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kolbaska11 [484]
3 years ago
15

Find a formula for an arithmetic sequence that begins 45, 40, 35, 30, ...

Mathematics
1 answer:
-BARSIC- [3]3 years ago
8 0

Answer:

C) 9n=45-5(n-5)....…...….......

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Find the common diffirence in the arithmetic sequence 1, 6/4, 10/4, 3...​
Temka [501]

Answer:

See explanation

Step-by-step explanation:

6/4 = 1.5

10/4 = 2.5

Therefore the sequence is 1, 1.5, 2.5, 3

So the common difference is either 0.5 or 1

8 0
3 years ago
How to make 4/7 into a rational number?
MrMuchimi
4/7 is a rational number. A rational number is a number that can be turned into or is a fraction.
8 0
3 years ago
15% of the content of a 454 g can of dog food is protein. How many grams of protein are in the can?
mote1985 [20]

Answer:

521.1

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one
diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0
4 years ago
What is 2 times 3 times n times n = 54
kolezko [41]

If I'm reading this correctly, it is 2*3*n*n = 6n^2

6n^2 = 54  Divide by 6

n^2 = 54/6

n^2 =  9      Take the square root of both sides.

sqrt(n^2) = sqrt(9)

n = +/- 3

n = + 3 or n = -3. Both will solve the equation above.

Edit

2 * 3 * n * n

Let n = 3

2 * 3 * 3 * 3 = 6 * 9 = 54

The other value is n = - 2

2 * 3 * n * n

2 * 3 * (-3) * (-3)   A minus times a minus is a plus.

2 * 3 * 9 = 6 * 9 = 54

both + 3 and - 3 for n will give 54.

8 0
3 years ago
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