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Goryan [66]
3 years ago
10

Astronomers measure large distances in light years. one light year is the distance that can travel in one year, or approximately

5.88 x 10^12 miles. suppose a star is 9.8 x 10^1 light years from earth. in scientific notation approximately how many miles is it?
A. 5.88 x 10^13 miles
B. 5.76 x 10^14 miles
C. 5.88 x 10^12 miles
D. 9.8 x 10^12 miles
Mathematics
1 answer:
Nadya [2.5K]3 years ago
7 0
The answer is b. i hope this helps.
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4 years ago
Read 2 more answers
Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th
Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312

Probability of having an emergency locator and not being discovered:

20% of 28%. So

P(A cap B) = 0.2*0.28 = 0.056

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

P(A) = 1 - 0.5312 = 0.4688

Probability of not having an emergency locator and being discovered:

34% of 72%. So

P(A \cap B) = 0.34*0.72 = 0.2448

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

P(A \cap B) = 0.66*0.72 = 0.4752

Probability of being discovered, given that it has the emergency recorder:

p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946

This question asks for P(X = 7) when n = 10. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064

0.064 = 6.4% probability that 7 of them are discovered.

8 0
3 years ago
Equivalent ratios what’s the answer <br> 3:6=
MAVERICK [17]

Step-by-step explanation:

you multiply both sides by the same factor, so that the main message or information, the ratio, stays the same :

3:6 × 2/2 = 6:12

3:6 × 1/3 / 1/3 = 1:2

3:6 × 4/3 / 4/3 = 4:8

3:6 × 100/100 = 300:600

so, these are all equivalent ratios. and there are infinitely more, as you can see.

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2 years ago
What two numbers make 68
Sergio039 [100]

Answer:67 and 1

Step-by-step explanation:

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4 years ago
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Andy has 9 math books and 6 reading books. If he wants to distribute them evenly among some bookshelves so that each has the sam
andrew11 [14]

Answer:

3 is the greatest number of book shelves Andy can use.

Step-by-step explanation:

You can solve this by finding the greatest common factor of 9 and 6, which turned out to be 3.

6/3 = 2 (reading books per shelf)

9/3 =3 (math books per shelf)

Every shelf there are 2 reading books and 3 math books.

4 0
3 years ago
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