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Scilla [17]
3 years ago
7

In trying to calculate the product of 5 and 12

Mathematics
2 answers:
Leno4ka [110]3 years ago
7 0

Answer:

60

Step-by-step explanation:

5x12 is 60.

shusha [124]3 years ago
6 0
60 hope that helps good luck:)
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Simplify the algebraic expression: 4(3x + y) – 2(x – 5y)
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C

Step-by-step explanation:

First use distribution

4(3x+y)= 12x+4y

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3 years ago
Write the missing power of 10 in each equation.
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2 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
3 years ago
Y+2 y+2 y+2 y+2 y+2<br> ^<br> 30
BartSMP [9]

Answer:

45

Step-by-step explanation:

5 0
3 years ago
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