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Maslowich
3 years ago
9

Choose the correct answer. If one balloon is larger than another, it has a greater _________________.

Mathematics
2 answers:
arsen [322]3 years ago
6 0
C. Volume.
The balloon with the larger volume of air will be biger than the balloon with less volume
Nimfa-mama [501]3 years ago
6 0
The correct answer is volume
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Evaluate 1/6 + 2/3 PLEASEEEEEE HELP
mart [117]

2/3 = 4/6.

1/6 + 4/6 = 5/6.

The answer is 5/6.

7 0
3 years ago
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What is the theoretical probability of Eric picking a black marble next time based on Eric's results what is the experimental pr
o-na [289]
The answer is a! i dont promise
5 0
3 years ago
A b c or d. use pic to solve
Neporo4naja [7]

Answer:

In Heather's solution to the problem, she wrote and solved an equation.

Her work is:

Step 1: 1.08(x +9.01 +0.98 +5.01) = 21.87

Solving like terms

1.08(x+15}=21.87

dividing both side by 1.08

we get

x+15=21.87/1.08

x+15=20.25

subtracting both side by 13.9198

x=20.25-15

x=5.25

According to correction

;

<u>Option 1st</u>

<u>yes,it is correct</u>

5 0
2 years ago
The data sets show the years of the coins in two collections. Derek's collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910
KATRIN_1 [288]

Answer:

Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

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

Mean is given by:

(1950+1952+ 1908+1902+1955+1954+1901+1910)/8

=1929

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

from the mean of absolute deviation gives the MAD of the data i.e.

(21+23+21+27+26+25+28+`9)/8

23.75

 

:arrange the given data to get the range and median

   1901   1902    1908   1910    1950  1952    1954   1955

The minimum value is: 1901

Maximum value is: 1955

Range is: Maximum value-minimum value

         Range=1955-1901

Range= 54

median is (1910+1950)/2

1930

   the lower set of data=

  1901   1902    1908   1910

first quartile becomes

1902+1908)/2

Q1=1905

and upper set of data is:

1950  1952    1954   1955

we find the median of the  upper quartile or third quartile is:

1952+1954)/2=1953

Q3-Q1=1953-1905=

IQR=48

 

Paul's collection:

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

Mean is given by:

1929+1935+ 1928+ 1930+ 1925+ 1932+1933+1920)/8

1929

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=0+6+1+1+4+3+4+9/8

28/8

3.5

arrange the data in ascending order we get:

1920   1925   1928   1929   1930   1932   1933   1935  

Minimum value= 1920

Maximum value= 1935

Range=  15 (  1935-1920=15 )

The median is between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920   1925   1928   1929  

the first quartile or upper quartile is

1925+1928/2

1926.5

and the upper set of data is:

1930   1932   1933   1935  

We have

1932+1933)/2

1932.5

IQR is calculated as:

Q3-Q1

6

7 0
3 years ago
Find the area of the shaded region round to the nearest tenth
Alenkinab [10]

Answer: 818.4

Step-by-step explanation:

Using the area of a sector formula, we get the area of the entire sector is (\pi)(27.8^{2})\left(\frac{150}{360} \right).

Using the formula A=\frac{1}{2} ab \sin C, we get the area of the triangle is \frac{1}{2}(27.8)(27.8)(\sin 150^{\circ}).

So, the area of the shaded sector is (\pi)(27.8^{2})\left(\frac{150}{360} \right)-\frac{1}{2}(27.8)(27.8)(\sin 150^{\circ}) \approx \boxed{818.4}

6 0
2 years ago
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