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rewona [7]
3 years ago
11

Will has 20 pounds of apples he makes to back to apple sauce that he's 4 pounds each one back to apple butter that use a 6 pound

s and he uses 3 pounds to make juice right any Quetion to represent how many pounds of apples will has left
Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0

Answer:

The equation representing pounds of apples will has left x=20-4-6-3.

Will has left with 7 pounds of apples.

Step-by-step explanation:

Given:

Total pounds of apple = 20 pounds.

Pounds of apple used for apple sauce = 4 pounds

Pounds of apple used for apple butter = 6 pounds

Pounds of apple used for making juice = 3 pounds

We need to write the equation to represent pounds of apples will has left.

Solution:

Let us assume pounds of apples will has left be 'x'.

So we can say that;

pounds of apples will has left can be calculated by subtracting Pounds of apple used for apple sauce and Pounds of apple used for apple butter and Pounds of apple used for making juice from Total pounds of apple.

framing in equation form we get;

x=20-4-6-3

Hence The equation representing pounds of apples will has left x=20-4-6-3.

On Solving the above equation we get;

x=7\ pounds

Hence Will has left with 7 pounds of apples.

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Step-by-step explanation:

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3 years ago
A television game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind t
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The answer to this question:
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I will focus answering the 3 doors probability since the 2nd door problem is solved in the previous problem. (brainly.com/question/5761449)

No car condition
1. 1st door consolation, 2nd door consolation=, 3rd door consolation= 4/6 * 3/5 * 2/4= 24/120
This was also can be found by: (4!/1!)/ (6!/3!) = 24/120

(At least one car probability)  is the opposite of (no car probability) In this case, the easier way is 
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One car probability is (At least one car probability) - (2 car probability). It will be easier to count the 2 car probability and subtract the (At least one car probability) 
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2.1st door car, 2nd door consolation, 3rd door car =2/6 * 4/5 * 1/4 = 8/120
3. 1st door consolation, 2nd door car, 3rd door car= 4/6 * 2/5 * 1/4= 8/120
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This was also can be found by: (2!) (4!/2!)/ (6!/3!) = 24/120

One car probability =  (At least one car probability) - (2 car probability)= 96/120-24/120= 82/120
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Step-by-step explanation:

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