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nasty-shy [4]
3 years ago
12

3 added to 5 times a number and the result is multiplied by 4. The final result is 72. Find the number

Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0
Hmm... I would say the equation would look like 3+5x=72/4. 
3+5x=72/4
3+5x=18
5x=18-3
5x=15
x=15/5
x=3

The number is 3

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Step-by-step explanation:

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Anyone know how to round 2.5678 round to the nearest half
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Read 2 more answers
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Yanka [14]
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8 0
3 years ago
Use the Pythagorean theorem to obtain the function​ D(x) for the length of the diagonal of a double square of width x.
Yuri [45]

Answer:

D(x)= x\sqrt{2}

Step-by-step explanation:

Given

Side = x

Required

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Substitute x for Side

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4 0
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:

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

Step-by-step explanation:

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

f_x(x)  = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0

Thus, the expected  total claim amount \mu =  1000

The variance of the total claim amount \sigma ^2  = 1000^2

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

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
2 years ago
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