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3 years ago

Which geometric series represents 0.4444... as a fraction?

2 answers:

6 0

If you look at each decimal place it is in decreasing powers of 10, ie...

after the decimal point in our number system you have columns with coefficient values of 10^-1, 10^-2, 10^-3, etc

1/10, 1/100, 1/1000 etc and in this case we have

4/10, 4/100, 4/1000, etc

So it would be C.

Technically the sequence is:

a(n)=(4/10)(1/10)^(n-1) if we are going to be talking about geometric sequences...

Lina20 [59]3 years ago

4 0

The answer is C. 4/10+4/100+4/1000+4/10000. Hope it help!

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Why is the lcm of two numbers never less than the larger number?

Lady bird [3.3K]

This involves a bit of logic in thinking about what LCM actually means.

LCM is the least common multiple. A common multiple is a multiple shared by two or more numbers. And by multiple, we mean some number multiplied by successive integers; this set contains multiples of 5 {5, 10, 15, 20, 25...}. You can see that the least (or lowest) common multiple is the lowest multiple shared by two numbers. Like for 6 and 9, you have {6, 12, 18, 24...} and {9, 18...}, so LCM(6,9) = 18.

Now, if an LCM must be shared by both numbers, and to get a multiple of the largest number, you have to multiply by an integer greater than or equal to 1, then the LCM of two numbers can never be less than the larger of the twi numbers. Generally, if x and y are positive real numbers, and x is greater than y, $LCM(x,y) \leq x$ .

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4 years ago

An equation will remain equal if the same thing is done to both sides of the equation.

Alenkasestr [34]

<span>False....................................................</span>

8 0

3 years ago

Read 2 more answers

Verify that fand g are inverse functions (both questions please)

Grace [21]

Answer:

hope it helps you..,.....,..........

3 0

2 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

1 point)

drek231 [11]

Ok it will be c so yea

4 0

3 years ago