The expression to find the number of notebooks James bought would be written as:
Number of notebooks = total spent / cost per item
Number of notebooks = (2y^2 + 6) / <span>(y^2 − 1)
</span><span>If y = 3, then the number of notebooks bought would be:
</span>Number of notebooks = (2y^2 + 6) / (y^2 − 1)
Number of notebooks = (2(3)^2 + 6) / (3^2 − 1)
Number of notebooks = 3 pieces<span>
</span><span>
</span>
Answer:
<em>An </em><em>echo</em>
Step-by-step explanation:
................
The answer is 5/72 you would just divide 5/6 by 12 hope this helped
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>
Let, the speed of boat = x
Speed of current = y
So, x + y = 280/7
x + y = 40 ---- first equation
x - y = 280/14
x - y = 20
x = 20 + y
Substitute this value of x in first equation,
20 + y + y = 40
2y = 40 - 20
y = 20/2
y = 10
Substitute this in first equation,
x + 10 = 40
x = 30
In short, Speed of Boat = 30 mph, & speed of current = 10 mph
Hope this helps!
