_121_ R7 or 121 7/13
13)1580
13
28
26
20
13
7
Answer:
Step-by-step explanation:
The distance the light traveled to the ground is the range of the plane expressed as:
R = U*√2H/g
U is he velocity
H is the max height
g is the acc. due to gravity
Let us get U first using the formula
H = u²sin²Ф/2g
1000 = u²sin²60/2(9.81)
1000*19.62 = 0.8660u²
19620 = 0.8660u²
u² = 19620/0.8660
u² = 22655.88
u = √22655.88
u = 150.52m/s
Next is to get the range
R = 150.52√2(1000)/9.81
R = 150.52 √2000/9.81
R = 150.52√203.87
R = 150.52*14.28
R = 2149.16m
Hence the distance the light traveled to the ground is 2,149.16m
Correction:
Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on
P(E∩E') = P(E)P(E').
Answer:
The case is not always true.
Step-by-step explanation:
Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.
And for any two mutually exclusive events, E and E',
P(E∩E') = 0
Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then
P(E)P(E') cannot be equal to zero.
So
P(E)P(E') ≠ 0
This makes P(E∩E') different from P(E)P(E')
Therefore,
P(E∩E') ≠ P(E)P(E') in this case.
Answer:
Step-by-step explanation:
<u>The distance is:</u>
<u>Return trip will take:</u>
- 480/75 = 6.4 hours = 6 hours and 24 min
5678(1+(0.045×6))
5678=P as it is the principal amount
