Vera examined a random sample of deodorants distributed by her company. She found the following probabilities:

The formula to calculate the difference between two standard deviations of populations n1 and n2 is:
sigma (difference)=√(sigma1/n1 + sigma2/n2). For this scenario:

sigma(d)= √(49/100 + 36/50)

Thus, the calculated standard deviation of the difference equals 1.1.

90kg of sand was divided into 3 bags, constituting a total of 2+3+7=12 sections.
90kg/12=7.5kg
The smaller bag comprises 2 sections, hence it weighs 7.5kg*2=15kg
The medium-sized bag consists of 3 sections, leading to a total weight of 7.5kg*3=22.5kg
The larger bag encompasses 7 sections, resulting in a weight of 7.5kg*7=52.5kg
Thus, the ratio 2:3:7 translates to 15kg:22.5kg:52.5kg
Verifying, 15kg+22.5kg+52.5kg equals 90kg

Answer:

We will mix 2.4 mL of water with 0.6 mL of the coloring solution.

The total solution volume amounts to 2.4 + 0.6, which equals 3 mL.

Therefore,

0.6 mL of coloring solution now contains 2.4 mL of water.

Alternatively,

1 mL of this coloring solution comprises 4 mL of water.

Thus,

We should prepare 3 mL of the coloring solution with the resultant water.

This 3 mL mixture will encapsulate 0.6 mL of the coloring solution.

Step-by-step explanation:

As stated:

The formula requires 0.6 mL of a coloring solution.

Using a 10-mL graduated cylinder that is divided from 2 to 10 mL in 1-mL increments:

We aim to apply the aliquot method to accurately measure 0.6 mL of the coloring solution.

Given that the graduated cylinder measures from 2mL to 10mL only, we'll blend 2.4 mL of water with 0.6 mL of coloring solution.

Thus, the total solution volume is 2.4 + 0.6, resulting in 3 mL.

Consequently,

This implies that 0.6 mL of coloring solution now carries 2.4 mL of water.

Additionally,

1 mL of coloring solution encapsulates 4 mL of water.

Hence,

We will quantify 3 mL of the final solution, which contains 0.6 mL of coloring solution.