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nadya68 [22]
4 years ago
11

What is the area of a triangle A=77mm B=5.9mm C=9.7mm

Mathematics
2 answers:
Masteriza [31]4 years ago
6 0
Could you specify if this is a right triangle?
USPshnik [31]4 years ago
3 0
Area = 1/2 base multiplied by height. I don't think you have the right dimensions. :(
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One serving of oatmeal has 8 grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily a
Y_Kistochka [10]

Answer:

24g

Step-by-step explanation:

8g / 0.33 = 24.242424g

5 0
3 years ago
A rug is 4 m wide. How much is this in centimeters?
Crank
It's very simple
1m=100cm
4m=4×100=400cm
6 0
3 years ago
Read 2 more answers
Pleaseeee helppp !!! Identify each polygon by its number of sides
lutik1710 [3]
Furthest to the left is a Hexagon. Middle is an Octagon. Furthest to the right is a Decagon.

4 0
4 years ago
Suppose you are interested in the effect of skipping lectures (in days missed) on college grades. You also have ACT scores and h
DIA [1.3K]

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
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
3 years ago
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