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KIM [24]
3 years ago
7

Round 49.87654321 to the nearest tenth

Mathematics
1 answer:
schepotkina [342]3 years ago
8 0
49.9 is the answer
hope it helps ya :D
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Does Y equals 3X minus to have a solution
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Graph the second equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis. At the point of intersection of the two equations x and y have the same values.

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A right rectangular prism has a length of 214 cm, width of 8 cm, and height of 2012 cm.
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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a stand
erastovalidia [21]

Answer:

z=0.524

And if we solve for a we got

a=5 +0.524*2=6.048 \approx 6.05 min

Step-by-step explanation:

Let X the random variable that represent the lenght time it takes to find a parking space at 9AM of a population, and for this case we know the distribution for X is given by:

X \sim N(5,2)  

Where \mu=5 and \sigma=2

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.3   (a)

P(X   (b)

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.524

And if we solve for a we got

a=5 +0.524*2=6.048 \approx 6.05 min

3 0
3 years ago
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