1. 6:3
2. 6:2
3. 3:18
4:5:6
5:13
6:x
7:x
8: check
9:check
10: x
Hey there!
When the value of something goes down over time- a house, a car, anything- it's called depreciation.
It would be expressed as exponential decay. This is because the value is going down exponentially as stated in the problem. Additionally, when it decays, the growth factor is less than one, so the value steadily decreases.
Let's rule out our other options:
B) Half-Life. This doesn't work because it's usually used for radioactive decay, which is usually a steady decrease of radioactivity and material that divides by two, hence the word half. For example:
Day 0) Started with 10 grams of plutonium
Day 1) 5 grams
Day 2) 2.5 grams
And so on.
C) Compound Interest. Like simple interest, but compunds usually at a certain point - yearly, monthly, every ten years, and so on. But, this is used for interest, mostly in a bank account or if you have a loan. It doesn't exhibit depreciation, and therefore it's not used to predict car value.
D) Exponential Growth- Exponential growth does not work for one main reason - it grows. That would mean that the value of the car goes up- and unless it's an old and very expensive car, that wouldn't be the case.
Hope this helps!
Answer:
7x+72xy
Step-by-step explanation:
7x+8x(9y)=7x+72xy
Answer:
$2000
Step-by-step explanation:
the profit would be determined by how many computers are involved, so that would vary
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>
