4)row 1. 20 15 10 row 2. 6
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:
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Step-by-step explanation:
Given that :
Mean = 265
standard deviation = 14
The formula for calculating the z score is
x = μ + σz
At middle of 50% i.e 0.50
The critical value for
From standard normal table
+ 0.67 or -0.67
So; when z = -0.67
x = μ + σz
x = 265 + 14(-0.67)
x = 265 -9.38
x = 255.62
when z = +0.67
x = μ + σz
x = 265 + 14 (0.67)
x = 265 + 9.38
x = 274.38
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Answer:
A. 8x10^7 is 4000 times larger tham 2x10^4
Step-by-step explanation:
The question doesn’t make much sense. Do you mean x=50-12? If so, x=38