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fgiga [73]
3 years ago
9

PLEEAASE HELP MEEEEEE

Mathematics
1 answer:
Gre4nikov [31]3 years ago
8 0

\mathsf{1.\; \frac{x - 6}{2} = x + 13}

⇒ x - 6 = 2x + 26

⇒ x = -32

⇒ This Equation has One Solution

So, It matches with A in Column B

2. 2 + 6x - 7 = 8 + 2x - 10 + 4x

⇒ 6x - 5 = 6x - 2

As, They are Equations of Parallel Lines. They never meet each other.

⇒ This Equation has No Solution

⇒ So, It matches with B in Column B

3. 15x + 25 = 15x + 10

As, They are Equations of Parallel Lines. They never meet each other.

⇒ This Equation has No Solution

⇒ So, It matches with B in Column B

4. 16 + 8x = 4(2x + 4)

⇒ 16 + 8x = 16 + 8x

As, Both Equations are Same. They Represent the Same Line.

⇒ This Equation has Infinite Number of Solutions

⇒ So, It matches with C in Column B

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The equatorial radius of the planet Jupiter is measured 40 times by a process that is practically free of bias. These measuremen
Nina [5.8K]

Answer:

The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.

Step-by-step explanation:

Sample size of 30 or larger, so we can use the normal distribution to find the confidence interval.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.93}{2} = 0.035

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Now, find M as such

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In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.81*\frac{28}{\sqrt{40}} = 8

The lower end of the interval is the sample mean subtracted by M. So it is 71492 - 8 = 71484 km.

The upper end of the interval is the sample mean added to M. So it is 71492 + 8 = 71500 km.

The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.

8 0
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