Answer:
(0,44)
Step-by-step explanation:
To answer this question, I would start by finding the equation of the second line. You can do that by plugging it into the formula y-y1=m(x-x1). You can take the slope from the first equation, which would be m=-13. So, y-5=-13(x-3). You can simplify that to get y=-13+44. Finally, to get a point, you can plug in almost any value for x and use the value of y to create a point. For example, if you choose x=0, you can get y=44 and the point (0,44)
Answer:
IQR = 5
The IQR covers 50%
Step-by-step explanation:
The interquartile range is the right line of the box minus the left side of the box
UQ - LQ = IQR
8 - 3 = 5
The interquartile range is 5
Now to find the percent in the interquartile range.
The Box divides the data in quartiles or 25%
The dot to the left line is 25%
The left line to the middle is 25%
The middle to the right line is 25%
The right line to the right dote is 25%
From the left line to the right line is 25% + 25% = 50 %
The IQR covers 50%
solution:
Since, it’s a standard cone opening up on the positive x – axis therefore,
X = √y²+z²
And because it has a vertex at (7,0,0)
Therefore the equation of surface is
X = 7 + √y²+z²
Answer:
2.15 times or 215%
Step-by-step explanation:
V(Luxor) = (646)² · 350/3
V(Tower) = 290 x 125 x 622
V(Luxor) ÷ V(Tower) = 2.15
Answer:
d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.
Step-by-step explanation:
Hello!
You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.
Be X₁: Chemical measurement in region 1 (ppm)
Sample 1
n= 12
981 726 686 496 657 627 815 504 950 605 570 520
μ₁= 678
σ₁= 164
Sample mean X[bar]₁= 678.08
X₂: Chemical measurement in region 2 (ppm)
Sample 2
n₂= 16
1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844
μ₂= 812
σ₂= 239
Sample mean X[bar]₂= 811.94
Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.
Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.
[(X[bar]₁-X[bar]₂)±
*
]

[(678.08-811.94)±1.648*
]
[-259.49;-8.23]ppm
Both bonds of the interval are negative, this means that with a 90% confidence level the difference between the population means of the chemical measurements of region 1 and region 2 may be included in the calculated interval.
You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.
I hope it helps!