You have before you two glasses of amontillado wine - one made this year and one made many years ago. you have claimed to be an expert wine connoisseur (now that you are 21 years old, of course) and able to "taste the age" of the wine. if you can determine which one is older, and its age, then you receive an entire cask of this wine! unfortunately, you have a cold and cannot taste any difference between the two. luckily, you have your handy-dandy pocket instrument that is capable of detecting (e. g., counting) tritium decay (tritium is a naturally-occurring, radioactive isotope of hydrogen). tritium undergoes beta-decay with a half-life of 12.26 years. the glass on the left has sixty-four times fewer tritium counts than the glass on the right. page 2 of 2 a. based on the information provided above, explain how the age of the wine in each glass can be determined, and determine the (numerical) age of each wine. b. explain what assumption(s) are necessary in order to be able to compare the tritium counts in the two glasses of wine.

A) One glass has the fresh wine; the one with the higher tritium concentration. To determine the age of the other one, we have to resort to a trick; we notice that 64 is 2 raised to the power 6. Hence, 6 half-lifes need to have passed so that the concentration of tritium becomes 1/64 of the initial concentration (every year the concentration halves, so after n years it is ). Thus, the age of the old wine is approximately 6*12.29 =73.5 years.
b) This is not an exact science. The major assumption that underlies this method is that initially each wine had the same tritium count. This can just be flat out wrong, since one variety of grapes could have been grown in soil that is more rich in tritium. Secondly, we need to be sure that the wine was not stored near tritum etc. It is also important to standardize for wine size; double a quantity of wine would give off double the titrium count, even if it were the same age. One can make further comments like that, but the basis of radiochronology (timing with radioactivity) is to know whether the two samples (one of known age, one of unknown) had the same concentration of radioactive materials initially or what that ratio of concentrations was; in this case we needed to assume that they were the same to do our calculation.

$527.7.if the following was the question  $527.7 is the answer.students who are trying to raise funds have an agreement with a local pizza chain. the chain has agreed to sell them pizzas at a discount, which the students can then resell to families in the local community for a profit. it is expected that of the 500 families in the community, at most 70% will buy pizza. based on a survey of their personal preferences, the students believe that they should order no more than 120 cheese pizzas, no more than 150 pepperoni pizzas, and no more than 100 vegetarian pizzas. they also want to make sure that at least 20% of the total pizzas are cheese and at least 50% of the pizzas are pepperoni. they make a profit of $1.45, $1.75, and $1.98, respectively, for each cheese, pepperoni, and vegetarian pizza they resell. how many pizzas of each type should they buy?

Can you choose anything?