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3 years ago

At what rate is the amount of water in the pool changing according to the table below?

Mathematics

1 answer:

Georgia [21]3 years ago

8 0

the answer is b because the rate would not change

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Help please asap i will give brainlist

7nadin3 [17]

28°

a triangle = 180°
180°-140°= 40°
180°-68°= 112°
180°-112°-40°= 28°

3 0

3 years ago

Assume X is normally distributed with a mean of 5 and a standard deviation of 2. Determine the value for x that solves each of t

mihalych1998 [28]

Answer:

U = 5

S = 4

1.) P(X>x) = 0.5

Prob = 1-0.5 = 0.5

We have z = 0, that is the z score with the probability of 0.5

X = u + z(s)

= 5+0*4

= 5

2.) 1-0.95 = 0.05

Z score having this probability

Z = -1.64

X = 5-1.64*4

= 5-6.56

= -1.56

3.) P(z<1.0) - p(X<x) = 0.2

0.841345-0.2 = .641345

We find the z score given this probability

Z= 0.36

X = 5+0.36*4

= 5+1.44

= 6.44

4.) P(X<x)-P(Z<-.5)

0.95 = p(X<x)-0.308538

p(X<x) = 0.308538 + 0.95

= 1.258538

There is no x value here, given that the probability is more than 1.

5. 1-0.99/2 = 0.005

We get the z score value

= -2.58

U - 5 = 5-5 = 0

-x = 0-2.58(4)

X = 10.32

5 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

Which view is not correct for the figure above?

Bezzdna [24]

Answer:

Front

Step-by-step explanation:

4 0

3 years ago

To pass a course with a C grade, a student must have an average of 70 or greater. A student's grades on three tests are 69, 7

mario62 [17]

Answer:

Step-by-step explanation:

8 0

3 years ago