Answer:
a) <u>0.4647</u>
b) <u>24.6 secs</u>
Step-by-step explanation:
Let T be interval between two successive barges
t(t) = λe^λt where t > 0
The mean of the exponential
E(T) = 1/λ
E(T) = 8
1/λ = 8
λ = 1/8
∴ t(t) = 1/8×e^-t/8 [ t > 0]
Now the probability we need
p[T<5] = ₀∫⁵ t(t) dt
=₀∫⁵ 1/8×e^-t/8 dt
= 1/8 ₀∫⁵ e^-t/8 dt
= 1/8 [ (e^-t/8) / -1/8 ]₀⁵
= - [ e^-t/8]₀⁵
= - [ e^-5/8 - 1 ]
= 1 - e^-5/8 = <u>0.4647</u>
Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>
<u></u>
b)
Now we find t such that;
p[T>t] = 0.95
so
t_∫¹⁰ t(x) dx = 0.95
t_∫¹⁰ 1/8×e^-x/8 = 0.95
1/8 t_∫¹⁰ e^-x/8 dx = 0.95
1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t = 0.95
- [ e^-x/8]¹⁰_t = 0.96
- [ 0 - e^-t/8 ] = 0.95
e^-t/8 = 0.95
take log of both sides
log (e^-t/8) = log (0.95)
-t/8 = In(0.95)
-t/8 = -0.0513
t = 8 × 0.0513
t = 0.4104 (min)
so we convert to seconds
t = 0.4104 × 60
t = <u>24.6 secs</u>
Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>
You would do 172 -8 and keep the i when you cancel the numbers out
85 x 0.07 = 5.95
Total = 85 + 5.95 = 90.95%
Answer: 15.7 minutes
Step-by-step explanation:
Let x be the time in the beginning (in minutes).
Given: The track team is trying to reduce their time for a relay race.
First they reduce their time by 
2.1 minutes.
Then they are able to reduce that time by 
10
If their final time is 3.96 minutes, then
Hence, their beginning time was 15.7 minutes.
Answer:
45 segundos.
Step-by-step explanation:
Un tren pasa por delante de un puente en 15 segundos; si el puente tuviera el doble de longitud, le tomaría el doble de tiempo en cruzarlo, si tuviera el triple de longitud, le tomaría el triple de tiempo y así sucesivamente.
En este caso, la pregunta es ¿En cuánto tiempo cruzaría el puente si tuviera el triple de su longitud? Por lo tanto, si cruza el puente en 15 segundos, teniendo el triple de longitud le tomaría 3 (15) = 45 segundos en cruzarlo.
