Answer:
the answer is D) 5(30 - 5) + 40; 165 stars
Step-by-step explanation:
because you multiply the number of levels between Level 30 and Level 5 by 5 stars per level. Add the number of stars needed in Level 5.
5(30 - 5) + 40
5(25) + 40
165 stars
Let's assume initial population is in 1999
so, production of passenger cars in Japan is 3.94 million in 1999
so, P=3.94 million
and the production of passenger cars in Japan is 6.74 million in 2009
so, A=6.74 million in t=2009-1999=10 years
now, we can use formula
here , r is interest rate
so, we can plug values
now, we can solve for
so,
the geometric mean annual percent increase is 5.516%............Answer
Answer:
$0
Step-by-step explanation:
A part-time landscaper made $8996.32 last year.
She claimed herself as an exemption for $3650 and had a $5700 standard deduction
Exemption means not subject to taxation.
Deduction means taking some amount of your income for the year, and not have to pay taxes on it.
So, 
Since Her income is lower than the exemption and the standard deduction.
So, her taxable income last year was $0.
Thus Option D is correct.
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>
