Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
Answer:
the distance bettwen 0 and -5 on a number line
Step-by-step explanation:
1.5x + 0.2y = 2.68....multiply by 0.3
1.6x + 0.3y = 2.98...multiply by - 0.2
------------------------
0.45x + 0.06y = 0.804 (result of multiplying by 0.3)
- 0.32x - 0.06y = - 0.596 (result of multiplying by - 0.2)
----------------------add
0.13x = 0.208
x = 0.208/0.13
x = 1.6
1.5x + 0.2y = 2.68
1.5(1.6) + 0.2y = 2.68
2.4 + 0.2y = 2.68
0.2y = 2.68 - 2.4
0.2y = 0.28
y = 0.28/0.2
y = 1.4
solution (they intersect at) (1.6,1.4)
The answer if you need helo with math like this come to me -25
Answer:
Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the 
percentile for the television weights, use the formula:
, where 
is the average of the set, 
is some constant relevant to the percentile you're finding, and 
is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute 
, 
, and 
:
Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:
The difference between these two weights is 
.
