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sergey [27]
2 years ago
15

Please help ill give brainliest

Mathematics
1 answer:
Pavlova-9 [17]2 years ago
3 0

Answer:

I forgot the answer but hey I go free points!

Step-by-step explanation:

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True or false: In (r, theta), the value of r can be negative.
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The answer to the mathematics question presented above is 'True'. It is correct to say that in (r theta), the value of r can be negative. A negative radius can be used when it comes to graphing a "polar'' function. Thus, the answer to the question is 'true'. 
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3 years ago
What is the eighth term in the arithmetic sequence defined by the explicit formula
RideAnS [48]

Answer:

Find out the what is the eighth term in the arithmetic sequence defined by the explicit formula.

a_{n} = 2n + 7

To prove

As given the explicit formula for the arithmetic sequence .

a_{n} = 2n + 7

Put n = 8

a_{8} = 2\times 8 + 7

a_{8} = 16 + 7

a_{8} = 23

Therefore the eighth term in the arithmetic sequence is 23 .




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3 years ago
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A square has a side of 6.25 feet. What is the area in square feet? Round to the correct significant digits
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3 years ago
How do I do number 4?
Galina-37 [17]
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3 years ago
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EasyElectrics supermarket orders light bulbs from suppliers, AA Electronics and AAA Electronics. EasyElectrics purchases 30% of
saw5 [17]

Answer: (a) 0.006

               (b) 0.027

Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

P(D/AA) = 0.02 and P(D/AAA) = 0.03

Thus P(D'/AA) = 1 - 0.02 = 0.98

and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

                                                       = P(AA) x P(D/AA)

                                                       = 0.3 x 0.02 = 0.006

(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

                         = P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

                         = 0.3(0.02) + 0.70(0.03)

                         = 0.027

3 0
3 years ago
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