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AnnyKZ [126]
4 years ago
11

PLEASE HELP ME ANSWER ASAP!

Mathematics
2 answers:
Blizzard [7]4 years ago
8 0
Either twice or more than twice. But I’d go with twice
SIZIF [17.4K]4 years ago
3 0

Answer:

C

Step-by-step explanation:

The modified version we need to use is one of Kepler's Laws modified. Without knowing this formula makes the problem impossible to determine-- or at best it is dependent of a good guess. The formula we need is

\dfrac{T^2}{R^3} =\dfrac{4*\pi^2}{G*M_{central}}

So doubling T still leaves you with a large problem. What happens to R? The answer is that it more than doubles the radius. I make the answer C

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hram777 [196]

Answer:

The predicted GPA is then y = 0.149(15) + 0.89 = 3.1

Step-by-step explanation:

Although you don't specifically say so, the equation you provide here is probably a "best fit" equation based upon data:  GPA versus the number of hours of study per week.

y = 0.149x + 0.89 and the number of study hours of interest is 15.

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Tiffany answered 80% of the questions on her math test correctly. There were 40 questions. How many questions did Tiffany get co
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Consider this expression.
il63 [147K]
I got -2(2x-3) but it should be C.-4x-2
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3 years ago
5. ¿A qué fracción es equivalente el número 27
nika2105 [10]

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Step-by-step explanation:

8 0
3 years ago
Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls
murzikaleks [220]

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222

Hence,

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684

Part b

P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}

Hence,

P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}

4 0
3 years ago
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