Is finding a car's depreciation over a 10 year period considered as univariate or bivariate?
1 answer:
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I already answered that question, here's the link: brainly.com/question/10534199.
But if you still want the explanation, here it is:
"So if we want to measure probability, you have to add, then divide, like this:
First
, which gives us 18.
Then we see, that 6 is 1/3 of 18, so we conclude. The answer is 33.33%.
Hope this helped!"
But if you still want the explanation, here it is:
"So if we want to measure probability, you have to add, then divide, like this:
First
Then we see, that 6 is 1/3 of 18, so we conclude. The answer is 33.33%.
Hope this helped!"
Answer:
10 in
Step-by-step explanation:
Old proportions:
2 x 3
New proportions
h x 15
Assuming the painting has been increased by a constant scale factor.
Where x is the unknown scale factor, and h is the height after it has been enlarged.
Rearrange second equation:
Substitute:
Answer:
Step-by-step explanation:
An incandescent light bulb filament is approximated by a black body radiator, and in the case of a 60W bulb the filament temperature is around 2500˚C which is 2870˚K.
There are a couple of standard black body results that we can use. Firstly the total irradiance emitted per unit area of black body is equal to:
Ed=σT4
where σ=5.67×10−8 Wm−2K−4 is the Stephan-Boltzmann constant. For our bulb we get:
Ed=5.67×10−8⋅28704=3.85×106 Wm−2
As this is a 60W bulb then it has a total irradiance of 60W. Therefore the equivalent black body surface area is:
603.85×106=1.56×10−5 m2
which is 15.6mm2 of filament area.
Secondly we have that the total number of photons emitted per second per unit area of black body is equal to:[1]
Qd=σQT3
where σQ=1.52×1015 photons.sec−1m−2K−3. For our bulb this is:
Qd=1.52×1015⋅28703=3.59×1025 photons.sec−1m−2
Multiplying by the bulb’s equivalent black body surface area gives the result we require:
3.59×1025⋅1.56×10−5=5.6×1020 photons/sec
As a sanity check we know that these photons have a total energy of 60 joules per second, so the average energy per photon is:
605.6×1020=1.1×10−19 joules
A photon of wavelength λ has energy E=hcλ, and so the average energy corresponds with a photon of wavelength:
λ=hc1.1×10−19=1.9μm
Here’s a chart of the power distribution by wavelength, with the average photon wavelength shown as the dashed line, and visible wavelengths highlighted:
emember that there are a higher proportion of photons for the longer, lower energy wavelengths, so the average is weighted to the right.
We can also see from the original calculation that the general case is:
QdWEd=σQT3WσT4=2.68×1022WT photons/sec
for a bulb of wattage W watts and filament temperature T ˚K. So the photon emission rate is inversely proportional to the filament temperature. As a somewhat counter-intuitive example, a 60W halogen bulb with 3200˚K filament only emits photons at 90% of the rate of the standard 60W bulb, despite being visibly brighter.
The reason is that as the temperature increases then an increased proportion of shorter wavelength photos are emitted and therefore the average energy per photon increases, decreasing the number emitted per second. However at the same time an increased proportion of the photons are visible rather than infra-red, making the bulb appear brighter. Here’s the power distribution chart with the 60W halogen curve added for comparison:
