Multiple im pretty sure- Well, a multiple is any number obtained by multiplying other numbers together. Multiples are most commonly discussed in the context of integers.
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>
1 and 19, 19 is a prime number.
Answer:
C
Step-by-step explanation:
A modified box plot does not include the outliers in the whiskers, instead they are points outside of the whiskers
5,25,33,34,34,37,37,40,42,45,45,46,46,49,73
This data has 2 outliers 5 and 73 so we have 2 choices for a modified box plot D or D
The lowest value outside of the outliers is 25 , so C would be the logical choice
D has the lower end of the whisker too close to the outlier
Two ratios that are equivalent are the same in value. We call these equivalent ratios. To find an equivalent ratio multiply or divide both of the numbers by a common number (or same number), this would be the exact same when finding equivalent fractions as well.
