The square root of thirty lies between the numbers 5 and 6.
In fact, this problem belongs to the chemistry section. Recall that many other sciences require mathematical calculations. The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.
Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute. Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.
The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g. Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.
9.1 yards because it’s just the answer because i know it’s the answer
There are 8 possible outcomes for a marble being drawn and numbered.
{1,2,3,4,5,6,7,8}
There are 4 possible outcomes for a card being selected from a standard deck.
{ <span>hearts, diamonds, clubs, spades}
So the number of outcomes in the sample space would be 8 x 4 = 32.
In the event "an even number is drawn", there are only 4 possible outcomes for a marble being drawn, {2,4,6,8}, whereas there are still 4 possible outcomes for a suit. So the number of outcomes in the event is 4 x 4 = 16.
</span><span>In the event "a number more than 2 is drawn and a red card is drawn", there are 6 possible outcomes for the marble being drawn, {3,4,5,6,7,8}, whereas there are only two possible suits for a card being selected as red, {heart, diamond}. So the number of outcomes in this event is 6 x 2 = 12.
In the event </span><span>"a number less than 3 is drawn or a club is not drawn", the number drawn could be 1 or 2 whereas a spade/heart/diamond could be selected. So the number of outcomes is 2 x 3 = 6.</span><span>
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