponding iterated integral.
Let's choose the dy dx order. The limits of integration for the outer integral are x = -7 to x = 7, and for each x, the limits of integration for the inner integral are y = x - 42 to y = sqrt(x).
Now, we can evaluate the double integral.
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Determine which of the following numbers could not be probabilities, and why? Select all that apply. A. The number 175% could not be a probability because it is larger than 100%. B. The number 5.91 could not be a probability because it is larger than 1. C. The number 0.66 could not be a probability because it is smaller than 1. D. The number 0.002 could not be a probability because it is extremely small. E. The number negative 110% could not be a probability because it is negative.
Answer:
A Not a probability B.Probability C.not a probability d.Probability E propability F.not a propability
Step-by-step explanation:
A propability of any event is a number between zero and one
A)175% converted 1.75 so lies outside zero and one not a propability
B)5.91 lies outside the zero and one rage not a propability
C)0.66 lies in the zero and one rage therefore a propability
D)0.002 Lies within the zero and one range therefore a propability
E) -110% converted equals -1.1 so does not lies in this range not a propability
The following numbers could not be probabilities:
A. The number 175% could not be a probability because it is larger than 100%.
B. The number 5.91 could not be a probability because it is larger than 1.
E. The number negative 110% could not be a probability because it is negative.
Probabilities must be between 0 and 1, inclusive. This is because a probability represents the likelihood of an event occurring, and there is no event that can be more likely than certain (1) or less likely than impossible (0).
The numbers 0.66 and 0.002 are valid probabilities.
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A mechanical assembly consists of a rod with a bearing on each end. The three parts are manufactured independently, and all vary a bit from part to part. The length of the rod has mean 23 centimeters (cm) and standard deviation 0.18 millimeters (mm). The length of a bearing has mean 2 cm and standard deviation 0.03 mm. What are the mean and standard deviation of the total length of the assembly? (Round your standard deviation answer to four decimal places.)
Answer:
Total mean= 27cm
Total standard deviation = 0.1849mm
Step-by-step explanation:
The mechanical assembly has three parts consisting of the length of the rod (the this be called x), the length of the bearing (be called y). You might be wondering where the third part is but the assembly usually has two bearings. Then the second length of the bearing (be called z).
From the problem,
μx = 23, μy =2, μz = 2
σx = 0.18, σy = 0.03, σz = 0.03
We can find the total Mean by adding all the means,
μxyz = 23 + 2 + 2 = 27cm
Since the length of the assembly are independent,
To find the total standard deviation, we must first find the total variance(square of standard deviation)
σ² = (0.18)² + (0.03)² + (0.03)² = 0.0342
Now, we find the standard deviation (square root of the variance)
σ = √0.0342 = 0.18493242
σ ≈ 0.1849mm (you can change to cm by multiplying by 10)
The mean total length of the mechanical assembly is 27 cm and the standard deviation is approximately 0.0183 cm when rounded to four decimal places.
Explanation:The question is asking for the mean and standard deviation of the total length of a mechanical assembly with a rod and two bearings. To find the mean total length, we simply add the mean lengths of the rod and two bearings.
The mean length of the rod is 23 cm and of each bearing is 2 cm, which sums up to a mean total length of 27 cm for the assembly since there are two bearings (23 + 2 + 2).
For standard deviation, since the parts are manufactured independently, we apply the rule of variances: The variance of the sum of independent variables is the sum of their variances. The standard deviation of the rod is 0.18 mm (or 0.018 cm), and of each bearing is 0.03 mm (or 0.003 cm).
Thus, the total variance for the assembly is the sum of the variances: (0.0182 + 0.0032 + 0.0032). After calculating, we take the square root to find the total standard deviation, which is approximately 0.0183 cm, rounded to four decimal places.
Based on the provided information about the characteristic roots and the right hand side function g(t), determine the appropriate form of a particular solution to be used with the undetermined coefficient method.
(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)
(b) r1=r2=0; r3=1 g(t)= t^2 +2t + 3
Answer:
Yp = t[Asin(2t) + Acos(2t)]
Yp = t²[At² + Bt + C]
Step-by-step explanation:
The term "multiplicity" means when a given equation has a root at a given point is the multiplicity of that root.
(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)
As you can notice the multiplicity of this equation is 1 since the roots r1 = 2i and r2 = 2i appear for only once.
The form of a particular solution will be
Yp = t[Asin(2t) + Acos(2t)]
where t is for multiplicity 1
(b) r1=r2=0; r3=1 g(t)= t² +2t + 3
As you can notice the multiplicity of this equation is 2 since the roots r1 = r2 = 0 appears 2 times.
The form of a particular solution will be
Yp = t²[At² + Bt + C]
where t² is for multiplicity 2
The Giordanos buy $7300 worth of furniture for their new home. They pay $1000 down. The store charged 10% add-on interest. The Giordanos will pay off the furniture in 18 monthly payments(1 1/2 years). Find the monthly payment.
Answer: the monthly payment is $379.1
Step-by-step explanation:
The balance to be paid would be
7300 - 1000 = $6300
We would apply the periodic interest rate formula which is expressed as
P = a/[{(1+r)^n]-1}/{r(1+r)^n}]
Where
P represents the monthly payments.
a represents the amount to be paid
r represents the annual rate.
n represents number of monthly payments. Therefore
a = $6300
r = 0.1/12 = 0.0083
n = 12 × 1.5 = 18
Therefore,
P = 6300/[{(1+0.0083)^18]-1}/{0.0083(1+0.0083)^18}]
6300/[{(1.0083)^18]-1}/{0.0083(1.0083)^36}]
P = 6300/{1.16 -1}/[0.0083(1.16)]
P = 6300/(0.16/0.009628)
P = 6300/16.618
P = $379.1
Final answer:
The Giordanos will have to pay $385 every month for 18 months to fully pay off the furniture with the 10% add-on interest taken into account.
Explanation:
To calculate the monthly payment for the furniture bought by the Giordanos, we'll start by subtracting the down payment from the total cost to find the amount that will be subject to interest. The total cost of the furniture is $7,300 and they paid $1,000 down, leaving $6,300 to be financed.
With a 10% add-on interest, the total interest would be 10% of $6,300, which is $630. Therefore, the total amount to be repaid is the financed amount plus the interest, which is $6,300 + $630 = $6,930.
The Giordanos plan to pay off this amount in 18 monthly payments. Dividing the total amount by 18 gives us the monthly payment.
Monthly Payment Calculation:
Total amount to be repaid: $6,930
Number of monthly payments: 18
Monthly payment: $6,930 ÷ 18 = $385
The Giordanos will have to pay $385 every month for 18 months to fully pay off the furniture.