PLS HELP WRITE ABSOLUTE VALUE EQUATION FOR GRAPH
Step-by-step explanation:
When x = 0 the value is 1
when x = -1 value is 0
- | - x | +1
determine a lower bound of the series solution for the radius of convergence about the point x0 = −1, x0 = 0, x0 = 1.
The lower bound of the series solution for the radius of convergence about the point x0 = −1 is -2 < x < 0, about the point x0 = 0 is -1 < x < 1, and about the point x0 = 1 is 0 < x < 2.
To determine a lower bound of the series solution for the radius of convergence about the point x0 = −1, x0 = 0, and x0 = 1, we can use the formula for the radius of convergence:
[tex]R = 1/lim sup (|an|^{(1/n)})[/tex]
where an is the nth coefficient of the power series.
For x0 = -1, we consider the power series centered at x0 = -1.
Let the power series be:
∑an(x+1)ⁿ
Then, we can use the ratio test to find the lim sup:
lim sup |an(x+1)ⁿ / a(n-1)(x+1)ⁿ⁻¹| = |x+1|
Therefore, the radius of convergence is:
[tex]R = 1/lim sup (|an|^{(1/n)}) = 1/lim sup (|x+1|^{(1/n)}) = 1[/tex]
So the series converges for all x such that |x+1| < 1, or -2 < x < 0.
For x0 = 0, we consider the power series centered at x0 = 0.
Let the power series be:
∑anxⁿ
Then, we can use the ratio test to find the lim sup:
lim sup |anxⁿ / a(n-1)xⁿ⁻¹| = |x|
Therefore, the radius of convergence is:
[tex]R = 1/lim sup (|an|^{(1/n)}) = 1/lim sup (|x|^{(1/n)}) = 1[/tex]
So the series converges for all x such that |x| < 1.
For x0 = 1, we consider the power series centered at x0 = 1.
Let the power series be:
∑an(x-1)ⁿ
Then, we can use the ratio test to find the lim sup:
lim sup |an(x-1)ⁿ / a(n-1)(x-1)ⁿ⁻¹| = |x-1|
Therefore, the radius of convergence is:
[tex]R = 1/lim sup (|an|^{(1/n)}) = 1/lim sup (|x-1|^{(1/n)}) = 1[/tex]
So, the series converges for all x such that |x-1| < 1, or 0 < x < 2.
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If AB=6 and BC=⅓(AB) what is the area of the rectangle
Answer:
Final answer is 12
Step-by-step explanation:
I have taken this class before and here is my explanation
If the demand function for city bus rides is P = 100 - 10Q and the present price of a ride is 60, then A. Raising prices will increase city revenue (note: remember that revenue = P*Q)
B. Raising prices will decrease city revenue
C. Raising prices will not change city revenue
D. From the information given it is not clear what would happen to city revenue if price is increased.
The correct option is B., that is, Raising prices will decrease city revenue.
To find out what would happen to city revenue if prices are raised, we need to consider the demand function and revenue equation.
The demand function given is P = 100 - 10Q, where P is the price and Q is the quantity demanded.
The revenue equation is R = P*Q, where R is the total revenue earned.
If the current price of a ride is 60, we can find the corresponding quantity demanded by setting P = 60 in the demand function and solving for Q:
60 = 100 - 10Q
10Q = 40
Q = 4
So currently, the city is selling 4 bus rides at a price of 60, which gives a total revenue of:
R = P*Q = 60*4 = 240
Now let's consider what would happen if the price is raised.
For example, if the price is raised to 70, then the demand function becomes:
70 = 100 - 10Q
10Q = 30
Q = 3
So at a price of 70, the city would sell 3 bus rides, which gives a total revenue of:
R = P*Q = 70*3 = 210
Comparing this to the current revenue of 240, we can see that raising prices would decrease city revenue.
Therefore, the correct answer is B. Raising prices will decrease city revenue.
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Find the length of the curvey=ln(x), 1 ≤ x ≤ √(3)arc length = _____?
The length of the curve y=ln(x) from x=1 to x=√(3) is approximately 0.732.
To find the length of the curve y=ln(x) from x=1 to x=√(3), we need to use the formula for arc length:
L = ∫ [1,√(3)] √[1 + (dy/dx)²] dx
First, we need to find dy/dx by taking the derivative of y=ln(x):
dy/dx = 1/x
Now we can substitute this into the formula for arc length and integrate:
L = ∫ [1,√(3)] √[1 + (1/x)²] dx
Using a trig substitution of x=tanθ, we can simplify the integrand:
dx = sec²θ dθ
√[1 + (1/x)²] = √[1 + sec²θ] = tanθsecθ
Substituting these back into the integral, we get:
L = ∫ [0,π/3] tanθsecθ sec²θ dθ
L = ∫ [0,π/3] tanθsec³θ dθ
Using a u-substitution of u=secθ, we can simplify this integral:
du/dθ = secθtanθ
tanθdθ = du/u²
Substituting these back into the integral, we get:
L = ∫ [1,√(3)] u du/u³
L = ∫ [1,√(3)] u⁻² du
L = [-u⁻¹] [1,√(3)]
L = -(√(3)⁻¹ - 1⁻¹)
L = -1 + √(3)
Therefore, the length of the curve y=ln(x) from x=1 to x=√(3) is approximately 0.732.
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An amount of P dollars is borrowed for the given length of time at an annual interest rate of r. Find the simple interest that is owed. (Round your answer to the nearest cent.)P = $3800, r = 3.0%, 9 months
Simple interest is a type of interest that is calculated based on the principal amount of a loan or investment and a fixed rate of interest over a specific period of time.
To find the simple interest owed for a borrowed amount of P dollars at an annual interest rate of r for a given length of time, you can use the formula:
Simple Interest = P × r × t
where P is the principal amount ($3800), r is the annual interest rate (3.0% or 0.03 as a decimal), and t is the time in years. Since the time given is 9 months, we need to convert it to years:
9 months = 9/12 = 0.75 years
Now plug in the values into the formula:
Simple Interest = $3800 × 0.03 × 0.75
Simple Interest = $114
The simple interest that is owed is $114.
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find the t valuelower tail area of .05 with 50 degrees of freedomthe answer is -1.676I'm confused how this is? what do you have to calculate in order to get this answer? I have the t table chart but it only goes to 30 degrees so how would I find 50 degrees without a chart?
The t-value associated with a lower tail area of 0.05 and 50 degrees of freedom is -1.676.
To find the t-value for a lower tail area of 0.05 with 50 degrees of freedom, you would typically consult a t-distribution table.
Since your table only goes up to 30 degrees of freedom, you can use online tools or statistical software to find the required value.
Here are the steps to find this value without a chart:
1. Use an online t-distribution calculator, statistical software, or a spreadsheet program that has built-in statistical functions.
2. Input the necessary information:
degrees of freedom (50) and the tail area (0.05 for a one-tailed test).
3. The calculator or software will provide the t-value, which in this case is -1.676.
Remember that the negative sign indicates that the t-value falls in the lower tail of the distribution.
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a hotelier thinks to welcome 565 guests but has overstated by 20 percent. how many guests does he actually welcome
well, he didn't welcome 565, he's off by 20%, that means he really welcomed 80% of 565, because 100% - 20% = 80%, so
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{80\% of 565}}{\left( \cfrac{80}{100} \right)565}\implies \text{\LARGE 452}[/tex]
You are driving on a highway and are about 195 miles from a state border. You set your cruise control at 60 miles per hour and plan to turn it off within 35 miles of the border on either side. Find the minimum and maximum numbers of hours you plan to have cruise control on.
The minimum duration with cruise control activated is approximately 2.67 hours. The maximum time with cruise control activated is around 2.08 hours.
How to find the minimum and maximum numbers of hours you plan to have cruise control on.Divide the maximum distance by the pace at which you are traveling to find the maximum time with cruise control on:
2.08 hours = 125 miles at 60 miles per hour
Hence, the maximum time with cruise control activated is around 2.08 hours.
To calculate the minimum time with cruise control turned on, multiply 195 miles by 60 miles per hour, which is 3.25 hours.
Subtract the time it would take you to drive 35 miles on either side of the border:
3.25 hours minus 0.58 hours (35 miles per hour x 60 miles per hour) equals 2.67 hours
Hence, the minimum duration with cruise control activated is approximately 2.67 hours.
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(a) Prove that the symbol < defines a relation on Z that is transitive but not reflexive and not symmetric. (b) Is < an antisymmetric relation? Prove your answer.
The relation R is transitive even if it is neither reflexive nor symmetric.
(a) If A=[5,6,7], then define a relation R on A as R=(5,6),(6,5).
The reflexivity of Relation R differs from that of (5,5),(6,6),(7,7)/R).
As a result of (5, 6)R and (6, 5)R, R is now symmetric.
On the other hand, (5,5)/R/R is not transitive.
R is hence symmetric but neither reflexive nor transitive: "(5,6), "(6,5)".
(b) Consider the relation R in the statement R, which is written as R=(a,b):ab.
We have (a,a) / R for any a because a cannot be strictly less than an itself. In reality, a=a.
R has no reflex.
Right now, (1,2)R(as12)
But two is not one less than one.
There is no symmetry in the ratio (2,1)/R.
Now, let (a,b),(b,c)R.
"A,B, and C" is a transitive verb.
As a result, relation R is transitive even if it is neither reflexive nor symmetric.
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Kejuan's square garden has an area of 196 square feet. He needs to replace the fence along two sides of his garden. How much fencing will he need? (Include your units in your answer.)
Answer:
28 ft of fence
Step-by-step explanation:
Area of square = 196 ft^2
Area of square = Length of one side ^2
Each side of Square = sqrt 196
Each side = 14 ft
2 sides of fence = 2 x 14
= 28 ft
To apply the Central Limit Theorem to the sampling distribution of the sample mean, the required sample is typically large enough if: A) n is greater than 50 C) n is less than 30 B) nis 50 or less D) nis 30 or larger
The correct option is D) n is 30 or larger.
What is the required sample size to apply the Central Limit Theorem to the sampling distribution of the sample mean?To apply the Central Limit Theorem (CLT) to the sampling distribution of the sample mean, the required sample size depends on the underlying population distribution.
Specifically, the CLT states that as the sample size (n) increases, the sampling distribution of the sample mean becomes approximately normal regardless of the population distribution.
However, there are some general rules of thumb that can be used to determine if the sample size is large enough to apply the CLT:
If the population is normally distributed, the sample size can be small (less than 30) and still follow a normal distribution.Therefore, the answer to the question is D) n is 30 or larger.
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Alex painted 178 ft2 of his apartment’s walls with one-third gallon of paint. He has 2 gallons of paint in all. If he wants to cover 1,000 ft2 of his apartment, does he have enough paint? Complete a true statement.
Answer:
One-third gallon of paint covers 178 ft2 of walls, so 2 gallons of paint will cover:
2 gallons * (178 ft2 / one-third gallon) = 11,880 ft2
Since Alex wants to cover 1,000 ft2, we can write the following statement:
1,000 ft2 ≤ 11,880 ft2
This statement is true, so Alex has enough paint to cover his apartment
2 gallons of paint will cover [ 1,068 ] ft, and Alex will needs to cover 1,000 ft, so he [ will ] have enough paint
Refer to the Lincolnville school District bus data. Select the variable referring to the number of miles traveled since the last maintenance, and then organize these data into a frequency distribution.What is a typical amount of miles traveled? What is the range?Comment on the shape of the distribution. Are there any outliers in terms of miles driven?Draw a cumulative relative frequency distribution. Forty percent of the buses were driven fewer than how many miles? How many buses were driven less than 10,500 miles?Draw a cumulative relative frequency distribution. Forty percent of the buses were driven fewer than how many miles? How many buses were driven less than 10,500 miles?
(1) the typical amount of miles traveled is 10932.1 miles.
(2) the range is from 9915 up to 11983 miles.
(3) there are no such values in our data, so there is no outlier
What is the average?
This is the arithmetic mean and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.
a-1) The typical amount of miles traveled can be given by measure of the central tendency of data.
As the mean is an unbiased estimator of the central tendency, so we will use 'Mean' as the point estimate of the central tendency representing the typical number of miles traveled.
Use the 'AVERAGE' function in Excel to get the mean of data.
For example, if the values are stored in cell range A1 to A80, then use the formula -
=AVERAGE(A1:A80)
This gives us the point estimate = 10932.1
Thus, the typical amount of miles traveled is 10932.1 miles.
-----------------------------------------------------
a-2)
Range is maximum and minimum values within which all the data lies.
As minimum value of data = 9915
And maximum value of data = 11983
So, the range is from 9915 up to 11983 miles.
a-3)
Use the following Excel functions to get the five-point summary of data -
Minimum Value =MIN(A1:A80)
First Quartile = Q1 =QUARTILE.EXC(A1:A80,1)
Median =MEDIAN(A1:A80)
Third Quartile = Q3 =QUARTILE.EXC(A1:A80,3)
Maximum Value =MAX(A1:A80)
This should give the following values -
Minimum Value 9915
First Quartile = Q1 10400
Median 10919
Third Quartile = Q3 11371
Maximum Value 11983
Then the interquartile range is -
IQR = Q3 - Q1
= 11371 - 10400
= 971
A value is said to be an outlier if it lies below (Q1 - 1.5*IQR) or above (Q3 + 1.5*IQR).
So, the boundary points are -
Q1 - 1.5*IQR = 10400 - 1.5(971)
= 8943.5
And, Q3 + 1.5*IQR = 11371 + 1.5(971)
= 12827.5
So, any value less than 8943.5 or greater than 12827.5 would be an outlier.
As there are no such values in our data, so there is no outlier.
Hence, (1) the typical amount of miles traveled is 10932.1 miles.
(2) the range is from 9915 up to 11983 miles.
(3) there are no such values in our data, so there is no outlier.
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Determine the probability P(1 or fewer) for a binomial experiment with n=8trials and the success probability p=0.3. Then find the mean, variance, and standard deviation.
1) Determine the probability P(1 or fewer). Round the answer to at least four decimal places.
2)Find the mean. If necessary, round the answer to two decimal places.
3)Find the variance and standard deviation. If necessary, round the variance to two decimal places and standard deviation to at least three decimal places.
The following can be answered by the concept of Probability.
1. The probability of getting 1 or fewer successes in 8 trials with a success probability of 0.3 is 0.2590.
2. The mean is 2.4.
3. The variance is 1.68 and the standard deviation is 1.296.
1) To determine the probability P(1 or fewer), we need to calculate the probability of getting 0 successes and the probability of getting 1 success, and then add them together.
Using the formula for binomial probability:
P(X = k) = (n choose k) × p^k × (1-p)^(n-k)
Where X is the number of successes, n is the number of trials, p is the probability of success on each trial, and (n choose k) is the binomial coefficient.
For k=0:
P(X=0) = (8 choose 0) × 0.3⁰ × 0.7⁸ = 0.0576
For k=1:
P(X=1) = (8 choose 1) × 0.3¹ × 0.7⁷ = 0.2014
So P(1 or fewer) = P(X=0) + P(X=1) = 0.2590
Therefore, the probability of getting 1 or fewer successes in 8 trials with a success probability of 0.3 is 0.2590.
2) To find the mean, we use the formula:
μ = np
Where μ is the mean, n is the number of trials, and p is the probability of success on each trial.
Plugging in the values, we get:
μ = 8 × 0.3 = 2.4
Therefore, the mean is 2.4.
3) To find the variance, we use the formula:
σ² = np(1-p)
Where σ² is the variance, n is the number of trials, and p is the probability of success on each trial.
Plugging in the values, we get:
σ² = 8 × 0.3 × 0.7 = 1.68
To find the standard deviation, we take the square root of the variance:
σ = √(1.68) = 1.296
Therefore, the variance is 1.68 and the standard deviation is 1.296.
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I had $370. 0. My Mom gave $150. 0. My Dad gave $150. 0. My Aunt and Uncle gave me $100. 0. I had another $30. 0. How much did I have?
You had a total of 800.0 after receiving all the contributions.
To determine how much you have, simply add the amounts given by your mom, dad, aunt and uncle, and the additional 30.0.
Start with your initial amount:
370.0
Add the amount given by your mom:
150.0
Add the amount given by your dad:
150.0
Add the amount given by your aunt and uncle:
100.0
Add the additional 30.0
Now, let's calculate:
370.0 + 150.0 + 150.0 + 100.0 + 30.0 = 800.0
You had a total of 800.0 after receiving all the contributions.
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How large should nn be to guarantee that the Simpson's rule approximation to ∫10ex2 dx∫01ex2 dx is accurate to within 0.000010.00001?
By Simpson's rule approximation, n should be at least 17 to guarantee that the Simpson's rule approximation is accurate to within 0.00001.
To guarantee that the Simpson's rule approximation to the integral ∫₀¹ e^(x²) dx is accurate to within 0.00001, you need to consider the error bound formula for Simpson's rule:
Error ≤ (K * (b - a)⁵) / (180 * n⁴)
In this case, a = 0, b = 1, and the desired error bound is 0.00001. The function to integrate is f(x) = e^(x²). To find the value of K, you need to determine the maximum value of the fourth derivative of f(x) on the interval [0, 1].
After calculating the fourth derivative, you'll find that K is less than or equal to 12 (K ≤ 12). Plug these values into the error bound formula:
0.00001 ≥ (12 * (1 - 0)⁵) / (180 * n⁴)
Solve for n:
n⁴ ≥ (12 * 1⁵) / (180 * 0.00001)
n⁴ ≥ 66666.67
n ≥ ∛√66666.67
n ≥ 16.10
Since n must be an integer, round up to the nearest whole number. Thus, n should be at least 17 to guarantee that the Simpson's rule approximation is accurate to within 0.00001.
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The pdf of x is f(x) = 0.1, 3 < x < 13. Find P(5 < X < 8).
The probability of X falling between 5 and 8 is 0.3, and this probability is proportional to the length of the interval.
To find P(5 < X < 8), we need to integrate the probability density function (PDF) of X over the interval [5, 8]. Since the PDF of X is given by f(x) = 0.1, 3 < x < 13, we know that the PDF is zero outside this interval.For more such question on probability
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A national science foundation in a certain country collects data on science and engineering (S&E) degrees awarded and publishes the results in a journal. During one year, 72.1% of S&E degrees awarded were for Bachelor's degrees and 35.1% of S&E degrees were Bachelor's degrees awarded to women. What percentage of S&E Bachelor's degrees were awarded to women?
The percentage of S&E Bachelor's degrees awarded to women is also 25.31%.
To find the percentage of S&E Bachelor's degrees awarded to women, follow these steps:
Step 1: Calculate the total number of S&E Bachelor's degrees awarded to women.
If 35.1% of S&E degrees are Bachelor's degrees awarded to women, and we know that 72.1% of S&E degrees are Bachelor's degrees, we can set up a proportion:
Step 2: Solve for the percentage of S&E Bachelor's degrees awarded to women.
To solve for the percentage, simply multiply both sides of the equation by 72.1%:
Percentage of S&E Bachelor's degrees awarded to women = 35.1% * 72.1%
Step 3: Calculate the percentage.
Percentage of S&E Bachelor's degrees awarded to women = 0.351 * 0.721 = 0.253071
Step 4: Convert the decimal to a percentage.
0.253071 * 100 = 25.31%
So, 25.31% of S&E Bachelor's degrees were awarded to women.
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A line segment has endpoints at (-12,19) and (13,- 11).
What is the length of the line segment rounded to the nearest whole number?
[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-12}~,~\stackrel{y_1}{19})\qquad (\stackrel{x_2}{13}~,~\stackrel{y_2}{-11})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(~~13 - (-12)~~)^2 + (~~-11 - 19~~)^2} \implies d=\sqrt{(13 +12)^2 + (-11 -19)^2} \\\\\\ d=\sqrt{( 25 )^2 + ( -30 )^2} \implies d=\sqrt{ 625 + 900 } \implies d=\sqrt{ 1525 }\implies d\approx 39[/tex]
Estimate ∫10cos(x2)dx∫01cos using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with n=4. Give each answer correct to five decimal places.
(a) T4=
(b) M4=
(c) By looking at a sketch of the graph of the integrand, determine for each estimate whether it overestimates, underestimates, or is the exact area.
Underestimate Overestimate Exact 1. M4
Underestimate Overestimate Exact 2. T4
(d) What can you conclude about the true value of the integral?
A. T4<∫10cos(x2)dx
B. T4>∫10cos(x2)dxand M4>∫10cos(x2)dx
C. M4<∫10cos(x2)dx
D. No conclusions can be drawn.
E. T4<∫10cos(x2)dx and M4<∫10cos(x2)dx
a)Using the Trapezoidal Rule with n=4: T4 = 1.06450
b)Using the Midpoint Rule with n=4: M4 = 1.14750
c)M4 overestimates the area while T4 underestimates the area
d) The true value of the integral is T4<∫10cos(x2)dx and M4<∫10cos(x2)dx
What is Trapezoidal Rule?
The Trapezoidal Rule is a numerical integration method that approximates the area under a curve by approximating it with a series of trapezoids and summing their areas.
According to the given information:
(a) Using the Trapezoidal Rule with n=4:
Δx = (1-0)/4 = 0.25
f(0) = cos(0) = 1
f(0.25) = cos(0.0625) ≈ 0.998
f(0.5) = cos(0.25) ≈ 0.968
f(0.75) = cos(0.5625) ≈ 0.829
f(1) = cos(1) ≈ 0.540
T4 = Δx/2 * [f(0) + 2f(0.25) + 2f(0.5) + 2f(0.75) + f(1)]
≈ 0.25/2 * [1 + 2(0.998) + 2(0.968) + 2(0.829) + 0.540]
≈ 1.06450
(b) Using the Midpoint Rule with n=4:
Δx = (1-0)/4 = 0.25
x1 = 0 + Δx/2 = 0.125
x2 = 0.125 + Δx = 0.375
x3 = 0.375 + Δx = 0.625
x4 = 0.625 + Δx = 0.875
f(x1) = cos(0.015625) ≈ 0.999
f(x2) = cos(0.140625) ≈ 0.985
f(x3) = cos(0.390625) ≈ 0.921
f(x4) = cos(0.765625) ≈ 0.685
M4 = Δx * [f(x1) + f(x2) + f(x3) + f(x4)]
≈ 0.25 * [0.999 + 0.985 + 0.921 + 0.685]
≈ 1.14750
(c) Looking at a sketch of the graph of the integrand, it appears that the function is decreasing on the interval [0,1], so the area under the curve should be decreasing. The Midpoint Rule tends to overestimate the area under a decreasing curve, while the Trapezoidal Rule tends to underestimate it. Therefore, the answers are:
M4 overestimates the area
T4 underestimates the area
(d) We can conclude that the true value of the integral is between the estimates given by the Trapezoidal Rule and the Midpoint Rule, since the Trapezoidal Rule underestimates and the Midpoint Rule overestimates. Therefore, we can say:
E. T4<∫10cos(x2)dx and M4<∫10cos(x2)dx
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When building a house, the number of days required to build varies inversely with with the number of workers. One house was built in 19 days by 28 workers. How many days would it take to build a similar house with 7 workers?
The number of days it would take 7 workers to be able to build a similar house would be 76 days.
How to find the number of days ?Proportionally speaking, more builders means a house will take less time to be built. If the number of workers reduces therefore, the number of days for the house to be built will increase.
We need k which is the constant of proportionality:
k = 19 x 28 = 532
The number of days it would take 7 workers is:
532 = d x 7
d = 532 / 7
= 76 days
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a water bottling facility has a mean bottling rate of 35.2 thousand bottles per hour with a standard deviation of 2.04 thousand bottles. a nearby cola bottling facility has a mean bottling rate of 26.9 thousand bottles per hour with a standard deviation of 1.51 thousand bottles. one wednesday from noon to 1:00 p.m., the water bottling facility bottled 37.4 thousand bottles of water, and the cola bottling facility bottled 28.8 thousand bottles of cola. which facility increased their efficiency more during that hour?
During that hour, the cola bottling facility increased their efficiency more compared to the water bottling facility.To determine which facility increased their efficiency more during that hour, we need to calculate the deviation from the mean for each facility.
For the water bottling facility, the deviation is calculated as:
(37.4 - 35.2) / 2.04 = 1.08
For the cola bottling facility, the deviation is calculated as:
(28.8 - 26.9) / 1.51 = 1.26
Since the deviation for the cola bottling facility is higher, this means that they had a larger increase in efficiency during that hour compared to the water bottling facility.
To determine which facility increased their efficiency more during that hour, we will calculate the number of standard deviations away from the mean for each facility's performance.
1. Calculate the deviations for each facility:
Water bottling facility:
Deviation = (Actual bottles - Mean bottles) / Standard deviation
Deviation = (37.4 - 35.2) / 2.04
Deviation ≈ 1.08
Cola bottling facility:
Deviation = (Actual bottles - Mean bottles) / Standard deviation
Deviation = (28.8 - 26.9) / 1.51
Deviation ≈ 1.26
2. Compare the deviations:
The cola bottling facility has a higher deviation (1.26) than the water bottling facility (1.08).
Conclusion:
During that hour, the cola bottling facility increased their efficiency more compared to the water bottling facility.
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the cumulative distribution function of random variable v is fv (v) = 0 v < −5, (v + 5)2/144 −5 ≤v < 7, 1 v ≥7. (a) what are e[v ] and var[v ]?
For cumulative distribution function;
e[v] = 1.25.
var[v] = 53.02.
How to find e[v] and var[v]?we need to integrate v*fv(v) over the entire range of v?
e[v] = ∫v*fv(v) dv from -∞ to ∞
= ∫v*0 dv from -∞ to -5 + ∫v*(v+5)²/144 dv from -5 to 7 + ∫v*1 dv from 7 to ∞
= 0 + [(v³/36 + 5v²/24 + 25v/72) from -5 to 7] + 0
= [(7³/36 + 5*7²/24 + 25*7/72) - (-5³/36 + 5*(-5)²/24 + 25*(-5)/72)]
= 1.25
Therefore, e[v] = 1.25.
To find var[v], we need to first find e[v²]:
e[v²] = ∫v²*fv(v) dv from -∞ to ∞
= ∫v²*0 dv from -∞ to -5 + ∫v²*(v+5)²/144 dv from -5 to 7 + ∫v²*1 dv from 7 to ∞
= 0 + [(v⁴/48 + 5v³/36 + 25v²/144) from -5 to 7] + ∞
= [(7⁴/48 + 5*7³/36 + 25*7²/144) - (-5⁴/48 + 5*(-5)³/36 + 25*(-5)²/144)]
= 54.86
Therefore, e[v²] = 54.86.
Now we can find var[v] using the formula:
var[v] = e[v²] - (e[v])²
= 54.86 - (1.25)²
= 53.02
Therefore, var[v] = 53.02.
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find the probability that a plant of this species will live longer than 126 days. (round your answer to three decimal places.)
We can't say the probability that a plant of this species will live longer than 126 days.
To answer this question, we need to know more information about plant species. Without this information, it is impossible to calculate the probability of a plant living longer than 126 days.
We need to know factors such as the average lifespan of the species, environmental conditions, and any potential diseases or predators that may impact the plant's survival. Please provide more details so I can assist you further.
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Let y=f(x) be the particular solution to the differential equation dydx=ex−1ey with the initial condition f(1)=0. What is the value of f(−2) ?
For differential equation dy/dx=e^x−1e^y, the value of f(-2) is ln(2-e^-2) - 2.
To get the value of f(-2), first solve the above differential equation and locate the specific solution y = f(x) that meets the initial condition f(1) = 0.
The variables in the differential equation can be separated to yield:
(e^y - 1)dx = (e^x - 1)dx
When both sides are combined, the following results:
e^y = e^x - x + C
where C is the integration constant. We can solve for C using the beginning condition f(1) = 0.
e^0 = e^1 - 1 + C
C = 1 - e
By reintroducing this value of C into the equation for ey, we obtain:
ey = e^x - x + 1 - e
We get the following when we take the natural logarithm of both sides and solve for y:
y = ln(e^x - x + 1 - e)
We can now calculate the value of f(-2) by entering x = -2:
f(-2) = ln(e^(-2) + 2 - e) - 2
Using the properties of exponents to simplify the formula inside the natural logarithm, we get:
f(-2) = ln(2 - e^-2) - 2
This is the definitive answer to the question of the value of f(-2).
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Complete question - Let y=f(x) be the particular solution to the differential equation dy/dx=e^x−1e^y with the initial condition f(1)=0. What is the value of f(−2) ?
When would you use the t-distribution procedure to find the confidence interval for the population mean?
Select one:
a. When you do not know the standard deviation of a normally distributed population.
b. When the only thing that you know about a population is its size.
c. When you are working with a population that does not have a normal distribution.
d. Only when you have the standard deviation and the mean of a normally distributed population
Your answer: a. When you do not know the standard deviation of a normally distributed population.
a. When you do not know the standard deviation of a normally distributed population, you would use the t-distribution procedure to find the confidence interval for the population mean. This is because the t-distribution allows for the estimation of the population standard deviation based on the sample standard deviation.
In statistics, the standard deviation is a measure of the variability or spread of an outcome. [1] A low standard deviation indicates that the value is close to the mean of the cluster (also called the expected value), while a high standard deviation indicates that the results are very interesting.
The standard deviation of can be abbreviated as SD and is often used in mathematics and equations with the Greek letter σ (sigma) for population standard deviation or the Latin letter s for different sample sizes.
The standard deviation of a variable, such as a population, a data set, or a probability, is the basis of its variance. Algebraically it is easier than the mean absolute difference, but in practice, it means a lower absolute difference. The useful feature of standard deviation is that it is expressed in the same unit as the data, not the difference.
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The table below shows that the number of miles driven by Jamal is directly proportional to the number of gallons he used.
Gallons Used
Gallons Used
Miles Driven
Miles Driven
14
14
525
525
43
43
1612.5
1612.5
47
47
1762.5
1762.5
How many gallons of gas would he need to travel
296.25
296.25 miles
Jamal would need approximately 7.9 gallons of gas to travel 296.25 miles.
We can use the concept of direct variation to solve this problem. Direct variation means that two quantities are related by a constant ratio. In this case, the number of miles driven is directly proportional to the number of gallons used.
To find the constant of proportionality, we can use the given data. From the table, we can see that when Jamal used 14 gallons, he drove 525 miles. So we can write:
14/525 = k
where k is the constant of proportionality.
Solving for k, we get:
k = 14/525
Now we can use this value of k to find how many gallons Jamal would need to travel 296.25 miles. Let x be the number of gallons he would need. Then we can write:
x/296.25 = k
Substituting the value of k, we get:
x/296.25 = 14/525
Solving for x, we get:
x = (296.25 × 14) / 525
x ≈ 7.9
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find the value of x please !!!
Answer:
the value of x is 11.333 aproximate to 11.3
Step-by-step explanation:
71° = (9X + 40) /2
71° ×2 = 9X + 40 ........ crisscros it
142° = 9X + 40
142-40 = 9X
9X = 102
X = 102/ 9
X = 11.333 ~ 11.3
Consider the following concentration-time data for the decomposition reaction AB→A+B.Time (min) 0 20 40 120 220[AB] 0.206 0.186 0.181 0.117 0.036Determine the value of the rate constant, k.Express your answer using three significant figures. (units= M/s)
To determine the value of the rate constant, we can use the integrated rate law for a first-order reaction, which is: ln([AB]₀/[AB]) = kt.
Where [AB]₀ is the initial concentration of AB, [AB] is the concentration at time t, k is the rate constant, and t is the time.
We can rearrange this equation to solve for k: k = (1/t) * ln([AB]₀/[AB]), We can use any set of data points to calculate k, but it's best to choose a set that gives a straight line when ln([AB]₀/[AB]) is plotted against time.
Let's use the first and second data points:
ln([AB]₀/[AB]) = ln(0.206/0.186) = 0.099
t = 20 min
k = (1/20 min) * 0.099 = 0.00495 min⁻¹
We can also use the other sets of data points to calculate k and check if the values are similar:
ln([AB]₀/[AB]) = ln(0.206/0.181) = 0.135
t = 40 min
k = (1/40 min) * 0.135 = 0.00338 min⁻¹
ln([AB]₀/[AB]) = ln(0.206/0.117) = 0.613
t = 120 min
k = (1/120 min) * 0.613 = 0.00511 min⁻¹
ln([AB]₀/[AB]) = ln(0.206/0.036) = 1.763
t = 220 min
k = (1/220 min) * 1.763 = 0.00801 min⁻¹
The values of k calculated using different sets of data points are similar, which indicates that the reaction is first-order. The average value of k is: k = (0.00495 + 0.00338 + 0.00511 + 0.00801) / 4 = 0.00536 min⁻¹
Therefore, the value of the rate constant is 0.00536 min⁻¹ or 5.36 × 10⁻³ min⁻¹ (rounded to three significant figures). The units of k are min⁻¹ because the time is in minutes and the concentration is in M. We can also express k in units of s⁻¹ by multiplying it by 60, which gives 0.322 s⁻¹ (rounded to three significant figures).
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