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pshichka [43]
4 years ago
12

Saturn is 8.867 × 108 miles away from the Sun. Uranus is 1.787 × 109 miles away from the Sun. Approximately how many times farth

er is Uranus from the Sun than Saturn is?
A. 0.2 times
B. 2 times
C. 20 times
D. 200 times
Mathematics
2 answers:
mart [117]4 years ago
6 0

Answer:

a =  0.2 times

ollegr [7]4 years ago
3 0

Answer:

Uranus is approximately 5 times farther from the sun than Saturn. The smallest zero is x = -10 as its the left-most value on a number line.

Step-by-step explanation:

Saturn:8.867 x 108 = 957. 636

Uranus: 1.787 x 109 = 194.783

b. 2 times

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Using the information given, select the statement that can deduce the line segments to be parallel. If there are none select non
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Answer:

Option (2). None

Step-by-step explanation:

A quadrilateral ABCD has been given with a property,

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Option (1). AB║ DC

For AB║DC, angle 7 and angle 3 should measure the same.

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Therefore, by the given property we can not deduce AB║DC.

Option (3). AD║BC

For AD║BC, angle 7 and angle 3 must be equal in measure.

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4 years ago
There are 14 apples in a basket. 6 of these apples are green. the rest of them are red.
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3 years ago
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in a final exam you have four multliple choice questions left to do. each questions has five suggested answers and only one of t
kompoz [17]

Answer:

The probability that you get zero questions correct is 0.4096

The probability that you get one questions correct is 0.4096

The probability that you get three questions correct is 0.0256

Step-by-step explanation:

These probability can be describe with a Binomial Distribution. These distribution can be used when we have n identical and independent situations in which there is a probability p or probability of success and a probability q or probability of fail. Additionally q is equal to 1 - p. The probability of x for a situation in which we can apply binomial distribution is:

P( x,n,p) = n Cx * p^{x } * q^ {n-x}

Where x is the variable that says the number of success in the n situations

And nCx is calculate as:

nC x = \frac{ n!}{ x! (n-x)! }

From the question we can identify that:

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  • p is 1/5 or 0.2, the probability of get one question correct
  • q is 4/5 or 0.8, the probability of get one question incorrect

Then the probability of get zero questions correct of 4 questions is:

P( 0, 4, 0.2) = 4 C0 * p^{0 } * q^ {4-0}= 0.4096

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P( 1, 4, 0.2) = 4 C1 * p^{1 } * q^ {4-1}= 0.4096

The probability of get three questions correct of 4 questions is:

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8 0
4 years ago
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