Animals in an experiment are to be kept under a strict diet. Each animal should receive 50 grams of protein and 12 grams of fat. The laboratory technician is able to purchase two food mixes: Mix A has 10% protein and 8% fat; mix B has 50% protein and 5% fat. How many grams of each mix should be used to obtain the right diet for one animal?

100 grams of A mixed with 80 grams of B will provide the daily diet for a single animal.

Step-by-step explanation:

Step 1

The requirement for each animal is given as 50g of protein  and 12g of fat

let mix A  be represent as   containing  0.10A protein and 0.08A fat

and mix B be represented as containing 0.5B protein and 0.05B  fat

Therefore for Protein , we have   0.10A + 0.5B= 50 Equation 1

And for fat , we have   0.08A + 0.05B= 12Equation 2

Step 2-- Solving

0.10A + 0.5B= 50 Equation 1

0.08A + 0.05B= 12Equation 2

multiplying equation 1 by 10 and equation 2 by 100 we have

1A+ 5B=500Equation 3

8A+ 5B= 1200Equation 4

Subtracting equation 3 from equation 4

7A= 700

A = 700/7= 100

Puting value of A = 100 IN  EQUATION 3

1(100) + 5B= 500

5B= 500-100

B=400/5= 80

Therefore 100 grams of A mixed with 80 grams of B will provide the daily diet for a single animal.

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05