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Vadim26 [7]
3 years ago
15

The force of gravity on the moon is approximately one sixth that of earth. the direct variation equation for weight on earth com

pared to wait on the moon is e=6m, where the e weight on the earth and m= weight on the moon what would be the weight of a 180 pound man on the moon
Mathematics
1 answer:
Inga [223]3 years ago
8 0
Plug in 180 for e and solve for m
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Wha it is 2,200 divided by 30 and 2,300 divided by 30
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2,200/30=73.3333333

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Step-by-step explanation:

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Pete bought a 5-ounce box of cereal for $5.45. What was the cost per ounce?
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Step-by-step explanation:5.45 ÷ 5=1.108

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Can someone help me please:
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18. An urn contains 3 red balls, 2 green balls and 1 yellow ball. Three balls are selected at random and without replacement fro
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27.77%

Step-by-step explanation:

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Suppose you are interested in the effect of skipping lectures (in days missed) on college grades. You also have ACT scores and h
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Answer:

a) For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got colGPA=2.52

b) This value represent the effect into the ACT scores in the GPA, we know that:

\hat \beta_{ACT} = 0.015

So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable

c) If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: \beta_i = 0

Alternative hypothesis: \beta_i \neq 0

Or in other wouds we want to check if an specific slope is significant.

The significance level assumed for this case is \alpha=0.05

Th degrees of freedom for a linear regression is given by df=n-p-1 = 45-3-1 = 41, where p =3 the number of variables used to estimate the dependent variable.

In order to test the hypothesis the statistic is given by:

t=\frac{\hat \beta_i}{SE_{\beta_i}}

And replacing we got:

t = \frac{-0.5}{0.0001}=-5000

And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .

Step-by-step explanation:

For this case we have the following multiple regression model calculated:

colGPA =2.52+0.38*HSGPA+0.015*ACT-0.5*skip

Part a

(a) Interpret the intercept in this model.

For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got colGPA=2.52

(b) Interpret \hat \beta_{ACT} from this model.

This value represent the effect into the ACT scores in the GPA, we know that:

\hat \beta_{ACT} = 0.015

So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable

(c) What is the predicted college GPA for someone who scored a 25 on the ACT, had a 3.2 high school GPA and missed 4 lectures. Show your work.

For this case we can use the regression model and we got:

colGPA =2.52 +0.38*3.2 +0.015*25 - 0.5*4 = 26.751

(d) Is the estimate of skipping class statistically significant? How do you know? Is the estimate of skipping class economically significant? How do you know? (Hint: Suppose there are 45 lectures in a typical semester long class).

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: \beta_i = 0

Alternative hypothesis: \beta_i \neq 0

Or in other wouds we want to check if an specific slope is significant.

The significance level assumed for this case is \alpha=0.05

Th degrees of freedom for a linear regression is given by df=n-p-1 = 45-3-1 = 41, where p =3 the number of variables used to estimate the dependent variable.

In order to test the hypothesis the statistic is given by:

t=\frac{\hat \beta_i}{SE_{\beta_i}}

And replacing we got:

t = \frac{-0.5}{0.0001}=-5000

And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .

7 0
3 years ago
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