ABC is congruent to FGH, AB=8.3, BC=10.4, and CA=4.2. What is the measure of segment GH?
2 answers:
For a better understanding of the explanation provided here please go through the diagram in the attachment given below.
As we can see from the diagram of the congruent triangles ABC and FGH, the vertices A,B,C of correspond to the vertices F,G,H of
, so the corresponding sides must correspond too. Thus, AB corresponds to FG, BC corresponds to GH and CA corresponds to GF.
Now, we know that BC=10.4. Therefore, it's corresponding side GH must be 10.4 too because they are corresponding sides of a congruent triangle.
Thus, GH=10.4 is the correct 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:
7 quarters are spent for throwing 350 food pebbles in pond.
Step-by-step explanation:
Given:
Reggie receives 50 food pebbles for every quarter he spends.
Number of food pebbles he has = 350
Let the number of quarters spent for 350 pebbles be 'x'.
Now, using proportion and cross multiplication method, we can find 'x'.
<u><em>1 quarter is equivalent to 50 pebbles thrown in pond.</em></u>
<u><em>So, 'x' quarters is equivalent to 350 pebbles thrown in pond.</em></u>
So, 7 quarters are spent for throwing 350 food pebbles in pond.