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iragen [17]
4 years ago
12

Given that l is parallel to line m, find x a. 1 b. 25 c. 3 d. 35

Mathematics
1 answer:
expeople1 [14]4 years ago
6 0
I WOULD SAY IT IS B>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
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Helpppppppppppppppppppppppppppppp
Rzqust [24]

Answer:

5

Step-by-step explanation:

11-6=10-x

11-6-10=-x

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11-6=10-5

8 0
3 years ago
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Ashley and Kris are saving money for a vacation. Ashley currently has
USPshnik [31]

Answer:

Step-by-step explanation:

1) y = 50x + 2500

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8 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}

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3 years ago
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Elanso [62]
The answer to 20+1 is 21
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3 years ago
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solmaris [256]
D. 28 cents per year
answer ^^^^^^^^^^^^
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