Answer:
129 points
Step-by-step explanation:
Let L and C represent the scores of Luke and Caleb, respectively.
L = 2C -15 . . . . . Luke scored 15 less than twice the number Caleb did
L +C = 201 . . . . . they scored 201 points altogether
Add twice the second equation to the first:
2(L +C) + (L) = 2(201) + (2C -15)
3L +2C = 387 +2C . . . . simplify
3L = 387 . . . . . . . . . . . . .subtract 2C
L = 387/3 = 129 . . . . . . .divide by 3
Luke scored 129 points.
Answer:
x = 4/3
Step-by-step explanation:
2x + x + 33 = 90
3x + 33 = 90
3x = 57
x = 19
Angle 1 = 38°
Angle 2 = 52°
Answer:
Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.
The value of p is 0 .00233. The result is significant at p < 0.10.
Step-by-step explanation:
1) Let the null and alternate hypothesis be
H0: μboys − μgirls > 0
against the claim
Ha: μboys − μgirls ≤ 0
2) The significance level is set at 0.01
3) The critical region is z ≤ ± 1.28
4) The test statistic
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
Here p1= 397/768= 0.5169 and p2= 331/745=0.4429
pc = 397+331/768+745
pc= 0.4811
qc= 1-pc= 1-0.4811=0.5188
5) Calculations
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
z= 0.5169-0.4429/√ 0.4811*0.5188( 1/768+ 1/745)
z= 2.82
6) Conclusion
Since the calculated value of z= 2.82 does not lie in the critical region the null hypothesis is accepted and it is concluded that the sample data support the authors' conclusion that the proportion of the country's boys who listen to music at high volume is greater than this proportion for the country's girls.
7)
The value of p is 0 .00233. The result is significant at p < 0.10.
Answer:
Plane T
Step-by-step explanation:
The given points A, D, and E are shown as being on the plane T as follows
Point a lays on the boundary of the boundary between plane S and plane T
Points D and E are points on a line DE that is on the plane T
Therefore, given that the line DE and the point A all lay on the same plane T, the plane that contains points A, D, and E, is plane T.
