1 square yard What is the area? 10 square yards 8 square yards 12 square yards

I'll give brainliest to the correct answer. Here are some answer choices:

Delvig [45]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

You’ve bought a cylindrical thermos 8 in. high with a base 6 in. across. How much coffee can you put in the thermos?

lana66690 [7]

Answer:

3.706 liters

Step-by-step explanation:

Hello

assuming that these measures correspond to the internal measures of the thermos,

we can calculate the volume as, the volume of a cylinder, which is given by

$V=\pi r^{2} h$

where r is the radius of the base, and h the height

also

Diameter=2r

r=Diameter/2

Step 1

let

h=8 in

base = diameter = 6 inch

r=diameter/2

r=6/2 inch,r=3 inch

Step 2

Put the values into the equation

$V=\pi r^{2} h\\V=\pi (3in)^{2} (8 in)\\V=\pi *72 \ in^{3} \\\\V=72\pi\ in^{3} \\\\$

Step 3

Now, the unit (cubic inches) is not an usual unit for measuring coffee, let´s convert cubic inches into liters by knowing

1 inch = 2.54 cm

1 litro = 1000 cubic centimeters

$if\ 1\ inch\ = 2.54\ cm\\(1\ inch)^{3} =(2.54\ cm)^{3}\\\\1\ inch^{3}=16.3870\ cm^{3}$

1 cubic inch = 16.3870 cubic centimeters

$1\ cubic\ inch\ = 16.3870\ cubic\ centimeters\\\\\\72\pi =x?\ cubic\ centimeters\\\\x=\frac{72\pi \cubic \ inch\ \*16.3870 }{1\ cubic\ inch} \\\\x=3706.65\ cubic\ centimeters$

Step 4

Finally, convert cubic centimeters into liters $1000\ cubic\ centimeters =1\ liter \\\\\\3706.65\ cubic\ centimeters = x?\ liters\ centimeters\\\\x=\frac{3706.65 \cubic \ centimeters\ \*1 liter}{1000 cubic centimeters} \\\\x=3706.65\ cubic\ centimeters\\x=3.706 Liters$

I can put 3.706 liters in the thermos

Have a good day

4 0

4 years ago

What is the mode of this data? 1.7 8.9 3 5.9 3.2 4.1 6.6 0.7

ExtremeBDS [4]

Answer:

The value of Mode is 2.43

Step-by-step explanation:

To find the mode of the given data first we have to arrange it in a increasing order then find out mean and median of the given data
0.7,1.7,3,3.2,4.1,5.9,6.6,8.9 is in increasing order
For finding the median we need to take the average of 4th and 5th terms because we have the no of terms in the sequence is even not odd so we need to take the average the 4th term=3.2 and the 5th term =4.1
so average =(3.2+4.1)/2=3.65
so the median is equal to 3.65
For mean we have to take the average of the data
so mean= sum of all data /no of data
mean =(0.7+1.7+3+3.2+4.1+5.9+6.6+8.9)/8=4.26
so by using the formula we can get mode
Mode=3×Median-2×Mean
Mode=3×3.65-2×4.26=2.43
∴The value of Mode is given as 2.43

5 0

4 years ago

50 POINTS pleaseee help its URGENT!!!! Ill give brainlist and if there all right i might send u 20 dollars!!! Pleasee help me!!

Sonbull [250]

Answer:

A. 22°

E. 15

I. 39°

F. 71°

C. 25.6

H.12.9

K.55°

Step by Step Explanation:

1. $cos^{-1}(52/56)=21.7867893 or 22$ degrees

2. 22sin43=15

3. $tan^{-1}(34/42) or 38.990999404 or 39 degrees$

4. $sin^{-1}(51/54)=70.81186355 or 71 degrees$

5. 20/(tan38)=25.58883264 or 25.6

6. 20sin40=12.85575219 or 12.9

7. $cos^{-1}(26/45)=54.70560532 or55 degrees$

Hope that helps!:)

8 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

4 years ago