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pshichka [43]
3 years ago
5

Josh wants to determine the number of loaves of bread he can bake using 13 1/3 cups of flour he has. If each loaf requires 3 4/5

cups of flour, how many loaves can josh bake?
Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
4 0

Answer:

4 loaves of bread.....

Sorry if i get it wrong i really tried :(

If it get it right comment down below please! I really want to know!

zhuklara [117]3 years ago
3 0

Answer:

B. 4 loaves of bread

Step-by-step explanation:

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The table below represents the displacement of a horse from its barn as a function of time:
katovenus [111]

A. The y-intercept (b) of a linear equation is obtained when x = 0. Therefore from the given table,

y - intercept = 8

Since at time zero the displacement is 8 ft, this means that the horse was already outside the barn initially.

B. The average rate of change of the function represents the slope of the linear equation (m). This can be calculated using the formula:

average rate of change = m = (y2 – y1) / (x2 – x1)

m = (158 – 58) / (3 – 1)

m = 50

<span>C.  Since we have determine the y-intercept and the slope, we can formulate the linear equation:</span>

y = m x + b

y = 50 x + 8

The domain is the value of x. When y = 508, x is equivalent to

508 = 50 x + 8

<span>x = 10 hrs</span>

6 0
3 years ago
Find the discriminant of the following equation.<br> 4x2 + 16x + 16 = 0 ...?
kenny6666 [7]

Discriminant = b^2 - 4ac

a = 4

b = 16

c = 16

16^2 - 4(4 x 16)

256 - (4x64)

256 - 256 = 0

Therefore it only has one root

7 0
3 years ago
Which of the following rules could represent the function shown in the table?
kvasek [131]
ANSWER

The rule is given by the relation,

y = 2x + 1


EXPLANATION

We need to check and see if there is a constant difference between the y-values.


1 -  - 1 = 2 = 3 - 1


We can see that, there is a constant difference of 2.

This means that the table represents a linear relationship.


Let the rule be of the form,
y = mx + c


Then the points in the table should satisfy the above rule.


So let us plug in


(0,1)


This implies that,


1 =m (0) + c


1 = 0 + c


c = 1



Our rule now becomes,


y = mx + 1 -  - (1)


We again plug in another point say, (-1,-1) in to equation (1) to get,



- 1 = m( - 1)   + 1

we solve for m now to obtain,

- m=-1-1


- m =  - 2


m = 2
We now substitute back in to equation (1) to get

y = 2x  + 1
5 0
3 years ago
Read 2 more answers
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
Please help algebra test tomorrow
user100 [1]
Dont open the links
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