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VladimirAG [237]
2 years ago
13

there are 53 students on a field trip. One chaperone is needed for every  6  students.How many chaperones are needed.

Mathematics
2 answers:
igomit [66]2 years ago
5 0
8 but there will be 5 left over so you would need someone to look after them 5 which means you will need 9
eimsori [14]2 years ago
4 0
First you have to split up the 53 students into groups of  6 .

53/6  =  8  with reminder  5 .

There are 8 groups of 6 students, so we know immediately that we need at least 6 chaperones.

But what to do about the 5 students left over ?

We can either let them go unchaperoned, or else hire one more chaperone to keep an eye on that group even though there are less than 6 students in it.

I think the teachers and parents, and maybe even the students themselves, would all agree that the right thing to do is get one more chaperone for the 5 students.

So all together, they need <em>7 chaperones</em> on the trip.
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Please Help!!<br><br> What should be done 10x ^ 2 - 8x to create a perfect square ?
Pani-rosa [81]

Answer: 10x^2-8x+16

Step-by-step explanation:

Use the formula (b/2)^2

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ans: 10x^2-8x+16

hope this helps

4 0
2 years ago
I need help asap!!!!
kolbaska11 [484]

Answer:

24: x = 31

25: x = 15

Step-by-step explanation:

Remark

24 is supplementary which means that the two angles add to 180o which is on the right hand side of the equation

25 is complementary which means that the two angles add to 90o which is on the left hand side of the equation

Twenty Four

3x + 31 +2x - 6 =  180              Collect the like terms

3x +2x + 31 - 6 = 180               Do the adding and subtracting.

5x + 25 = 180                          Subtract 25 from both sides

5x + 25 - 25 = 180 - 25

5x = 155                                   Divide by 5

x = 155/5

x = 31

Check

3x + 31 = 3*31 + 31

3x + 31 = 93 + 31

3x + 31 = 124

2x - 6 = 2*31 - 6 = 62 - 6 = 56

Total 124 + 56 = 180 as it should.

Twenty Five

Equation

3x+ 4x -  15 = 90

Solution

7x - 15 = 90        Like terms have been collected on the left.

7x = 90 + 15       15 was added to both sides

7x = 105              Divide by 7      

x = 15                       I'll leave the check to you


5 0
3 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
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