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yan [13]
3 years ago
13

Given the equation P2 = A3, what is the orbital period, in years, for the planet Saturn? (Saturn is located 9.5 AU from the sun.

)
Mathematics
1 answer:
algol [13]3 years ago
5 0

Answer: 29.28 years

Explanation:

From Kepler's third law the square of orbital period of revolving celestial body is proportional to the cube of semi -major axis from the body  it is revolving about.

P² =A³

Where, P is the orbital period in years and A is the semi-major axis in AU (Astronomical units)

It is given that, For Saturn, A = 9.5 AU. We need to find P

⇒P² = (9.5 AU)³

⇒P² = 857.38

⇒P = 29.28 years

Thus, the orbital period of Jupiter is 29.28 years around the Sun.

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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
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