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How far away is a star with a parallax of 2 arc sec in parsecs? In AU?

Vesna [10]

Answer:

ear

Step-by-step explanation:

8 0

2 years ago

That you're positive that you should be trying out these difficult math questions, let’s get right to it! The answers to these q

Lemur [1.5K]

Answer:

that you're positive that you should be trying out these difficult math questions, let’s get right to it! The answers to these questions are in a separate section below, so you can go through them all at once without getting spoiled.

#1:

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#2:

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#3:

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#4:

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#5:

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#6:

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#7:

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#8:

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#9:

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#10:

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#11:

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#15:

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Step-by-step explanation:

5 0

2 years ago

Which number has the greatest value

irina1246 [14]

9.31 is the answer, in the decimal 31 has the greatest value

3 0

2 years ago

Read 2 more answers

F(-2) if f (x) = -3x+7

katrin [286]

1 is the awnser to this problem

4 0

2 years ago

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310 3

Mariana [72]

Answer:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310 383 285 300 300

We can calculate the sample mean and deviation with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=310.9$ represent the sample mean

$s=31.09$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =300$ represent the value to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0

2 years ago