Answer:
e = 4796
Step-by-step explanation:
given,
mean of five distinct positive number = 1000
median of the number = 100
100 is median means two number will be less than 100 and two number will be greater than 100.
let five number be
a , b, c, d, e
'e' should be the largest number
As 100 is median so 'c' = 100.
'a' and 'b' should be as small as possible and d should be the number nearest to 100.
As all the number are distinct so the least number be equal to 1 and 2
now d will be equal to 101 (nearest to 100)
now,
sum of the five number = 5 x 1000 = 5000
a + b + c + d + e = 5000
1 + 2 + 100 + 101 + e = 5000
e = 5000 - 204
e = 4796
hence, the largest number will be equal to e = 4796
Add all the sides together
10+15+7+8+8+2=50 ft
Y coordinate on solving both equations comes out to be 0
-2x+3y=-6
3y = -6+2x
put the value of 3y in equation 2nd
5x-2(-6+2x) =15
5x+12-4x = 15
x=3
put value of X in 3y = -6+2x
3y = -6+2*3 = -6+6 = 0
thus y = 0
He makes it up by 1/3=.333%=$75*1.333=$99.98
he then discounts it by 20%=1/5=99.98/5=$19.99
If the original profit is $99.98-$75=$24.98
and then discounted by $19.99
it would be $24.98-$19.99
=
$4.99
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>
