Answer:
Distance= 6.6 miles
Bearing= N 62.854°W
Step-by-step explanation:
Let's determine angle b first
Angle b=20° (alternate angles)
Using cosine rule
Let the distance between the liner and the port be x
X² =8.8²+2.4²-2(8.8)(2.4)cos20
X²= 77.44 + 5.76-(39.69)
X²= 43.51
X= √43.51
X= 6.596
X= 6.6 miles
Let's determine the angles within the triangle using sine rule
2.4/sin b = 6.6/sin20
(2.4*sin20)/6.6= sin b
0.1244 = sin b
7.146= b°
Angle c= 180-20-7.146
Angle c= 152.854°
For the bearing
110+7.146= 117.146
180-117.146= 62.854°
Bearing= N 62.854°W
Answer:
196
Step-by-step explanation:
A. X Z Y
B. P N M
C. F E H G D
The answer to your question is 22 degrees Celsius. To figure out the problem, all you do is take the minutes( 58 and 2 and 35 sec.) and convert to seconds(3,480 and 145) Then divide 3,480 by 145 ,which equals 24. Then take 24 and add to the final temperature, which equals 22.
hopes this helps! <span />
Answer:
90
Step-by-step explanation:
The sum of squares of the deviation from mean=sum(x-xbar)²=?
x
12
6
15
9
18
12
xbar=sumx/n
xbar=(12+6+15+9+18+12)/6=72/6=12
x x-xbar (x-xbar)²
12 12-12=0 0
6 6-12=-6 36
15 15-12=3 9
9 9-12=-3 9
18 18-12=6 36
12 12-12=0 0
sum(x-xbar)²=0+36+9+9+36+0=90
So, the sum of squares of the deviations from the mean is 90.
