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swat32
3 years ago
5

When finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probabili

ty, assuming a confidence level of 86%? 0.14 0.86 0.93 0.99]?
Mathematics
2 answers:
Tju [1.3M]3 years ago
6 0
C. is the answer ( just took the quiz)
sergiy2304 [10]3 years ago
4 0
C. 0.93 I just toke the test
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Find the average rate of change for fx) = x2 – 3x – 10 from x = 4 to x = 6.
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Answer:

7

Step-by-step explanation:

The rate of change is defined as

f(x2) - f(x1)

--------------------

x2-x1

x2 = 6

f(6) = x^2 – 3x – 10 = 6^2 -3*6 -10 = 36 - 18-10=8

x1 = 4

f(4) = x^2 – 3x – 10 = 4^2 -3*4 -10 = 16 - 12-10=-6

Substitute the values into the expression

8 -  -6

--------------------

6-4

8+6

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2

14/2= 7

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3 years ago
Answer this please i don't really understand it so help me please lol.
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Answer:

11

Step-by-step explanation:

Step 1:

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Step 2:

6 + 20 ÷ 4     Subtract

Step 3:

6 + 5   Divide

Answer:

11    Add

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8 0
3 years ago
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Can anyone do this ??
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Answer:

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Step-by-step explanation:

7 0
4 years ago
Whic two transformations are applied to pentagon ABCDE to create A'B'C'D'E?
Blababa [14]
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3 years ago
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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
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