I guess the "5" is supposed to represent the integral sign?
[tex]I=\displaystyle\int_1^4\ln t\,\mathrm dt[/tex]
With [tex]n=10[/tex] subintervals, we split up the domain of integration as
[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]
For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to
[tex]\ell_i=1+\dfrac{3(i-1)}{10}[/tex]
and right endpoints are given by
[tex]r_i=1+\dfrac{3i}{10}[/tex]
where [tex]1\le i\le10[/tex].
a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, [tex]\dfrac{4-1}{10}=\dfrac3{10}[/tex], and "bases" equal to the values of [tex]\ln t[/tex] at both endpoints of each subinterval. The area of the trapezoid over the [tex]i[/tex]-th subinterval is
[tex]\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)[/tex]
Then the integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}[/tex]
b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of [tex]\ln t[/tex] at the average of the subinterval's endpoints, [tex]\dfrac{\ell_i+r_i}2[/tex]. The area of the rectangle over the [tex]i[/tex]-th subinterval is then
[tex]\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}[/tex]
so the integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}[/tex]
c. For Simpson's rule, we find a quadratic interpolation of [tex]\ln t[/tex] over each subinterval given by
[tex]P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}[/tex]
where [tex]m_i[/tex] is the midpoint of the [tex]i[/tex]-th subinterval,
[tex]m_i=\dfrac{\ell_i+r_i}2[/tex]
Then the integral [tex]I[/tex] is equal to the sum of the integrals of each interpolation over the corresponding [tex]i[/tex]-th subinterval.
[tex]I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt[/tex]
It's easy to show that
[tex]\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)[/tex]
so that the value of the overall integral is approximately
[tex]I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}[/tex]
The question asks to approximate the given integral using three numerical methods: the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule. These methods use simple geometric shapes to estimate the area under the curve. Due to the complexity of the integral in question, assistance from computer software or a graphing calculator will likely be necessary.
Explanation:The question is about using numerical methods to approximate a given integral using three methods: the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule. All of these methods are used to approximate the definite integral of a function over an interval. They divide the interval into n subintervals and then use simple geometric shapes to approximate the area under the curve of the function.
To compute these, you would follow these steps: 1. For the Trapezoidal Rule, average the end points and multiply by the width of each interval. 2. For the Midpoint Rule, evaluate the function at the midpoint of each interval, multiply by the width of each interval. 3. For Simpson's Rule, apply the specific weighted average formula that gives more weight to the midpoint
Please note, however, that due to the complexity of the integral of ln(t), you would likely need to use computer software or a graphing calculator to perform these approximations. Please consult with your teacher for the best approach based on what resources are available to you.
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please help asap!!!!!!
Answer:
103.62 cm
Step-by-step explanation:
Given : Diameter, D = 33 cm
circumference = πD = 3.14 x 33 = 103.62 cm
Hello, I believe your answer is C.
You can find your answer by plugging in the diameter into the circumference formula: 2πr² (You must divide your diameter in half to get the radius).
The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?
Answer:
624
Step-by-step explanation:
The sequence is 49, 47, 45,...., 7, 5, 3. This is an arithmetic sequence, because the difference between terms is the same.
The sum of the first n terms of an arithmetic sequence is:
S = n/2 (a₁ + an)
where a₁ is the first term and an is the nth term.
Here, we know that a₁ = 49 and an = 3. But we need to find what n is. To do that, we use definition of an arithmetic sequence:
an = a₁ + (n-1) d
where d is the common difference (in this case, -2)
3 = 49 + (n-1) (-2)
2(n-1) = 46
n - 1 = 23
n = 24
So there are 24 terms in the sequence.
The sum is:
S = 24/2 (49 + 3)
S = 12 (52)
S = 624
There are 624 bricks in the wall.
The total number of bricks in the wall is 624. This is a math problem that involves arithmetic sequence, where each term is obtained from the previous one by subtracting a fixed number (2, in this case), and concepts from algebra (equations).
Explanation:The problem describes a scenario where each row of a brick wall has two fewer bricks than the row below it, which characterizes a sequence in mathematics. More specifically, this is an arithmetic sequence, which is characterized by a common difference between terms, in this case, the difference is -2.
To solve the problem, we need to find the sum of an arithmetic sequence. The formula of the sum is given by:
S = n/2 * (a1 + an)
Where S is the sum, n the number of terms, a1 the first term, and an the last term. Here, a1 is 49 and an is 3. To find n, we use the formula n = (a1 - an) / d + 1, with d being the common difference which is -2. Solving the equation we find that n = 24.
We now plug these values into the sum formula and find that the sum S, which represents the total number of bricks in the wall is
S = 24/2 * (49 + 3) = 12 * 52 = 624.
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Use the definition of the Laplace transform to find L(f(t) if f(t)=t^5
Answer:
120/s^6
Step-by-step explanation:
There is an easy formula for this...
L(t^n)=n!/(s^(n+1))
Your n=5 here
L(t^5)=5!/(s^6)
L(t^5)=120/s^6
[tex]\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}[/tex]
So
[tex]\mathcal{L}\{f(t)\}=\dfrac{5!}{s^{5+1}}=\dfrac{120}{s^6}[/tex]
A market researcher obtains a list of all streets in a town. She randomly samples 10 street names from the list, and then administers survey questions to every family living on those 10 streets. What type of sampling is this?
Answer: Simple random sampling
Step-by-step explanation:
Given: A market researcher obtains a list of all streets in a town. She randomly samples 10 street names from the list, and then administers survey questions to every family living on those 10 streets.
Since she randomly samples street names , therefore the type of sampling is simple random sampling.
A simple random sample is a sample that is a subset of the population the researcher surveyed selected in a way such that all the individuals in the population has an equal chance to be selected.
Consider the given function and the given interval. f\(x\) = 2 sin\(x\) - sin\(2 x\) text(, ) [0 text(, ) pi] (a) Find the average value fave of f on the given interval. fave = Correct: Your answer is correct. (b) Find c such that fave = f(c). (Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes. Round the answers to three decimal places.)
a. The average value of [tex]f[/tex] on the given interval is
[tex]\displaystyle f_{\rm ave}=\frac1{\pi-0}\int_0^\pi(2\sin x-\sin2x)\,\mathrm dx=\boxed{\frac4\pi}[/tex]
b. Solve for [tex]c[/tex]:
[tex]\dfrac4\pi=2\sin c-\sin2c\implies\boxed{c\approx1.238\text{ or }c\approx2.808}[/tex]
Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48feet above the ground, the function h(t)=−16t2+32t+48 models the height, h, of the ball above the ground as a function of time, t. Find the times the ball will be 48feet above the ground.
[tex]\bf \stackrel{height}{h(t)}=-16t^2+32t+48\implies \stackrel{48~ft}{~~\begin{matrix} 48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=-16t^2+32t~~\begin{matrix} +48 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} \\\\\\ 0=-16t^2+32t\implies 16t^2-32t=0\implies 16t(t-2)=0\implies t= \begin{cases} 0\\ 2 \end{cases}[/tex]
t = 0 seconds, when the ball first took off, and t = 2, 2 seconds later.
Answer: [tex]t_1=0\\t_2=2[/tex]
Step-by-step explanation:
We know that the function [tex]h(t)=-16t^2+32t+48[/tex] models the height "h" of the ball above the ground as a function of time "t".
Then, to find the times in which the ball will be 48 feet above the ground, we need to substitute [tex]h=48[/tex] into the function and solve fot "t":
[tex]48=-16t^2+32t+48\\0=-16t^2+32t+48-48\\0=-16t^2+32t[/tex]
Factorizing, we get:
[tex]0=-16t(t-2)\\t_1=0\\t_2=2[/tex]
Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 4 + ln (t), y = t^2 + 6, (4, 7)
Answer:
y = 2x − 1
Step-by-step explanation:
By eliminating the parameter, first solve for t:
x = 4 + ln(t)
x − 4 = ln(t)
e^(x − 4) = t
Substitute:
y = t² + 6
y = (e^(x − 4))² + 6
y = e^(2x − 8) + 6
Taking derivative using chain rule:
dy/dx = e^(2x − 8) (2)
dy/dx = 2 e^(2x − 8)
Evaluating at x = 4:
dy/dx = 2 e^(8 − 8)
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
Now, without eliminating the parameter, take derivative with respect to t:
x = 4 + ln(t)
dx/dt = 1/t
y = t² + 6
dy/dt = 2t
Finding dy/dx:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (2t) / (1/t)
dy/dx = 2t²
At the point (4, 7), t = 1. Evaluating the derivative:
dy/dx = 2(1)²
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
To find the tangent to the curve represented by the parametric equations x = 4 + ln(t), y = t² + 6, both methods, eliminating and not eliminating the parameter t, yield the same result. The slope of the tangent line at the point (4, 7) is determined to be 2, thus the equation of the tangent is y - 7 = 2(x - 4).
To find the equation of the tangent to the given curve at the point (4, 7) with the parametric equations x = 4 + ln(t) and y = t² + 6, we can approach the problem in two ways: with and without eliminating the parameter t.
Firstly, without eliminating the parameter, we need to find the derivatives dx/dt and dy/dt, and then use them to find dy/dx which is the slope of the tangent at the given point. Since dx/dt = 1/t and dy/dt = 2t, at the point (4, 7), we have t = 1, making the slope dy/dx = (dy/dt)/(dx/dt) = 2 × 1 / (1/1) = 2. The equation of the tangent line can thus be written as y - y₁ = m(x - x₁), where m is the slope, and (x₁, y₁) is the point of tangency.
This gives us the equation y - 7 = 2(x - 4).
Federal Rent-a-Car is putting together a new fleet. It is considering package offers from three car manufacturers. Fred Motors is offering 5 small cars, 5 medium cars, and 10 large cars for $500,000. Admiral Motors is offering 5 small, 10 medium, and 5 large cars for $400,000. Chrysalis is offering 10 small, 5 medium, and 5 large cars for $300,000. Federal would like to buy at least 700 small cars, at least 600 medium cars, and at least 700 large cars. How many packages should it buy from each car maker to keep the total cost as small as possible?
Answer:
40 packages from Fred Motors20 packages from Admiral Motors40 packages from ChrysalisStep-by-step explanation:
I would formulate the problem like this. Let f, a, c represent the numbers of packages bought from Fred Motors, Admiral Motors, and Chrysalis, respectively. Then the function to minimize (in thousands) is …
objective = 500f +400a +300c
The constraints on the numbers of cars purchased are …
5f +5a +10c >= 700
5f +10a +5c >= 600
10f +5a +5c >= 700
Along with the usual f >=0, a>=0, c>=0. Of course, we want all these variables to be integers.
Any number of solvers are available in the Internet for systems like this. Shown in the attachments are the input and output of one of them.
The optimal purchase appears to be …
40 packages from Fred Motors20 packages from Admiral Motors40 packages from ChrysalisThe total cost of these is $40 million.
This is a linear programming problem that requires to minimize a cost function subject to several constraints about the total number of small, medium, and large cars in the fleet. It can be set up using the system of inequalities and then solved using methods like the Simplex one.
Explanation:This problem can be solved through linear programming, which involves creating a system of inequalities to represent the constraints of the problem, and then optimizing a linear function. To start, let's define the variables: x is the number of packages bought from Fred Motors, y is the number from Admiral Motors, and z is the number from Chrysalis.
The fleet requirements translate to the following constraints: 5x + 5y + 10z ≥ 700 (small cars), 5x + 10y + 5z ≥ 600 (medium cars), and 10x + 5y + 5z ≥ 700 (large cars).
Then, the cost to minimize is: $500,000x + $400,000y + $300,000z.
This is a linear programming problem and can be solved using various methods, such as the Simplex method or graphically. Exact solutions would require a more detailed analysis.
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Solve |z| > (1/2)
{-1/2, 1/2}
{z|(-1/2) < z < (1/2)}
{z|z < (-1/2) ∪ z > (1/2)}
Answer:
{[tex]z|z<-\frac{1}{2}[/tex]}U{[tex]z|z>\frac{1}{2}[/tex]}
Step-by-step explanation:
Given the inequality [tex]|z| > \frac{1}{2}[/tex] you need to set up two posibilities:
FIRST POSIBILITY : [tex]z>\frac{1}{2}[/tex]
SECOND POSIBILTY: [tex]z<-\frac{1}{2}[/tex]
Therefore, you got that:
[tex]z<-\frac{1}{2}\ or\ z>\frac{1}{2}[/tex]
Knowing this, you can write the solution obtained in Set notation. This is:
Solution: {[tex]z|z<-\frac{1}{2}[/tex]}U{[tex]z|z>\frac{1}{2}[/tex]}
Answer:
{z|z < (-1/2) ∪ z > (1/2)}
Step-by-step explanation:
I got it right. I would explain if I were better at making sense lol
Consider the vector field F=(x2+y2,4xy). Compute the line integrals ∫c1F⋅ds and ∫c2F⋅ds, where c1(t)=(t,t2) and c2(t)=(t,t) for 0≤t≤1. Can you decide from your answers whether or not F is a gradient vector field? Why or why not?
[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec s=\int_0^1(t^2+t^4,4t^3)\cdot(1,2t)\,\mathrm dt=\int_0^1(t^2+9t^4)\,\mathrm dt=\boxed{\frac{32}{15}}[/tex]
Integral over [tex]C_2[/tex]:[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1(2t^2,4t^2)\cdot(1,1)\,\mathrm dt=\int_0^16t^2\,\mathrm dt=\boxed{2}[/tex]
The value of the line integral depends on the path, so [tex]\vec F[/tex] is not a gradient vector field.
The line integrals ∫c1F⋅ds and ∫c2F⋅ds are computed by replacing x and y with the parametric representations, calculating ds, completing the dot product, and conducting the integration. If the results are identical, F is a gradient vector field.
Explanation:To compute the line integrals, ∫c1F⋅ds and ∫c2F⋅ds, where c1(t)=(t, t2) and c2(t)=(t,t) for 0≤t≤1 of the vector field F=(x2+y2,4xy), we can reduce each of them to an integral over t, the parameter of the path. In the case of c1(t), replace x and y by t and t² correspondingly, for calculation. Similarly, in the case of c2(t), replace x and y by t in calculations.
Let's consider ∫c1F⋅ds. Here, F = (t²+t⁴,4t³) and ds can be calculated using the Pythagorean theorem leading to sqr(1+4t²). The dot product F.ds is then calculated and integrated from 0 to 1. Repeat the process for ∫c2F⋅ds.
A vector field F is said to be a gradient vector field if integral from one point to another remains the same regardless of the path chosen to get from one point to the other. Comparing the obtained results will determine the truth of this statement.
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x = Temperature (°C) 1100 1200 1300 1100 1500 1200 1300 y = Porosity (%) 30.8 19.2 6 13.5 11.4 7.7 3.6 (a) Fit the simple linear regression model using the method of least squares. Find the least squares estimates of the intercept and slope in the simple linear regression model.
Answer:
intercept: 55.6256slope: -0.0341585Step-by-step explanation:
This sort of problem is best worked by a tool such as a graphing calculator or spreadsheet.
Suppose that 3% of all athletes are using the endurance-enhancing hormone EPO (you should be able to simply compute the percentage of all athletes that are not using EPO). For our purposes, a “positive” test result is one that indicates presence of EPO in an athlete’s bloodstream. The probability of a positive result, given the presence of EPO is .99. The probability of a negative result, when EPO is not present, is .90. What is the probability that a randomly selected athlete tests positive for EPO? 0.0297
Answer:
Step-by-step explanation:
So there is a 3% probability that an athlete is using EPO .
The probability of showing positive on a test when you've used it is 0.99.
3% x 0.99= 2.97%
The probability of a positive result without EPO is 0.1
97% x 0,1 = 9,7 %
My guess is that 2.97% + 9,7% = 12.67% or 0.1267.
I don't know i may be wrong because you've put as an answer 0.0297 but if you like you may take only the first part of the answer.
There is a 0.1267 = 12.67% probability that a randomly selected athlete tests positive for EPO.
A positive test can happen in two cases:
When EPO is present(3% of the time), with 0.99 probability.When EPO is not present(100 - 3 = 97% of the time), with 1 - 0.9 = 0.1 probability.Then, adding these probabilities:
[tex]p = 0.03(0.99) + 0.97(0.1) = 0.1267[/tex]
0.1267 = 12.67% probability that a randomly selected athlete tests positive for EPO.
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How many different three-digit numbers can be formed using the digits 1 comma 2 comma 9 comma 6 comma 4 comma 3 comma and 8 without repetition? For example, 664 is not allowed.
Answer:
129 468
Step-by-step explanation:
I'm not sure if you're supposed to reuse the same numbers just in a different Three digit number. Or if you're supposed to use the number one time and one time only. But if not here's there's some numbers that you could use
He basically all You have to do Is take the numbers and turning them into three digit numbers Without repetition.
Hope this helps you!
Also it's saying that 664 is not allowed Because it they are reusing the six When there's no Extra six to use. So remind you not to use the same number twice!
Explain why vertical lines are a special case in the definition of parallel lines.
Answer:
A vertical line has an infinite or undefined slope since the denominator is zero.
Step-by-step explanation:
Parallel lines by definition refers to lines that never intersect or meet since they have identical slopes. The slope of line is defined as;
(change in y)/(change in x)
For a vertical line, the y values are changing while the x values remain constant. The slope of this line will thus have a zero value in the denominator implying that its slope will not defined or will be infinity.
Answer:
A vertical line has an infinite or undefined slope since the denominator is zero.
Step-by-step explanation:
College algebra homework review... Having issues calculating this by hand and on TI-84 receiving errors like "8e12" when trying to calculate the actual quadratic equation it calls for in question B.... Please help
checking the vertex of this upside-down parabola, it has a vertex at (1000, 2000000), so that's the U-turn, when as the price "p" increases the revenue goes down.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{a}x^2\stackrel{\stackrel{b}{\downarrow }}{+b}x\stackrel{\stackrel{c}{\downarrow }}{+c} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
now, if we solve the quadratic using the value of 500000
[tex]\bf \stackrel{R(p)}{500000}=-2p^2+4000p\implies 250000=-p^2+2000p \\\\\\ p^2-2000p+250000=0[/tex]
and we run the quadratic formula on it, we get the values of x = 133.97 and x = 1866.03, one value is obviously when going upwards, the first one, and the other is when going downwards.
so we know that the R(p) is 500,000 at x = 133.97, and it keeps on going up, up to the vertex above at x = 1000, so we can say from x = [134, 1000] R(p) > 500000.
what is the measurement of angle p? Round your answer to the nearest degree.
A. 29°
B.42°
C.65°
D.78°
You can use the law sines, which states that in a triangle the ratio between one side length and the sine of the opposite angle is constant.
So, we have
[tex]\dfrac{PR}{\sin(Q)}=\dfrac{QR}{\sin(P)}=\dfrac{PQ}{\sin(R)}[/tex]
In particular, we can use
[tex]\dfrac{PR}{\sin(Q)}=\dfrac{QR}{\sin(P)}[/tex]
to write
[tex]\dfrac{68}{\sin(73)}=\dfrac{47.6}{\sin(P)} \iff \sin(P) = \dfrac{47.6\sin(73)}{68}\approx 0.66[/tex]
Which means
[tex]P\approx \arcsin(0.66)\approx 42[/tex]
Travel: One Cyclist drives at Six miles per hour faster then another Cyclist. Express the speed of the faster Cyclist in terms of the speed of the lowest Cyclist .....
Answer:
Faster cyclist: 6x
Slower: x
Step-by-step explanation:
x = mph
The speed of the faster cyclist is expressed as v + 6 mph, where v represents the speed of the slower cyclist.
To express the speed of the faster cyclist in terms of the speed of the slower cyclist, let's denote the speed of the slower cyclist as v mph. The problem states that the faster cyclist travels at a speed that is six miles per hour faster than the slower cyclist. Therefore, the speed of the faster cyclist can be expressed as v + 6 mph.
For example, if the slower cyclist is travelling at a speed of 9 mph, the faster cyclist would be traveling at 9 mph + 6 mph = 15 mph.
The table below shows the approximate height of an object x seconds after the object was dropped. The function h(x)= -16x^2 +100 models the data in the table. For which value of x would this model make it the least sense to use?
A. -2.75
B. 0.25
C. 1.75
D. 2.25
Answer:
the awnser is a -2.75
Step-by-step explanation:
If y = e2x is a solution to y''- 5y' + ky = 0, what is the value of k?
Answer:
The value of k is 6
Step-by-step explanation:
we need to find the value of k
Given : - [tex]y=e^{2x}[/tex] is the solution [tex]y''-5y'+ky=0[/tex]
[tex]y=e^{2x}[/tex] ........(1)
differentiate [tex]y=e^{2x}[/tex] with respect to 'x'
[tex]\frac{dy}{dx}=\frac{d}{dx}e^{2x}[/tex]
Since, [tex]\frac{d}{dx}e^{x} =e^{x}\frac{d}{dx}(x)[/tex]
[tex]\frac{dy}{dx}=e^{2x}\frac{d}{dx}(2x)[/tex]
[tex]\frac{dy}{dx}=e^{2x}\times 2[/tex]
[tex]\frac{dy}{dx}=2e^{2x}[/tex]
so, [tex]y'=2e^{2x}[/tex] ..........(2)
Again differentiation above with respect to 'x'
[tex]\frac{d}{dx}\frac{dy}{dx}=\frac{d}{dx}2e^{2x}[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\frac{d}{dx}(2x)[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=2e^{2x}\times 2[/tex]
[tex]\frac{d^{2}y}{dx^{2}}=4e^{2x}[/tex]
so, [tex]y''=4e^{2x}[/tex] ........(3)
Now, put the value of [tex]y\ ,y' \ \text{and} \ y''[/tex] in [tex]y''-5y'+ky=0[/tex]
[tex]4e^{2x}-5(2e^{2x})+(e^{2x})k=0[/tex]
[tex]4e^{2x}-10e^{2x}+e^{2x}k=0[/tex]
[tex]-6e^{2x}+e^{2x}k=0[/tex]
add both the sides by [tex]6e^{2x}[/tex]
[tex]e^{2x}k=6e^{2x}[/tex]
Cancel out the same terms from left and right sides
[tex]k=6[/tex]
Hence, the value of k is 6
Solve the system y = -x + 7 and y= 0.5(x - 3)2
Answer:
The solutions of the system of equations are (-1,8) and (5,2)
Step-by-step explanation:
[tex]y=-x+7[/tex] -------> equation A (equation of a line)
[tex]y=0.5(x-3)^{2}[/tex] ----> equation B (vertical parabola open upward)
Solve the system of equations by graphing
Remember that the solution is the intersection points both graphs
using a graphing tool
The intersection points are (-1,8) and (5,2)
see the attached figure
therefore
The solutions of the system of equations are (-1,8) and (5,2)
Answer: (-1,8) and (5,2)
Step-by-step explanation: The person above me is correct. Give him five stars and a thanks!
for the following right triangle find the side length x
Since there is a right angle, you can use Pythagoras' Theorem:
So x = √(24² + 7²) = 25
---------------------------------------------------------
Answer:
25
Forty dash one percent of people in a certain country like to cook and 68% of people in the country like to shop, while 14% enjoy both activities. What is the probability that a randomly selected person in the country enjoys cooking or shopping or both?
Answer:
0.86 or 86%
Step-by-step explanation:
The data given represent 41% of people in a certain country like to cook and 68% of people in the country like to shop, while 14% enjoy both activities.
The probability that a randomly selected person in the country enjoys cooking or shopping or both.
People who like to cook P(C) = 41% = 0.40
People who like to shopping P(S) = 68% = 0.60
People who like cooking and shopping both P(C∩S) = 14% = 0.14
People who like cooking or shopping or both = P(C∪S)
= P(C) + P(S) - P(C∩S)
= 0.40 + 0.60 - 0.14
= 0.86
The probability that a randomly selected person in the country enjoys cooking or shopping or both is 0.86 or 86%
To calculate the probability that a selected person likes cooking or shopping or both, we add the probabilities of each individual event and subtract the overlapping probability. In this case, it's 94.1%.
Explanation:You want to find the probability that a randomly selected person in the country enjoys cooking, shopping, or both. To calculate this probability, you can use the principle of inclusion-exclusion for two sets A and B, where:
A is the event that someone enjoys cooking.B is the event that someone enjoys shopping.The formula for the probability that a randomly selected person enjoys either cooking or shopping (or both) is:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Given:
P(A) = 40.1%P(B) = 68%P(A ∩ B) = 14%Plug in the values:
P(A ∪ B) = 40.1% + 68% - 14%
= 94.1%
So, the probability that a randomly selected person enjoys cooking or shopping or both is 94.1%.
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Suppose that you believe that the probability you will get a grade of B or better in Introduction to Finance is .6 and the probability that you will get a grade of B or better in Introduction to Accounting is .5. If these events are independent, what is the probability that you will receive a grade of B or better in both courses?
Answer: Probability that he will receive a grade of B or better in both courses is 0.30.
Step-by-step explanation:
Since we have given that
Probability that he will get a grade of B or better in introduction to Finance say P(A) = 0.6
Probability that he will get a grade of B or better in introduction to Accounting say P(B) = 0.5
Since A and B are independent events.
We need to find the probability that he will receive a grade of B or better in both the courses.
So, it becomes,
[tex]P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=0.6\times 0.5\\\\P(A\cap B)=0.30[/tex]
Hence, Probability that he will receive a grade of B or better in both courses is 0.30.
The probability that you will receive a grade of B or better in both courses is 0.3.
The probability that you will receive a grade of B or better in both Introduction to Finance and Introduction to Accounting, given that the events are independent, is calculated by multiplying the individual probabilities of each event.
[tex]\( P(F) = 0.6 \) \( P(A) = 0.5 \)[/tex]
Since the events are independent, the probability of both events occurring is given by the product of their individual probabilities:
[tex]\( P(F \text{ and } A) = P(F) \times P(A) \)[/tex]
Substituting the given probabilities:
[tex]\( P(F \text{ and } A) = 0.6 \times 0.5 \) \( P(F \text{ and } A) = 0.3 \)[/tex]
The final answer is [tex]\(\boxed{0.3}\).[/tex]
Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 233 mg of Bismuth-210, how much remains after 8 days?
Approximately 85.87 mg of Bismuth-210 would remain after 8 days.
Explanation:To calculate the amount of Bismuth-210 remaining after 8 days, we need to apply the concept of radioactive decay. Bismuth-210 decays by about 13% each day, meaning that 13% of the remaining Bismuth-210 transforms into another atom (Polonium-210) each day.
Let's calculate the amount remaining:
Start with 233 mg of Bismuth-210.After the first day, 13% of the remaining Bismuth-210 will decay, leaving 87% of the original amount: 0.87 * 233 mg = 202.71 mg.Repeat this process for each subsequent day.After 8 days, the amount remaining would be: (0.87)^8 * 233 mg = 85.87 mg.Therefore, after 8 days, approximately 85.87 mg of Bismuth-210 would remain.
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After 8 days, approximately 76.69 mg of Bismuth-210 remains out of the initial 233 mg.
To calculate the amount of Bismuth-210 remaining after 8 days of radioactive decay, follow these steps:
Step 1:
Understand the decay rate.
Each day, 13% of the remaining Bismuth-210 decays into Polonium-210. This means that 87% of the Bismuth-210 remains each day.
Step 2:
Calculate the remaining amount each day.
Start with the initial amount of 233 mg of Bismuth-210.
After the first day: [tex]\( 233 \text{ mg} \times 0.87 = 202.71 \text{ mg} \)[/tex]
After the second day: [tex]\( 202.71 \text{ mg} \times 0.87 = 176.43 \text{ mg} \)[/tex]
Continue this process for 8 days.
Step 3:
Perform the calculations for 8 days.
[tex]\[ \text{Day 1: } 233 \text{ mg} \times 0.87 = 202.71 \text{ mg} \][/tex]
[tex]\[ \text{Day 2: } 202.71 \text{ mg} \times 0.87 = 176.43 \text{ mg} \][/tex]
[tex]\[ \text{Day 3: } 176.43 \text{ mg} \times 0.87 = 153.62 \text{ mg} \][/tex]
[tex]\[ \text{Day 4: } 153.62 \text{ mg} \times 0.87 = 133.67 \text{ mg} \][/tex]
[tex]\[ \text{Day 5: } 133.67 \text{ mg} \times 0.87 = 116.33 \text{ mg} \][/tex]
[tex]\[ \text{Day 6: } 116.33 \text{ mg} \times 0.87 = 101.28 \text{ mg} \][/tex]
[tex]\[ \text{Day 7: } 101.28 \text{ mg} \times 0.87 = 88.10 \text{ mg} \][/tex]
[tex]\[ \text{Day 8: } 88.10 \text{ mg} \times 0.87 = 76.69 \text{ mg} \][/tex]
Step 4:
Interpret the result.
After 8 days, approximately 76.69 mg of Bismuth-210 remains.
So, after 8 days, approximately 76.69 mg of Bismuth-210 remains.
How do I simply this radical expression?
Answer:
[tex]1000±100\sqrt{55}[/tex]
Step-by-step explanation:
To simplify that expression, first we need to find the largest common of the expression inside the radical, in this case: 2.200.000.
We know that 2.200.000 = 2 · 2 · 2 · 2 · 2 · 2 · 5 · 5 · 5 · 5 · 5 · 11 = [tex]2^{6}[/tex] ×[tex]5^{5}[/tex]× [tex]11[/tex]
Now, [tex]\sqrt{2^{6}5^{5}11} = 200\sqrt{55}[/tex].
Now we have: [tex]\frac{2000±200\sqrt{55}}{2}[/tex]
Dividing by 2: [tex]1000±100\sqrt{55}[/tex]
So the simplified expression is: [tex]1000±100\sqrt{55}[/tex]
Last year, Scott had 10,000 to invest. He invested some of it in an account that paid 7%
simple interest per year, and he invested the rest in an account that paid 9% simple interest per year. After one year, he received a total of $740 in interest. How much did he invest in each account?
Answer:
$8000 is invested for 7% interest and $2000 is invested for 9% interest
Step-by-step explanation:
Points to remember
Simple interest formula
I = PNR/100
P - Principle amount
N - Number of years
R - Rate of interest
To find the amount of investment
It is given that total amount = 10,000 and total interest = $740
Let 'x' be the amount invested at the rate of 7%
10,000 - x be the amount invested at the rate of 9%
I = PNR/100
740 = (x*1*7)/100 + (10000 - x)*1*9/100
740 = 7x/100 + 90000/100 - 9x/100
740 = 7x/100 + 900 - 9x/100
740-900 = -2x/100
-160 = -2x/100
x = 16000/2 = 8000
10000-8000 = 2000
Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest
Answer:
$8000 is invested for 7% interest and $2000 is invested for 9% interest
Step-by-step explanation:
Points to remember
Simple interest formula
I = PNR/100
P - Principle amount
N - Number of years
R - Rate of interest
To find the amount of investment
It is given that total amount = 10,000 and total interest = $740
Let 'x' be the amount invested at the rate of 7%
10,000 - x be the amount invested at the rate of 9%
I = PNR/100
740 = (x*1*7)/100 + (10000 - x)*1*9/100
740 = 7x/100 + 90000/100 - 9x/100
740 = 7x/100 + 900 - 9x/100
740-900 = -2x/100
-160 = -2x/100
x = 16000/2 = 8000
10000-8000 = 2000
Therefore $8000 is invested for 7% interest and $2000 is invested for 9% interest
I really hope it helped idiot noob die
Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100 and a standard deviation sigma equals 20. Find the probability that a randomly selected adult has an IQ less than 132. The probability that a randomly selected adult has an IQ less than 132 is?
Answer:
There is a 94.52% probability that a randomly selected adult has an IQ less than 132.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 20. This means that [tex]\mu = 100, \sigma = 20[/tex].
The probability that a randomly selected adult has an IQ less than 132 is?
This probability is the pvalue of Z when [tex]X = 132[/tex]. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{132 - 100}{20}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a pvalue of 0.9452.
This means that there is a 94.52% probability that a randomly selected adult has an IQ less than 132.
Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100 and a standard deviation sigma equals 20. The probability that a randomly selected adult has an IQ less than 132 is 0.9452 or 94.52%.
What is the probability?Let standardize the IQ value of 132 using the formula for standardization:
Z = (X - μ) / σ
Where:
Z= standardized value (Z-score)
X = IQ value
μ = mean= 100
σ = standard deviation =20
Let's calculate the Z-score for an IQ of 132:
Z = (132 - 100) / 20
Z = 32 / 20
Z = 1.6
Using a standard normal distribution table, the probability is 0.9452.
Therefore, the probability that a randomly selected adult has an IQ less than 132 is 0.9452 or 94.52%.
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Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²
The derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Step-by-step explanation:We are given a function f(x) as:
[tex]f(x)=5-x^2[/tex]
We have:
[tex]f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)[/tex]
( Since,
[tex](a+b)^2=a^2+b^2+2ab[/tex] )
Hence, we get:
[tex]f(x+h)=5-x^2-h^2-2xh[/tex]
Also, by using the definition of f'(x) i.e.
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
Hence, on putting the value in the formula:
[tex]f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x[/tex]
Hence, the derivative of the function f(x) is:
[tex]f'(x)=-2x[/tex]
Answer:
The derivative of given function is -2x.
Step-by-step explanation:
The first principle of differentiation is
[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
The given function is
[tex]f(x)=5-x^2[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x+h)^2-(5-h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-(x^2+2xh+h^2)-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{5-x^2-2xh-h^2-5+h^2}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2-2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-\frac{2xh}{h}[/tex]
[tex]f'(x)=lim_{h\rightarrow 0}\frac{-x^2}{h}-2x[/tex]
Apply limit.
[tex]f'(x)=\frac{-x^2}{0}-2x[/tex]
[tex]f'(x)=0-2x[/tex]
[tex]f'(x)=-2x[/tex]
Therefore, the derivative of given function is -2x.
Find the dimensions of the box described. The length is twice as long as the width. The height is 4 inches greater than the width. The volume is 48 cubic inches. Find the length, width, height
Answer:
Width of box= 2inches
Length of box= 4inches
Height of box= 6inches.
Step-by-step explanation:
Let width of box=x inches
Length of box = twice of width=[tex]2\times x[/tex]=[tex]2x[/tex]
Height of box= 4 inches greater than width= [tex]x+4[/tex]
Volume of box= 48 cubic inches
We know that the formula of volume of cuboid= [tex] length\times breadth\times height[/tex]
Apply the formula
Volume of box= [tex]x\times 2x\times (x+4)[/tex]
Volume of cube = [tex]2x^2(x+4)[/tex]
[tex]2x^2(x+4)=48[/tex]
[tex]x^2(x+4)=24[/tex]
[tex]x^3+x^2-24[/tex]
Apply inspection method to solve the equation
Put [tex]x=0[/tex]
Then we get [tex]-24\neq0 [/tex]
Hence, x=0 is not the solution of x
Put x=1 in the equation then we get
[tex]-22\neq 0[/tex]
Hence x=1 is not the solution of equation.
Put x=2 then we get
[tex](2)^3+(4)^2-24[/tex]
8+16-24=0
Hence, x=2 is the solution of equation .
[tex] (x-2)(x^2+6x+12)[/tex]=0
Now substitute equation [tex]x^2+6x+12[/tex]=0
Sum roots =6
Product of roots=12
When sum of roots is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.
Hence, the roots of equation [tex]x^2+6x+12=0[/tex] are imaginary.
Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.
Hence,[tex] x=2 [/tex] is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .
Widht of box=2 inches
Length of box = [tex]2\times2[/tex]=4inches
Height of box=[tex]x+4[/tex]=2+4=6 inches
The dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.
Explanation:
Let's use the given information to solve for the dimensions of the box:
Let the width of the box be represented by x inches.
The length of the box is twice as long as the width, so the length is 2x inches.
The height is 4 inches greater than the width, so the height is (x + 4) inches.
The volume of a box can be calculated by multiplying the length, width, and height. Since the volume of the box is given as 48 cubic inches, we can set up the equation: 2x * x * (x + 4) = 48.
Simplifying the equation, we get 2x^3 + 8x^2 - 48 = 0.
Factoring the equation, we find that (x - 2)(x + 4)(x + 6) = 0.
The possible solutions are x = 2, x = -4, or x = -6.
Since we are dealing with dimensions, the width cannot be negative, so we can disregard the negative solutions. The width, therefore, is 2 inches.
The length is twice as long as the width, so the length is 2 * 2 = 4 inches.
The height is 4 inches greater than the width, so the height is 2 + 4 = 6 inches.
Therefore, the dimensions of the box are: length = 4 inches, width = 2 inches, and height = 6 inches.
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For Mexican American infants born in Arizona in 1986 and 1987, the probability that a child's gestational age is less than 37 weeks is 0.142 and the probability that his or her birth weight is less than 2500 grams is 0.051. Furthermore, the probability that these two events occur simultaneously is 0.031. Please show work!a. are A and B independent?b. For a randomly selected Mexican American newborn, what is the probability that A or B or both occur?c.What is the probability that event A occurs given that event B occurs?
Answer: Hence, a) No, they are not independent
b) 0.193
c) 0.60
Step-by-step explanation:
Since we have given that
Probability that a child's gestational age is less than 37 weeks say P(A)= 0.142
Probability that his or her birth weight is less than 2500 grams say P(B) = 0.051
P(A∩B) = 0.031
We need to check whether it is independent or not.
Since ,
[tex]P(A).P(B)=0.142\times 0.051=0.0072[/tex]
and
[tex]P(A\cap B)=0.051[/tex]
So, we can see that
[tex]P(A).P(B)\neq P(A\cap B)[/tex]
So, it is not independent.
a) Hence, A and B are not independent.
b) P(A∪B) is given by
[tex]P(A\ or B\ or\ both)=P(A)+P(B)\\\\P(A\ or\ B\ or\ both)=0.142+0.051=0.193[/tex]
c) P(A|B) is given by
[tex]P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.031}{0.051}=0.60[/tex]
Hence, a) No,
b) 0.193
c) 0.60