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3 years ago

"without performing an inverse laplace transform, find the limiting value"

1 answer:

maks197457 [2]3 years ago

8 0

If your transformed function is F(s), the limiting value of f(t) as t goes to infinity is the limiting value of s*F(s) as s goes to zero.

Multiply your function by s, then evaluate for s=0. If the form is indeterminate, evaluate the limit as s approaches zero.

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Colt1911 [192]

Answer:

EDA by angle degree.

Step-by-step explanation:

Don't know if it's right or not.

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4 years ago

How metric units and place value units are alike and unlike

dem82 [27]

Well metric unit is for measurement and place value units is for numbers

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3 years ago

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Standard notation 42 * 10 6

Aleks [24]

$Answer:$ ☆ ⭐︎ ✩ ✶ ❊

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$4.2*10^{6}$

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3 years ago

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What are parallel lines cut by a transversal?

Neko [114]

In geometry, a transversal is a line that intersects two or more other (often parallel ) lines. In the figure below, line n is a transversal cutting lines l and m . When two or more lines are cut by a transversal, the angles which occupy the same relative position are called corresponding angles .

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3 years ago

The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students re

Vladimir79 [104]

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ = 0.95

The 95% of confidence intervals

$x^{-} ± t_{\alpha } \frac{S}{\sqrt{n} }$

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} },x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })$

$(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )$

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

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3 years ago