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Andrew [12]
3 years ago
5

A diet to contain at least 2400 units of vitamins, 1800 units of minerals, and 1200 calories two foods , Food A and Food B are t

o be purchased. Each unit of Food A provides 50 units of vitamins, 30 units of minerals and 10 calories. Each unit of food B provides 20 units of vitamins, 20 units of minerals and 40 calories. If Food A costs $2 per unit and Food B costs $1 per unit, how many units of each food should be purchased to keep costs at a minimum?
Computers and Technology
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

30 units of Food A and 45 units of Food B are to be purchased to keep costs at the minimum $105.

Explanation:

X = Amount of food A

Y = Amount of food B

Z= 2X+Y..... minimum cost equation

50X + 20Y > 2400 .................Vitamins .......(1)

30X + 20Y > 1800 ...................Minerals.......(2)

10X + 40Y > 1200 .................Calories ..........(3)

X  > 0

y  > 0

X=30  and Y = 45

Z= 2(30) + 45 = $105

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Write a Scheme function called "sum" which takes an input function func and a nonnegative number n and outputs the value
solong [7]

Answer:

function sum(number) {

      if (number == 1) {

             return 1;

      }

      return number + sum(number -1);

}

Explanation:

This is a recursive function, it means that is a function that calls itself for example: if you call the function with sum(5) the process is :

sum(5)

   |______ 5 + sum(4)

                           |_______ 4 + sum(3)

                                                       |______ 3 + sum(2)

                                                                                   |_____2 + sum(1)

                                                                                                        |_____ 1

                                                                                                 

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7 0
3 years ago
Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, a
ANTONII [103]

Answer:

a) the average CPI for machine M1 = 1.6

the average CPI for machine M2 = 2.5

b) M1 implementation is faster.

c) the clock cycles required for both processors.52.6*10^6.

Explanation:

(a)

The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4

= 1.6

The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4

= 2.5

(b)

The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6

Given 80MHz = 80 * 10^6

The average MIPS ratings for M1 = 80 x 10^6  / 1.6 x 10^6

= 50

Given 100MHz = 100 * 10^6

The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6

= 40

c)

Machine M2 has a smaller MIPS rating

Changing instruction set A from 2 to 1

The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)

and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.

7 0
3 years ago
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