Rationalize the denominator √7x/75
1 answer:
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Answer:
a)
X[bar]= 71.8cm
X[bar]= 72cm
b)
M.A.D.= 8.16cm
M.A.D.= 5.4cm
c) The data set for Soil A is more variable.
Step-by-step explanation:
Hello!
The data in the stem-and-leaf plots show the heights in cm of Teddy Bear sunflowers grown in two different types of soil (A and B)
To read the data shown in the plots, remember that the first digit of the number is shown in the stem and the second digit is placed in the leaves.
The two data sets, in this case, are arranged in a "back to back" stem plot, which allows you to compare both distributions. In this type of graph, there is one single stem in the middle, shared by both samples, and the leaves are placed to its left and right of it corresponds to the observations of each one of them.
Since the stem is shared by both samples, there can be observations made only in one of the samples. For example in the first row, the stem value is 5, for the "Soil A" sample there is no leaf, this means that there was no plant of 50 ≤ X < 60 but for "Soil B" there was one observation of 59 cm.
X represents the variable of interest, as said before, the height of the Teddy Bear sunflowers.
a) To calculate the average or mean of a data set you have to add all observations of the sample and divide it by the number of observations:
X[bar]= ∑X/n
For soil A
Observations:
61, 61, 62, 65, 70, 71, 75, 81, 82, 90
The total of observations is n= 10
∑X= 61 + 61 + 62 + 65 + 70 + 71 + 75 + 81 + 82 + 90= 718
X[bar]= ∑X
/n
= 218/10= 71.8cm
For Soil B
Observations:
59, 63, 69, 70, 72, 73, 76, 77, 78, 83
The total of observations is n= 10
∑X= 59 + 63 + 69 + 70 + 72 + 73 + 76 + 77 + 78 + 83= 720
X[bar]= ∑X
/n
= 720/10= 72cm
b) The mean absolute deviation is the average of the absolute deviations of the sample. It is a summary of the sample's dispersion, meaning the greater its value, the greater the sample dispersion.
To calculate the mean absolute dispersion you have to:
1) Find the mean of the sample (done in the previous item)
2) Calculate the absolute difference of each observation and the sample mean |X-X[bar]|
3) Add all absolute differences
4) Divide the summation by the number of observations (sample size,n)
For Soil A
1) X[bar]= 71.8cm
2) Absolute differences |X-X[bar]
|
|61-71.8|= 10.8
|61-71.8|= 10.8
|62-71.8|= 9.8
|65-71.8|= 6.8
|70-71.8|= 1.8
|71-71.8|= 0.8
|75-71.8|= 3.2
|81-71.8|= 9.2
|82-71.8|= 10.2
|90-71.8|= 18.2
3) Summation of all absolute differences
∑|X-X[bar]
|= 10.8 + 10.8 + 9.8 + 6.8 + 1.8 + 0.8 + 3.2 + 9.2 + 10.2 + 18.2= 81.6
4) M.A.D.=∑|X
-X[bar]
|/n
= 81.6/10= 8.16cm
For Soil B
1) X[bar]= 72cm
2) Absolute differences |X-X[bar]
|
|59-72|= 13
|63-72|= 9
|69-72|= 3
|70-72|= 2
|72-72|= 0
|73-72|= 1
|76-72|= 4
|77-72|= 5
|78-72|= 6
|83-72|= 11
3) Summation of all absolute differences
∑ |X-X[bar]
|= 13 + 9 + 3 + 2 + 0 + 1 + 4 + 5 + 6 + 11= 54
4) M.A.D.=∑ |X
-X[bar]
|/n
= 54/10= 5.4cm
c)
If you compare both calculated mean absolute deviations, you can see M.A.D. > M.A.D.
. As said before, the M.A.D. summary of the sample's dispersion. The greater value obtained for "Soil A" indicates this sample has greater variability.
I hope this helps!
Answer:
Yes;Each side of triangle PQR is the same length as the corresponding side of triangle STU
Step-by-step explanation:
You can observe the sides of both triangles to see if this property holds
Lets check the length of AB
A(0,3) and B(0,-1)
Now check the length of the corresponding side DE
D(1,2) and E(1,-2)
The side AB has the same length as side DE.This is also true for the remaining corresponding sides.
Answer:
The probability that it will still be working after one week is
Step-by-step explanation:
Given :
Total number of bulbs = 25
Number of bulbs which are good condition and will function for at least 30 days = 5
Number of bulbs which are partially defective and will fail in their second day of use = 10
Number of bulbs which are totally defective and will not light up = 10
To find : What is the probability that it will still be working after one week?
Solution :
First condition is a randomly chosen bulb initially lights,
i.e. Either it is in good condition and partially defective.
Second condition is it will still be working after one week,
i.e. Bulbs which are good condition and will function for at least 30 days
So, favorable outcome is 5
The probability that it will still be working after one week is given by,