Step-by-step explanation:
By using Ms Excel you should do this problem.
Kindly check the attachment for solution and cell reference for the clear understanding.
1- which of the following represents a radius of the circle
2 answers:
Answer:
Part 1) Option D. MN
Part 2) Option A TR
Part 3) Option D SRN
Part 4) Option A
Part 5) Option C NST
Step-by-step explanation:
Part 1- which of the following represents a radius of the circle
we know that
The radius of a circle is the distance from the center point to the edge of the circle
In this problem, the radius is equal to
therefore
the answer is option D
Part 2-which of the following represents a chord but not a diameter of the circle
we know that
The <u>chord</u> is a line segment that joins any two points of the circle. The endpoints of this line segments lie on the circumference of the circle.
The <u>diameter</u> is a chord that passes through the center. Is the longest chord of the circle
so
In this problem the diameter is equal to and a chord that not represent a diameter is
Part 3-Which of the following is a semicircle
we know that
A <u>semicircle</u> is a half circle
so
In this problem SRN represent a semicircle and STPN represent a semicircle too
Part 4-if the length of the diameter of the circle is 15 inches what is the length of the radius of the circle
we know that
The radius is half the diameter
so
we have
substitute
Part 5-which of these arcs is the largest
Verify each case
Arc NR --------> The length of the arc belong to the interval degrees
Arc NRS ------> The length of the arc is equal to degrees
Arc NST --------> The length of the arc belong to the interval degrees
Arc NP --------> The length of the arc belong to the interval degrees
therefore
the answer is the arc NST
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Answer:
Step-by-step explanation:
Hello!
You have the information about voltage readings at an old and a new manufacturing location obtained remotely.
a, b and c in the attachment.
Histograms for a and c:
To construct a frequency histogram you have to first arrange the data for both locations in a frequency table. For this, I'm going to determine 5 class interval for each location. To do so you need to calculate the width of the intervals. First, you calculate the range of the variable and then you have to divide it by the number of intervals you want to do.
Old location: Range= 10,55-8,05= 2,5 → Class width: 2,5/5= 0,5
New Location: Range= 10,12-8,51= 1,61 → Class width: 1,61/5= 0,322
Starting from the minimum value you add the calculated width and create the intervals:
Old Location:
8,05-8,55
8,55-9,05
9,05-9,55
9,55-10,05
10,05-10,55
New Location
8,51-8,83
8,83-9,15
9,15-9,48
9,48-9,80
9,80-10,12
Stem and Leaf diagram for b:
To construct this diagram first I've ordered the data from leat to greatests. Then I've used the integer to form the stem 8,- 9.- and 10.- and the decimals are placed in the leafs of the diagram.
Comparing it to the histogram and stem and leaf diagram for the readings of the Old Location, the histogram stem, and leaf diagram show better where most of the readings lie.
d.
Comparing both histograms, it looks like the readings in the new location are more variable than the readings in the old location but more uniformly distributed. I would say that the readings in the new location are better than the readings in the old location.
e.
To calculate the mean you have to apply the following formula:
X[bar]= (∑xi'fi)/n
X[bar]OLD=(∑xi'fi)/n= (8.3*1+8.8*3+9.3*0+9.8*17+10.3*9)/30= 294/30= 9.8
X[bar]NEW=(∑xi'fi)/n= (8.67*6+8.99*2+9.315*7+9.64*8+9.96*7)/30= 282.045/30= 9.4015≅9.40
First you have to calculate the position of the median:
For both data sets the PosMe= 30/2=15
Now you arrange the data from least to highest and determine wich observation is in the 15th position:
Old Location
8,05 , 8,72 , 8,72 , 8,8 , 9,55 , 9,7 , 9,73 , 9,8 , 9,8 , 9,84 , 9,84 , 9,87 , 9,87 , 9,95 , 9,97 , 9,98 , 9,98 , 10 , 10,01 , 10,02 , 10,03 , 10,05 , 10,05 , 10,12 , 10,15 , 10,15 , 10,26 , 10,26 , 10,29 , 10,55
MeOLD= 9.97
New Location
8,51 , 8,65 , 8,68 , 8,78 , 8,82 , 8,82 , 8,83 , 9,14 , 9,19 , 9,27 , 9,35 , 9,36 , 9,37 , 9,39 , 9,43 , 9,48 , 9,49 , 9,54 , 9,6 , 9,63 , 9,64 , 9,7 , 9,75 , 9,85 , 10,01 , 10,03 , 10,05 , 10,09 , 10,1 , 10,12
MeNEW= 9.43
The mode is the observation with more absolute frequency.
To determine the mode on both data sets I'll use the followinf formula:
Md= Li + c [Δ₁/(Δ₁+Δ₂)]
Li= Lower bond of the interval with most absolute frequency (modal interval)
c= amplitude of the modal interval
Δ₁= absolute frequency of the modal interval minus abolute frequency of the previous interval
Δ₂= absolute frequency of the modal interval minus the absolute frequency of the next interval
Modal interval OLD
9,55-10,05
Δ₁= 17-0= 17
Δ₂= 17-9= 8
c= 0.5
Li= 9.55
MdOLD= 9.55 + 0.5*[17/(17+8)]= 9.89
Modal interval NEW
9,48-9,80
Δ₁= 8-7= 1
Δ₂= 8-7= 1
c= 0.32
Li= 9.48
MdNEW= 9.48+0.32*[1/(1+1)]= 9.64
f.
OLD
Mean 9.8
SE 0.45
X= 10.50
Z= (10.50-9.8)/0.45= 1.56
g.
NEW
Mean 9.4
SE 0.48
X=10.50
Z= (10.50-9.4)/0.48=2.29
h. The Z score for the reading 10.50 for the old location is less than the Z score for the reading 10.50 for the new location, this means that the reading is closer to the mean in the old location than in the new location.
The reading 10.50 is more unusual for the new location.
i. and k. Boxplots attached.
There are outliers for the readings in the old location, none in the readings for the new location.
j. To detect outliers using the Z- score you have to "standardize every value of the data set using the corresponding values of the mean and standard deviation. Observations that obtained a Z-score greater than 3 or less than -3 are outliers.
The data set for the new location has no outliers, To prove it I've calculated the Z-scores for the max and min values:
Min: Z=(8.51-9.4)/0.48= -1.85
Max: Z=(10.12-9.4)/0.48= 1.5
The records for the old location show, as seen in the boxplot, outliers:
To find them I'll start calculating values of Z from the bottom and the top of the list until getting a value Z≥-3 and Z≤3
Bottom:
1) 8,05 ⇒ Z=(8.05-9.8)/0.45= -3.89
2) 8,72 ⇒ Z= (8.72-9.8)/0.45= -2.4
Top
1) 10,55⇒ Z= (10.55-9.8)/0.45= 1.67
m. As mentioned before, the distribution for the new location seems to be more uniform and better distributed than the distribution for the old location. Both distributions are left-skewed, the distribution for the data of the old location is severely affected by the presence of outliers.
I hope this helps!
Answer:
The total of ways the buildings can be painted are = 4 3
2
2
1 = 48 ways.
Step-by-step explanation:
i) color five buildings that are in a row.
ii) any two neighboring buildings must be different colors
iii) the first third and fifth buildings must be different colors.
iv) the number of ways to paint the fist building are 4.
v) the number of ways to paint the second building are 3.
vi) the number of ways to paint the third building are 2.
vii) the number of ways to paint the fourth building are 2.
viii) the number of ways to paint the last building are 1.
ix) therefore the total of ways the buildings can be painted are
= 4 3
2
2
1 = 48 ways.