Answer:
20%
Step-by-step explanation:
150 and 120 have a difference of 30
30 ÷ 150
.2 then multiply by 100
20 there is your answer
Hope this helped :)
Answer:
6
Step-by-step explanation:
Since the x coordinate is the same, we only have to look at the y coordinate
3 - -3 = 3 +3 = 6
The distance is 6
Answer:
I'm pretty sure that the answers would go, Wrong, increased, 9 1/2 points, 6 5/8 points, dropped, 12 3/4 points.
Step-by-step explanation:
The reason why it would be wrong is because the amount of points lost was less than the total amount of points gained.
In week one and two the company gained 9 1/2 points and 6 5/8 points. When you add them together, you get 16 1/8 points.
In week three the company lost 12 3/4 points. If you multiply the fraction to change the denominator to 8, the mixed fraction becomes 12 6/8.
16 1/8 - 12 6/8 = 3 3/8
As you can see, there are still points left over.
In the fourth week, they ended up losing all of their points and then some more, but from my understanding it's just asking about the third week.
Hope that this helps!
Answer:
False ; the answer to that equation is 56
Step-by-step explanation:
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>
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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>
