If the results of an experiment contradict the hypothesis, you have falsified the hypothesis.
A hypothesis is a proposed explanation for a scientific phenomenon. It is based on observations, prior knowledge, and logical reasoning. When conducting an experiment, scientists test their hypothesis by collecting data and analyzing the results.
If the results of the experiment do not support or contradict the hypothesis, meaning they go against what was predicted, then the hypothesis is considered to be falsified. This means that the hypothesis is not a valid explanation for the observed phenomenon.
Falsifying a hypothesis is an important part of the scientific process. It allows scientists to refine their understanding of the phenomenon under investigation and develop new hypotheses based on the evidence. It also helps prevent bias and ensures that scientific theories are based on reliable and valid data.
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Write six different iterated triple integrals for the volume of the tetrahedron cut from the first octant by the plane xyz. Evaluate the first integral. Question content area bottom Part 1
Using triple integration, the volume of tetrahedron cut from the plane 2x + y + z = 4 is [tex]\frac{16}{3}[/tex].
A tetrahedron is nothing but a three dimensional pyramid.
To find the volume of tetrahedron cut from the plane 2x + y + z = 4, we need to first take one of the three dimension as base. Let as take xy plane as base.
XY as plane implies z = 0, equation becomes 2x + y = 4. To find the limits of X and Y, we put y = 0.
Thus, 2x + 0 = 4 , implying, x = 2.
Thus the range of x is : [0,2]
Putting the value of x in the given equation, the range of y is [0, 4 - 2x]
Similarly, range of z becomes: [0, 4 - 2x - y]
Since z is dependent upon y and x, and, y is dependent on x, Therefore the order of integration must be z, then y and then x.
The volume of tetrahedron becomes:
[tex]=\int\limits^0_2 \int\limits^{4-2x}_0 \int\limits^{4-2x-y}_0 {1} \, dz \, dy \, dx \\\\=\int\limits^0_2 \int\limits^{4-2x}_0 4-2x-y \, dy \, dx \\\\=\int\limits^0_2[ (4-2x)y - \frac{y^2}{2}]^{4-2x}_0 dx\\ \\=\int\limits^0_2 (4-2x)^2 - \frac{1}{2} (4-2x)^2 dx\\\\[/tex]
[tex]=\int\limits^2_0 {\frac{1}{2}(16+4x^2-16x )} \, dx \\\\=\int\limits^2_0(8+2x^2-8x)dx\\\\=[8x+\frac{2}{3} x^3-4x^2]^2_0\\\\=\frac{16}{3}[/tex]
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The complete question is given below:
Use triple integration to find the volume of tetrahedron cut from the plane 2x + y + z = 4.
Assume the following for this question. Lower and Upper specification limits for a service time are 3 minutes and 5 minutes, respectively with the nominal expected service time at 4 minutes. The observed mean service time is 4 minutes with a standard deviation of 0.2 minutes. The current control limits are set at 3.1 and 4.9 minutes respectively.
The observed mean service time falls within the current control limits. We can conclude that the process is stable, the service time is in control, and it meets the required specifications.
1. Calculate the process capability index (Cpk) using the formula: Cpk = min((USL - mean)/3σ, (mean - LSL)/3σ), where USL is the upper specification limit, LSL is the lower specification limit, mean is the observed mean service time, and σ is the standard deviation.
2. Plug in the values: USL = 5 minutes, LSL = 3 minutes, mean = 4 minutes, σ = 0.2 minutes.
3. Calculate Cpk: Cpk = min((5-4)/(3*0.2), (4-3)/(3*0.2)) = min(0.556, 0.556) = 0.556.
4. Since the calculated Cpk is greater than 1, the process is considered capable and the service time is in control.
5. The current control limits (3.1 and 4.9 minutes) are wider than the specification limits (3 and 5 minutes) and the observed mean (4 minutes) falls within these control limits.
6. Therefore, the process is stable and meets the specifications.
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Figure 10.5
Coverage
garage and other structures
loss of use
personal property
percent coverage
10%
20%
50%
Replacement value: $270,000; Coverage: 80%
Problem:
a. Amount of insurance on the home
b. Amount of coverage for the garage
c. Amount of coverage for the loss of use
d. Amount of coverage for personal property
Answers:
The amount of Insurance on the home as $216,000, but the amounts of coverage for the garage, loss of use, and personal property cannot be determined without additional information.
To calculate the amounts of coverage for the different components, we need to use the given replacement value and coverage percentages.
a. Amount of insurance on the home:
The amount of insurance on the home can be calculated by multiplying the replacement value by the coverage percentage for the home. In this case, the coverage percentage is 80%.
Amount of insurance on the home = Replacement value * Coverage percentage
Amount of insurance on the home = $270,000 * 80% = $216,000
b. Amount of coverage for the garage:
The amount of coverage for the garage can be calculated in a similar manner. We need to use the replacement value of the garage and the coverage percentage for the garage.
Amount of coverage for the garage = Replacement value of the garage * Coverage percentage for the garage
Since the replacement value of the garage is not given, we cannot determine the exact amount of coverage for the garage with the information provided.
c. Amount of coverage for the loss of use:
The amount of coverage for the loss of use is usually a percentage of the insurance on the home. Since the insurance on the home is $216,000, we can calculate the amount of coverage for the loss of use by multiplying this amount by the coverage percentage for loss of use. However, the percentage for loss of use is not given, so we cannot determine the exact amount of coverage for loss of use with the information provided.
d. Amount of coverage for personal property:
The amount of coverage for personal property can be calculated by multiplying the insurance on the home by the coverage percentage for personal property. Since the insurance on the home is $216,000 and the coverage percentage for personal property is not given, we cannot determine the exact amount of coverage for personal property with the information provided.
the amount of insurance on the home as $216,000, but the amounts of coverage for the garage, loss of use, and personal property cannot be determined without additional information.
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Suppose that in a particular sample, the mean is 50 and the standard deviation is 10. What is the z score associated with a raw score of 68?
The z-score associated with a raw score of 68 is 1.8.
Given mean = 50 and standard deviation = 10.
Z-score is also known as standard score gives us an idea of how far a data point is from the mean. It indicates how many standard deviations an element is from the mean. Hence, Z-Score is measured in terms of standard deviation from the mean.
The formula for calculating the z-score is given as
z = (X - μ) / σ
where X is the raw score, μ is the mean and σ is the standard deviation.
In this case, the raw score is X = 68.
Substituting the given values in the formula, we get
z = (68 - 50) / 10
z = 18 / 10
z = 1.8
Therefore, the z-score associated with a raw score of 68 is 1.8.
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