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Goshia [24]
3 years ago
12

I'm trying to figure out the formula for figuring out percentages. For example, what percentage of 3175 is 510?

Mathematics
1 answer:
aivan3 [116]3 years ago
8 0

Answer: There is no one specific formula

Step-by-step explanation:

The math to determine a percentage is to divide the numerator (the number on top of the fraction) by the denominator (the number on the bottom of the fraction), then multiply the answer by 100. For example, the fraction 6/12 turns into a decimal like this: 6 divided by 12 (which equals . 5) times 100 equals 50 percent

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The product of a particular number and a second value, which is given by the sum of 8 and the original number, is 65. What is th
Marina86 [1]

Given:

The product of a particular number is 65.

Second value is given by the sum of 8 and the original number.

To find:

The unknown number.

Solution:

Let x be the unknown number.

Then, second number = x+8

The product of a particular number is 65.

x\times (x+8)=65

x^2+8x=65

x^2+8x-65=0

Splitting the middle term, we get

x^2+13x-5x-65=0

x(x+13)-5(x+13)=0

(x-5)(x+13)=0

Using zero product property, we get

x-5=0 and x+13=0

x=5 and x=-13

So, the unknown number is either 5 or -13.

Therefore, the correct option is B.

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2 years ago
Please solve 3x = 81
sasho [114]

Answer: x = 27

Step-by-step explanation:

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3 years ago
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Phantasy [73]

Your answer is y=-3/4x+7 because when you find rise over run it's -9/8 which rounds down to -3/4 which is your slope and 7 is the

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3 years ago
A factory is to be built on a lot measuring 150 ft by 200 ft. A local building code specifies that a lawn of uniform width and e
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2 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
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