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Rama09 [41]
1 year ago
10

A satellite travels from Earth to the moon (238,905 miles), then from the moon to Mars (48,340,732 miles) and finally from Mars

to Venus (71,948,832 miles). How far did the satellite travel in all?
Mathematics
1 answer:
hodyreva [135]1 year ago
8 0

The total distance the satellite traveled in all is 120,528,469 miles

<h3>Calculating total distance</h3>

From the question, we are to determine how far the Satellite traveled in all

From the given information,

A satellite travels from Earth to the moon (238,905 miles),

and

It traveled from the moon to Mars (48,340,732 miles)

and

It finally traveled from Mars to Venus (71,948,832 miles)

The total distance the satellite traveled in all = 238,905 miles + 48,340,732 miles + 71,948,832 miles

∴The total distance the satellite traveled in all = 120,528,469 miles

Hence, the total distance the satellite traveled in all is 120,528,469 miles

Learn more on Calculating distance here: brainly.com/question/24295081

#SPJ1

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Answer:

(3, -6)

General Formulas and Concepts:

<u>Pre-Algebra</u>

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  5. Division
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<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define Systems</u>

y = 4x - 18

y = -5x + 9

<u>Step 2: Solve for </u><em><u>x</u></em>

<em>Substitution</em>

  1. Substitute in <em>y</em> [2nd Equation]:                                                                           4x - 18 = -5x + 9
  2. [Addition Property of Equality] Add 5x on both sides:                                   9x - 18 = 9
  3. [Addition Property of Equality] Add 18 on both sides:                                   9x = 27
  4. [Division Property of Equality] Divide 9 on both sides:                                  x = 3

<u>Step 3: Solve for </u><em><u>y</u></em>

  1. Substitute in <em>x</em> [1st Equation]:                                                                           y = 4(3) - 18
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  3. Subtract:                                                                                                            y = -6
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in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok
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Answer:

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

\hat p=0.849 estimated proportion of adults were no longer smoking after one month

p_o=0.80 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:  

Null hypothesis:p=0.8  

Alternative hypothesis:p \neq 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

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