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Sonja [21]
3 years ago
12

Help with proofs please

Mathematics
1 answer:
alexandr402 [8]3 years ago
8 0

Answer:

D. Subtraction Property of Equality

Step-by-step explanation:

Hope it helps.

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Find the sale of the item. Round to two decimal places if necessary.
Jet001 [13]
First multiply the original price by the discount. (250.99 x .78 = 195.7722) Then subtract that number from the original. (259.99 - 195.7722 = 55.2178) Then round to 2 decimals because dealing with money. Answer: $55.22
6 0
3 years ago
Please help me. i don’t want to fail
rosijanka [135]

Answer: i’m pretty sure the answer is wx, xy, yw

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP
34kurt
I believe that in this problem, we are asked for the hypotenuse of the triangle that can be formed when we make a right triangle out of the planes and the intersection of their paths. To solve for the hypotenuse of a right triangle, we make use of the equation,
                                   h² = 64.2² + 85.7²
The value of h from the equation is equal to 106.0 miles. 
5 0
3 years ago
The area of a square mirror is 64 in.² what is the length of the mirror?
Molodets [167]

Answer:

8 or B

Step-by-step explanation:

\sqrt{64 which is equal to 8

3 0
4 years ago
Suppose you just received a shipment of twelve televisions. Three of them are defective. Assuming that two televisions are rando
lukranit [14]
<span>The probability that both televisons work is 18/33.
</span><span>The probability that at least one does not work is 15/33.
</span>
Let x = number of shipments, x = 12
P(defective) = 3/12
P(working) =  9/12
The probability that two randomly selected televisions are both working is:
9/12 * 8/11 = 72 / 132 or 18/33
P(2 working) = 18/33 
8/11 is used for the second selection since the number of televisions has been reduced due to the first selection.

The probability of 1 not working among the two can be solved through"
P(at least one defective among the 2) = 1 - <span>P(2 working) 
</span><span>P(at least one defective among the 2)  = 15/ 33
</span>

Thank you for posting your question. I hope you found you were after. Please feel free to ask me another.


4 0
4 years ago
Read 2 more answers
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