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Bingel [31]
3 years ago
12

Find the mean median mode and range 2,5,9,4,3

Mathematics
2 answers:
snow_lady [41]3 years ago
3 0
Mean: 4.6 (basically the average
Median: 4 (the number in the middle
Mode: there isn’t one (it’s the number repeated the most)
Range: 7 (largest and smallest number subtracted
timurjin [86]3 years ago
3 0

list the following data set from least to greatest:

2, 3, 4, 5, 9

range: highest value - lowest value = answer

Range: 9 - 2 = 7

Mode: there is no mode in this data set, (mode represents how many times a number is repeated)

(for example: 2, 5, 5, 6, 14, the following mode in this data set is 5)

Mode: none

mean:

(just add up all the numbers and divide it by 5 since there are 5 numbers in the following data set, or the listed numbers.)

2 + 5 + 9 + 4 + 3 = 23/5 = 4.6

mean: 4.6

range: 7

Mode: None, there are no mode in the data set or listed numbers.

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If I'm correct 40.5 meters

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3 years ago
The equation y = 0.035x shows the relationship between the grams and ounces.
scoundrel [369]
The equation y = 0.035x where x is in grams, and y is ounces, then 22 grams would be 0.035 x 22 = 0.77 ounces
4 0
3 years ago
Please any help is necessary
Ugo [173]

Answer:

Option D. 50 ft

Step-by-step explanation:

we know that

The area of the figure is equal to the area of three rectangles

so

A=18x+(48-36)(32-11)+(18)(32)

A=18x+(12)(21)+(18)(32)\\A=18x+828

Remember that the area is given

A= 1,728\ ft^2

so

18x+828=1,728

solve for x

18x=1,728-828\\18x=900\\x=50\ ft

3 0
3 years ago
I need some help with this question
yarga [219]

Answer:

-7/15

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-4/5 translates to -12/15 and 1/3 translates to 5/15.

When you add -12/15+5/15 you get

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5 0
3 years ago
Suppose a certain manufacturing company produces connecting rods for 4- and 6-cylinder automobile engines using the same product
gregori [183]

Answer:

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder  to  W_4 or  0 is  

     x_4 \le W_4 *  s_4

    x_4 \le 5000 *  s_4

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

     x_6 \le W_6 *  s_6

     x_6 \le 8,000 *  s_6

Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

     s_4 + s_6 = 1

The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

Step-by-step explanation:

The cost for the four cylinder production line is  C_4 =  \$2,100

The cost for the six cylinder production line is  C_6 = \$3,500

The manufacturing cost for each four cylinder is  M_4= \$13

 The manufacturing cost for each six cylinder is M_6= \$16

  The weekly production capacity for 4 cylinder connecting rod is W_4 = 5,000

   The weekly production capacity for 6 cylinder connecting rod is W_6 = 8,000

Generally the constraint that sets next week are shown below

Generally the constrain that sets next week maximum production of connecting rod for 4 cylinder  to  W_4 or  0 is  

     x_4 \le W_4 *  s_4

    x_4 \le 5000 *  s_4

Generally the constrain that sets next week maximum production of connecting rod for 6 cylinder  to  W_6 or  0  is  

     x_6 \le W_6 *  s_6

     x_6 \le 8,000 *  s_6

Generally the constrain that limits the production of connecting rods  for both 4 cylinder and 6 cylinders  is

     x_4 \le W_4 *  s_6

=>   x_4 \le 5000 *  s_6

     x_4 \le W_6 *  s_4

=>    x_4 \le 8000 *  s_4

     s_4 + s_6 = 1

The minimum cost of production for next week is  

   U  =  M_4 *  x_4 + M_6 * x_6 + C_4 * s_4 + C_6 * s_6

=>  U  =  13x_4 + 16x_6 + 2000 s_4 + 3500 s_6

3 0
3 years ago
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