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4 years ago

How can you find approximations (estimations) of square roots?

2 answers:

6 0

By finding these tables, you can use them to locate the square root of numbers 1 to 100. Choose or make a method to find an approximate square root of each number. Then Round your answer to the nearest tenth.

Luda [366]4 years ago

3 0

To find approximations of square roots, you should find the nearest prefect square. For example if your looking to approximate the square rot of 10, the closest perfect square is 9 and the square root of that is three. Thus, you can say that the square root is about 3. Hope this helps and please mark brainiest. :D

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Which is equivalent to 1.04?

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Step-by-step explanation:

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3 years ago

Find the global maximum and global minimum values (if they exist) of x 2 + y 2 in the region x + y = 1. If there is no global ma

Grace [21]

Given that $x+y=1$ , we have $y=1-x$ , so that

$f(x,y)=x^2+y^2\iff g(x)=x^2+(1-x)^2$

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Take the second derivative and evaluate it at the critical point:

$g''(x)=4>0$

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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

7 0

3 years ago