Mathematics: 10 Find the indicated measure in CABCD.

10 Find the indicated measure in CABCD., №15167780, 05.04.2023 11:40

Mathematics

Circles and segments create the many relationships addressed in the theorems throughout this unit. use those theorems and the information given for the following diagram to find each indicated measure. show all your work. note that the importance of justifications for your answers is equal to or greater than the importance of simply the answers in the earning of points. b) find ki if lm =5. c) find ep if pl = 6, pf = 4, and op = 3. e) find cj if ck = 6. f) find fh if af = 10 and ao = 12. g) find ae if af = 10, al = 16 and ah = 14.

Step-by-step answer

P Answered by Specialist

5

In your questions where the segment and circles create many relationship addressed in the theorem throughout this unit. So the following are the answers:
b.KI=5
c. EP = 2
e. CJ = 6
f.FH=14.4
g.AE=8.75

I hope you are satisfied with my answer

Mathematics

Circles and segments create the many relationships addressed in the theorems throughout this unit. use those theorems and the information given for the following diagram to find each indicated measure. show all your work. note that the importance of justifications for your answers is equal to or greater than the importance of simply the answers in the earning of points. b) find ki if lm =5. c) find ep if pl = 6, pf = 4, and op = 3. e) find cj if ck = 6. f) find fh if af = 10 and ao = 12. g) find ae if af = 10, al = 16 and ah = 14.

Step-by-step answer

P Answered by Master

5

In your questions where the segment and circles create many relationship addressed in the theorem throughout this unit. So the following are the answers:
b.KI=5
c. EP = 2
e. CJ = 6
f.FH=14.4
g.AE=8.75

I hope you are satisfied with my answer

Mathematics

Use the law of sines. find the indicated measure to the nearest tenth. in triangle rst, m∠r =78, m∠t =39, and ts with line on top = 19 in. find rs. a. 12.2 in c. 15.8 in b. 10.7 in d. 20.7 in

Step-by-step answer

P Answered by PhD

2

RS = 12.2 in ⇒ answer (a)

Step-by-step explanation:

∵ m∠R = 78° , m∠T = 39° , TS = 19

∵ By using sin Rule:

∵ ST/sin∠R = RS/sin∠T

∴ 19/sin78 = RS/sin39

∴ RS = (19 × sin39) ÷ sin 78 = 12.2 in ⇒ answer (a)

Mathematics

Use the law of sines. find the indicated measure to the nearest tenth. in triangle rst, m∠r =78, m∠t =39, and ts with line on top = 19 in. find rs. a. 12.2 in c. 15.8 in b. 10.7 in d. 20.7 in

Step-by-step answer

P Answered by PhD

2

RS = 12.2 in ⇒ answer (a)

Step-by-step explanation:

∵ m∠R = 78° , m∠T = 39° , TS = 19

∵ By using sin Rule:

∵ ST/sin∠R = RS/sin∠T

∴ 19/sin78 = RS/sin39

∴ RS = (19 × sin39) ÷ sin 78 = 12.2 in ⇒ answer (a)

Mathematics

Plez i need this for homework now the data sets show the years of the coins in two collections. derek s collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910 paul s collection: 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920 find the indicated measures of center and the measures of variation for each data set. round your answer to the nearest hundredth, if necessary. derek s collection paul s collection find the mean median range iqr mad for derek s collection and paul s collection

Step-by-step answer

P Answered by PhD

10

for Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

for Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

Derek's collection:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

$Mean=\dfrac{1950+1952+1908+1902+1955+1954+1901+1910}{8}\\\\\\Mean=\dfrac{15432}{8}\\\\\\Mean=1929$

Now absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

and the mean of these absolute deviation gives the MAD of the data i.e.

$MAD=\dfrac{21+23+21+27+26+25+28+19}{8}\\\\\\MAD=23.75$

Now, on arranging the data in increasing order we get:

1901 1902 1908 1910 1950 1952 1954 1955

The least value is: 1901

Maximum value is: 1955

Range is: Maximum value-Least value

Range=1955-1901

Range= 54

Also, the median lie between 1910 and 1950 and is calculated as:

$Median=\dfrac{1910+1950}{2}\\\\\\Median=\dfrac{3860}{2}\\\\\\Median=1930$

Also, the lower set of data is:

1901 1902 1908 1910

and the median of lower set of data also known as first quartile or lower quartile is:

$Q_1=\dfrac{1902+1908}{2}\\\\\\Q_1=\dfrac{3810}{2}\\\\\\Q_1=1905$

and upper set of data is:

1950 1952 1954 1955

and the median of upper set of data i.e. upper quartile or third quartile is:

$Q_3=\dfrac{1952+1954}{2}\\\\\\Q_3=\dfrac{3906}{2}\\\\\\Q_3=1953$

Hence, IQR is calculated as:

$IQR=Q_3-Q_1\\\\\\i.e.\\\\\\IQR=1953-1905\\\\\\IQR=48$

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

$Mean=\dfrac{1929+1935+1928+1930+1925+1932+1933+1920}{8}\\\\\\Mean=1929$

Now absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

$MAD=\dfrac{6+1+1+4+3+4+9}{8}\\\\\\MAD=\dfrac{28}{8}\\\\\\MAD=3.5$

Now, on arranging the data in increasing order we get:

1920 1925 1928 1929 1930 1932 1933 1935

Least value= 1920

Maximum value= 1935

Range= 15 ( Since, 1935-1920=15 )

The median lie between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920 1925 1928 1929

and median of lower set of data is the first quartile or upper quartile and is calculated as:

$Q_1=\dfrac{1925+1928}{2}\\\\\\Q_1=1926.5$

and the upper set of data is:

1930 1932 1933 1935

Hence, we get:

$Q_3=\dfrac{1932+1933}{2}\\\\\\Q_3=1932.5$

Hence, IQR is calculated as:

$IQR=Q_3-Q_1\\\\\\IQR=1932.5-1926.5\\\\\\IQR=6$

Mathematics

Derek s collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910 paul s collection: 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920 find the indicated measures of center and the measures of variation for each data set. round your answer to the nearest hundredth, if necessary. find the mean, median, range, iqr, mad for derek s collection and paul s collection.

Step-by-step answer

P Answered by Specialist

3

Derek:

Mean:1929

Median: 1930

range: 54

IQR: 48

MAD: 23.75

Paul:

Mean: 1929

Median: 1929.5

range: 15

IQR: 6

MAD: 3.5

Step-by-step explanation:

Mathematics

The tide is the regular rising or falling of the ocean’s surface. This is due in large part to the gravitational forces of the moon. The following table represents water level of the tide off the coast of Kings Point, N.Y., for a 24 hour period, February 9th through February 10th of 2009. Hour Measurement in feet above the average low tide 0 -0.98 2 5.83 4 7.78 6 7.34 8 2.33 10 -0.47 12 -0.21 14 5.32 16 8.83 18 8.31 20 3.80 22 -0.28 24 0.53 Plot the data points given in the table using hour as the independent variable. Does this data indicate that

Step-by-step answer

P Answered by PhD

5

Option A

Step-by-step explanation:

Points plotted.

See the graph attached.

Correct choice is:

A. Yes, in the scatterplot the wavelike behavior is apparent although not precise
The tide is the regular rising or falling of the ocean’s surface. This is due in large part to the g

The tide is the regular rising or falling of the ocean’s surface. This is due in large part to the g

Mathematics

Plez i need this for homework now the data sets show the years of the coins in two collections. derek s collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910 paul s collection: 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920 find the indicated measures of center and the measures of variation for each data set. round your answer to the nearest hundredth, if necessary. derek s collection paul s collection find the mean median range iqr mad for derek s collection and paul s collection

Step-by-step answer

P Answered by PhD

10

for Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

for Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

Derek's collection:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

$Mean=\dfrac{1950+1952+1908+1902+1955+1954+1901+1910}{8}\\\\\\Mean=\dfrac{15432}{8}\\\\\\Mean=1929$

Now absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

and the mean of these absolute deviation gives the MAD of the data i.e.

$MAD=\dfrac{21+23+21+27+26+25+28+19}{8}\\\\\\MAD=23.75$

Now, on arranging the data in increasing order we get:

1901 1902 1908 1910 1950 1952 1954 1955

The least value is: 1901

Maximum value is: 1955

Range is: Maximum value-Least value

Range=1955-1901

Range= 54

Also, the median lie between 1910 and 1950 and is calculated as:

$Median=\dfrac{1910+1950}{2}\\\\\\Median=\dfrac{3860}{2}\\\\\\Median=1930$

Also, the lower set of data is:

1901 1902 1908 1910

and the median of lower set of data also known as first quartile or lower quartile is:

$Q_1=\dfrac{1902+1908}{2}\\\\\\Q_1=\dfrac{3810}{2}\\\\\\Q_1=1905$

and upper set of data is:

1950 1952 1954 1955

and the median of upper set of data i.e. upper quartile or third quartile is:

$Q_3=\dfrac{1952+1954}{2}\\\\\\Q_3=\dfrac{3906}{2}\\\\\\Q_3=1953$

Hence, IQR is calculated as:

$IQR=Q_3-Q_1\\\\\\i.e.\\\\\\IQR=1953-1905\\\\\\IQR=48$

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

$Mean=\dfrac{1929+1935+1928+1930+1925+1932+1933+1920}{8}\\\\\\Mean=1929$

Now absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

$MAD=\dfrac{6+1+1+4+3+4+9}{8}\\\\\\MAD=\dfrac{28}{8}\\\\\\MAD=3.5$

Now, on arranging the data in increasing order we get:

1920 1925 1928 1929 1930 1932 1933 1935

Least value= 1920

Maximum value= 1935

Range= 15 ( Since, 1935-1920=15 )

The median lie between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920 1925 1928 1929

and median of lower set of data is the first quartile or upper quartile and is calculated as:

$Q_1=\dfrac{1925+1928}{2}\\\\\\Q_1=1926.5$

and the upper set of data is:

1930 1932 1933 1935

Hence, we get:

$Q_3=\dfrac{1932+1933}{2}\\\\\\Q_3=1932.5$

Hence, IQR is calculated as:

$IQR=Q_3-Q_1\\\\\\IQR=1932.5-1926.5\\\\\\IQR=6$

Mathematics

Derek s collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910 paul s collection: 1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920 find the indicated measures of center and the measures of variation for each data set. round your answer to the nearest hundredth, if necessary. find the mean, median, range, iqr, mad for derek s collection and paul s collection.

Step-by-step answer

P Answered by Specialist

3

Derek:

Mean:1929

Median: 1930

range: 54

IQR: 48

MAD: 23.75

Paul:

Mean: 1929

Median: 1929.5

range: 15

IQR: 6

MAD: 3.5

Step-by-step explanation:

Chemistry

The electrical conductivities of the following 0.100 m solutions were measured in an apparatus that contained a light bulb as the indicator of conductivity. rank the solutions in order of decreasing intensity (brightest to dimmest) of the light bulb. rank from brightest to dimmest bulb. to rank items as equivalent, overlap them. hf, ch₃oh, ki, al(no₃)₃.

Step-by-step answer

P Answered by PhD

8

Al(NO₃)₃ > KI > HF > CH₃OH

Explanation:

The electrical conductivities of the solutions will depend on the concentration of ions in solution.

Al(NO₃)₃ solution contains 0.1 M of Al³⁺ ions and 0.3 M of NO₃⁻ ions

KI solution contains 0.1 M of K⁺ ions and 0.1 M of I⁻ ions

HF solution contains less than 0.1 M of H⁺ ions and less then 0.1 M of F⁻ ions, because the HF acid will not dissociate completely

CH₃OH practically it does not dissociate, so in the solution will not be electrical conductive (comparative with the other solutions)

The solutions in order of decreasing intensity of the bulb are ranked as following:

Al(NO₃)₃ > KI > HF > CH₃OH

10
Find the indicated measure in CABCD.

Faq

Try asking the Studen AI a question.