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3 years ago

PLEASE ANSWER!!! WILL MARK BRAINLIEST :D

Mathematics

2 answers:

Gala2k [10]3 years ago

7 0

Answer:

Alina had at the beginning $\frac{10}{3}c$ cookies

Step-by-step explanation:

we know that

c-----> represent 30% of all the cookies Alina had

Let

x-----> the 100% of all the cookies Alina had

by proportion

$\frac{c}{30\%} =\frac{x}{100\%}\\ \\x=100*c/30\\ \\x=\frac{10}{3}c$

Dmitry [639]3 years ago

7 0

Answer: 3 1/3 c

Step-by-step explanation: We know that Aliana ate 30%.

Let's put it this way. Say there were 10 cookies. She would've ate 3 since it is 30%.

We need to divide 10 by 3, therefore saying that the answer is 3 1/3 c

Do not forget to label with c.

Hope this helps. ;D

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Please help with this acellus question!

Anna11 [10]

Answer:

The value of x is 5

Step-by-step explanation:

In the given triangle

∵ A segment joining two points lie on the two side

∵ This segment is parallel to the third side

∴ This segment divides the two sides into parts that have the same ratio

→ That means x over 2x - 3 equal to 10 over 14

∵ $\frac{x}{2x-3}$ = $\frac{10}{14}$

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→ Add 30 to both sides

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6 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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xxMikexx [17]

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