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ludmilkaskok [199]
3 years ago
14

4 Millimeters decrease to 3 millimeters what's the percent of decrease

Mathematics
2 answers:
Fiesta28 [93]3 years ago
5 0

Answer:

Now first we have to found the decrease length.

which is:4mm-3mm=1mm

Now

we have to use formula to calculate the percentage which is

= decrease length/total length ×100%

=. 1/4×100%

=. 25 %

Therefore,the decrease rate is 25%.

frutty [35]3 years ago
3 0
It’s a 75 percent decrease
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A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

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                                            \int \frac{dA}{rA+P} = \int  dt

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Multiply both sides by r.

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Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

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3 years ago
The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =
gizmo_the_mogwai [7]

Answer:

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Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy  is given as :

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Thus, the expected  total claim amount \mu =  1000

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P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})

P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})

P(X>1100n) = P(Z> \dfrac{10*100}{1000})

P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345

\mathbf{P(X>1100n) = 0.158655}

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is \mathbf{P(X>1100n) = 0.158655}

4 0
3 years ago
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