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valentinak56 [21]
3 years ago
8

4. Error Analysis Jacinta says that the product of a rational number and an irrational number is always irrational. Explain her

error.​
Mathematics
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

She did not consider the multiplicative inverse property of a real number

Step-by-step explanation:

Jacinta in her conclusion did not consider the Multiplicative Inverse property of real numbers.

Let x be an irrational number.

So, $ x.\frac{1}{x}= 1$

If x is an irrational number, then its reverse is also an irrational number. So by the Multiplicative Inverse property their product is 1 and 1 is a rational number. Thus, Jacinta was incorrect.

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-35 − (-12) = A) -47 B) -23 C) 23 D) 47
Alexus [3.1K]
B) -23 because a double negative is a positive so it’s -35 + 12 which is -23
5 0
3 years ago
X+3=15 what’s x because I keep on getting 18 and it says I’m wrong on Pearson
icang [17]

Answer:

x = 12

Step-by-step explanation:

Subtract 3 from both sides

p.s. it would be 18 if it was x-3 = 15

8 0
3 years ago
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How do you solve this
Orlov [11]
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8 0
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
Katyanochek1 [597]
The answer to this question:
One car probability 82/120
No car probability = 24/120
At least one car probability= 96/120

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 
100% - (no car probability) = 120/120 - 24/120= 96/120

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) 
Two car condition:
1. 1st door car, 2nd door car, 3rd door consolation = 2/6 * 1/5 * 4/4 =8/ 120
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
The total probability will be 8/120+ 8/120 + /120= 24/120
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
3 0
3 years ago
What is the perimeter of nadia's square ?
Nesterboy [21]
You forgot the picture of the square, please repost the question with the picture next time.
4 0
3 years ago
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