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allochka39001 [22]
4 years ago
5

HELP...

Mathematics
1 answer:
aivan3 [116]4 years ago
3 0
Hello!

Looking for a Genius Answer? Here It Is!

To get the solution, we are looking for, we need to point out what we know. 

1. We assume, that the number 150 is 100% - because it's the output value of the task. 
2. We assume, that x is the value we are looking for. 
3. If 100% equals 150, so we can write it down as 100%=150. 
4. We know, that x% equals 25 of the output value, so we can write it down as x%=25. 
5. Now we have two simple equations:
1) 100%=150
2) x%=25
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=150/25
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for 25 is what percent of 150

100%/x%=150/25
(100/x)*x=(150/25)*x       - we multiply both sides of the equation by x
100=6*x       - we divide both sides of the equation by (6) to get x
100/6=x 
16.6666666667=x 
x=16.6666666667

now we have: 
<span>25 is 16.6666666667% of 150
</span>

Hope this Helps! Took Me Time! Have A Wonderful Day! :)
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The area of the actual object is 282.24 mm square

Given

Shape: Square

Scale: 1 in : 8 mm

Length of Scale Drawing = 6 in

Required:

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- Area of Actual Object

Calculating the perimeter of the actual object

Provided that the shape is a square;

The perimeter of the actual object is calculated by

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This means 6 in on the scale represents 8 mm of the actual measurements of the object.

So, if 6 in ≈ 8 mm, then

6.8 in ≈ 6.8 * 8 mm

6.8 in = 16.8 mm

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Calculating the area of the actual object

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Here is the complete question.

1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?

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1).  \frac{1}{8}

2).  $8.

Step-by-step explanation:

1).

The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:

\left[\begin{array}{cccccc}(1,2)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\end{array}\right] \left[\begin{array}{cccccc}(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right]

Let F represent the event that the sum of both dice turns up is 5.

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