sat-dog

Choice B is correct. It’s given that t minutes after an initial observation, the 40,000 2 79t0 number of bacteria in a population is ^ h . This expression consists of the 40,000 279 t 0 initial number of bacteria, , multiplied by the expression . The time, in minutes, it takes for the number of bacteria to double is the increase in the value 279t0 of t that causes the expression to double. Since the base is 2, the expression 279t0 will double when the exponent increases by 1. Since the exponent of this 790 expression is 79t0, the exponent will increase by 1 when t increases by . Therefore, the time, in minutes, it takes for the number of bacteria in the 790 population to double is . Alternate approach: The initial number of bacteria in the population can be found 0 0 =40,000 2 790 by substituting 0 for t in the given function. This yields f^ h ^ h , or 0 =40,000 f^ h . Therefore, the initial number of bacteria present in the population is 40,000 =80,000 , so the bacteria population will have doubled when f^th . 35 SAT PRACTICE TEST #7 ANSWER EXPLANATIONS SAT ANSWER EXPLANATIONS n MATH: MODULE 1 80,000 80,000=40,000 2 79t0 Substituting for f^th in the given function yields ^ h . 40,000 2=279t0 21 =279t0 Dividing both sides of this equation by yields , or . It follows that 1= 79t0. Multiplying both sides of this equation by 790 yields 790= t. Therefore, the time, in minutes, it takes for the number of bacteria in the 790 population to double is . Choice A is incorrect. This is the base of the exponent, not the time it takes for the number of bacteria in the population to double. Choice C is incorrect. This is the number of minutes it takes for the population to double twice. Choice D is incorrect. This is the number of bacteria that are initially observed, not the time it takes for the number of bacteria in the population to double.