The graph of the equation y=-2 is a line Explain your answer 1) parallel to the x-axis 2)parallel to the y-axis 3)passing throug
h the origin 4) passing through the point (-2,0)
1 answer:
Answer:
<h2>Parallel to the x-axis</h2>Step-by-step explanation:
Given the graph of the equation y = -2. To determine how the line will look like on Cartesian plane, first we must understand the following;
The standard form of equation of a line is y = mx+c where m is the slope and c is the intercept. Comparing this standard form to the given equation shows that the equation is in slope-intercept form.
It can also written as y = 0x-2. This shows that slope m = 0 and the y-intercept c = -2. The graph can be graphed as y = 0x-2
The line will be an horizontal line drawn on the negative y-axis at the coordinate x = 0 and y = -2. <u>The line will therefore be parallel to the x axis </u>
<em>Check the attachment for the graph.</em>
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Answer:
95.44% of the grasshoppers weigh between 86 grams and 94 grams.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 90 grams and a standard deviation of 2 grams.
This means that
What percentage of the grasshoppers weigh between 86 grams and 94 grams?
The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 86. So
X = 94
has a p-value of 0.9772.
X = 86
has a p-value of 0.0228.
0.9772 - 0.0228 = 0.9544
0.9544*100% = 95.44%
95.44% of the grasshoppers weigh between 86 grams and 94 grams.
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:
