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juin [17]
1 year ago
9

(2x)/(25)=(0.4)/(0.15)

Mathematics
1 answer:
erastovalidia [21]1 year ago
8 0
For Solve the x, simplifying both sides of the equation so it would be x=33.3
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9 + 22 = x + 1<br><br> HALPP
rosijanka [135]

Hey!

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

Solution:

9 + 22 = x + 1

9 + 22 - x = x + 1 - x

31 - x = 1

31 - x  31 = 31 - 1

x = 30

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

Answer:

x = 30

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

Hope This Helped! Good Luck!

4 0
3 years ago
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The sum of the areas of two circles is 80π square meters. Find the length of a radius of each circle if one of them is twice as
aleksley [76]
Am very bad at explaining things but the answer is 8

5 0
3 years ago
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Need help on this, what is the correct name for this number?
Flauer [41]

Answer:

I believe its the third one

6 0
3 years ago
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2. Solve each given equation and show your work. Tell whether each equation has one solution, an infinite number of solutions, o
Ugo [173]
Alright,

(a) 2x+4(x-1)=2
2x+4x-4=2---- Distrubution of 4(x-1)
6x-4=2---- Combining like terms
6x=6---- Making -4 equal 0 by adding +4 to both sides
x=1---- Divide the coefficiant (6)

(b) 25-x=15-(3x+10)
25-x=15-3x-10----Distribute the -1
25-x=5-3x---- Combine like terms
25=5-2x---- Make -x equal 0 by adding +x to both sides
20=-2x---- Make 5 equal zero by subtracting 5 (or adding -5) to both sides
-10=x---- Divide by the coefficiant (-2)

(c) 4x= 2x+2x+5(x-x)
4x=2x+2x+5(0)---- Distribute. Alright this might look hard but realy it is 1x-1x so the answer is 0
4x=2x+2x---- 5 times 0 is 0
4x=4x---- Combine like terms
x=x ---- divide by the coefficiant

Alright! Hope this helps! ;)
4 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
3 years ago
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