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Lana71 [14]
3 years ago
7

The radius of Earth's orbit is approximately 15,000,000,000 meters. The radius of Mercury's orbit is 5.8 × 109 meters. What is t

he difference of their radii?
A.4.3 × 1010



B.4.3 × 1011



C.9.2 × 109



D.9.2 × 1010



E.9.2 × 1011
Mathematics
1 answer:
Ilya [14]3 years ago
6 0

15,000,000,000 = 15 × 10⁹

15 × 10⁹ - 5.8 × 10⁹ = 9.2 × 10⁹ m

Answer: C. 9.2 × 10⁹ m

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<u>Answer-</u>

a. Probability that  three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

\binom{19}{3} = \frac{19!}{3!16!} = 969

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

\binom{33}{6} = \frac{33!}{6!27!} =1107568

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

\binom{52}{9} = \frac{52!}{9!43!} = 3679075400

∴P(3 white candies) = \frac{1073233392}{3679075400} =0.29



Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}

=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})

=(969)(45)(7)(5)(15)=22892625

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

\binom{52}{9}=\frac{52!}{9!43!} =3679075400

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

\frac{22892625}{3679075400} = 0.006


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Step-by-step explanation:

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Answer:

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Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

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Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

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50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

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