Answer:
(6,2)
Step-by-step explanation:
Variable Definitions:
x= the number of commercials
y= the number of movies
Each commercial earns Emily $50, so x commercials would earn her 50x dollars in royalties. Each movie earns Emily $150, so y movies would earn her 150y dollars in royalties. Therefore, the total royalties 50x+150y equals $600:
50x+150y=600
Since Emily's songs were played on 3 times as many commercials as movies, if we multiply 3 by the number of movies, we will get the number of commercials, meaning x equals 3y.
x=3y
Write System of Equations:
50x+150y=600
x=3y
Solve for y in each equation:
1) 50x+150y=600
150y=−50x+600
y=-1/3x+4
2) x=3y
y=1/3x
The x variable represents the number of commercials and the yy variable represents the number of movies. Since the lines intersect at the point (6,2) we can say:
Emily's songs were played on 6 commercials and 2 movies.
4 books make 1 foot since there are 12 inches in a foot so 4 * 5 is 20
There are 20 books in the stack
Answer:
There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, <em>assuming a significance level 0.1</em>
Step-by-step explanation:
Let p1 be the proportion of West University students who involved in a car accident within the past year
Let p2 be the proportion of East University students who involved in a car accident within the past year
Then
p1=p2
p1≠p2
The formula for the test statistic is given as:
z=
where
- p1 is the <em>sample</em> proportion of West University students who involved in a car accident within the past year (0.15)
- p2 is the <em>sample</em> proportion of East University students who involved in a car accident within the past year (0.12)
- p is the pool proportion of p1 and p2 (
) - n1 is the sample size of the students from West University (100)
- n2 is the sample size ofthe students from East University (100)
Then we have z=
≈ 0.6208
Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.
<em>Assuming significance level 0.1</em>, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance