what are the minimum first quartile median third quartile and the maximum of the data set? 9,20,4,18,4,18,20,9
1 answer:
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A right triangle is a special triangle that has a right angle. In this case, we have to write a system of inequalities that defines a shaded region that looks like a right triangle. First of all, let's say that at the origin the triangle will have a right angle. To do so, we'd need to set:
So the shaded region in this first part will be the First Quadrant as indicated in the first figure below. So if the opposite side lies on the x-axis the adjacent side will lie on the y-axis or if the adjacent side lies on the x-axis the opposite side will lie on the y-axis. Everything is ok up to this point. We just need to define the hypotenuse, so we'd need to define a line. In order to have a right triangle, we need a line with negative slope and positive y-intercept shaded under the line. So, this inequality could be:
Finally, the system of inequalities would be:
And the final shaded region is the one shown in the second figure.
<h2>Learn more:</h2>Inequalities: brainly.com/question/2486051
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integrate to get
this only is true when x>0
when x turns negative, we get -ln(3) because of math
Answer:
Step-by-step explanation:
Hello!
To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:
Sample 1: Compressed soil
X₁: penetrability of a plot with compressed soil.
n₁= 20 plots
X[bar]₁= 2.90
S₁= 0.14
Sample 2: Intermediate soil
X₂: penetrability of a plot with intermediate soil.
n₂= 20 (with outlier) n₂= 19 plots (without outlier)
X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)
S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)
Outlier: 4.26
Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:
Parámeter of interest: μ₁-μ₂
Interval:
[(X[bar]₁-X[bar]₂)±*Sa
]
[(2.90-2.29)±2.715*0.20]
[0.436; 0.784]
I hope this helps!