Answer:
Therefore they are 734.106 miles apart.
Step-by-step explanation:
Given that ,
Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.
It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.
Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles
Then the distance between the ships = AC
We know
The sum of all angles = 180°
⇒∠A+∠B+∠C=180°
⇒∠A+135°40'+∠C=180°
⇒2∠A= 180°- 135°40' [ since ∠A=∠C]
⇒2∠A=44°60'
⇒∠A= 22°30'
Again we know that,
Taking last two ratio,
Putting the value of BC , AC ,∠A,∠B
≈734.106 miles
Therefore they are 734.106 miles apart.
Answer:
0.347
Step-by-step explanation:
n = 3
p = 1/6
r = 1
Use binomial probability:
P = nCr pʳ qⁿ⁻ʳ
P = ₃C₁ (1/6)¹ (5/6)³⁻¹
P = 0.347
Or, using a calculator:
P = binompdf(n, p, r)
P = binompdf(3, 1/6, 1)
P = 0.347
6 can go into 48 8 times. How I figured out?
Start multiplying from a small number like
6×6= 36
6×7=42
6×8= 48
until you see your number 48, or a number that is lower that 48, but can not go any higher than 48.
Answer:
Step-by-step explanation:
Here are the 3 answers you needed :)!!!
Answer:
B. 41.8° is the correct answer
Step-by-step explanation:
We are given that,
Length of Eduardo's flight path = 273 miles
Length of Paul's flight path = 357 miles
Distance between their destinations = 238 miles.
Now, using the law of cosines, we get,
i.e. 
i.e. 
i.e. 
i.e. 
i.e. 
i.e. θ = 41.8°
Hence, the angle between their flight paths is 41.8°.
