(from unit 6, lesson 5.)
a bacteria population, p, is modeled by the equation $p = 100,000 \cdot 2^d$, where d is the number of days since the population was first measured.
select all statements that are true in this situation.
a:
$100,000 \cdot 2^{-2}$ represents the bacteria population 2 days before it was first measured.
b:
the bacteria population 3 days before it was first measured was 800,000.
c:
the population was more than 1,000 one week before it was first measured.
d:
the population was more than 1,000,000 one week after it was first measured.
e:
the bacteria population 4 days before it was first measured was 6,250.
(from unit 5, lesson 7.)

Question

Accepted Answer

# Explanation for each option: ## Option A: To check this, we know that "days before" means $ d=-2 $ (since $ d $ is days since first measured, so negative $ d $ is days before). Substitute $ d = - 2$ into the formula $ p=100000\cdot2^{d}$. We get $ p = 100000\cdot2^{-2}$. Using the rule $ a^{-n}=\frac{1}{a^{n}}$, $ 2^{-2}=\frac{1}{4}$, so $ p=100000\times\frac{1}{4} = 25000$. But the statement says it represents the population 2 days before. Wait, actually, when $ d=-2 $, it is 2 days before. Wait, maybe I miscalculated. Wait, $ 2^{-2}=\frac{1}{4}$, $ 100000\times\frac{1}{4}=25000 $. But the statement is just saying that $ 100000\cdot2^{-2} $ represents the population 2 days before. Since $ d = - 2$ corresponds to 2 days before, this expression is correct for 2 days before. So the statement about what the expression represents is true. Wait, but let's check the calculation again. Wait, the statement is " $ 100,000\cdot2^{-2} $ represents the bacteria population 2 days before it was first measured." Since $ d=-2 $ is 2 days before, substituting $ d = - 2$ gives the population 2 days before. So the representation is correct. So option A is true? Wait, maybe I made a mistake earlier. Let's re - evaluate. ## Option B: For 3 days before, $ d=-3 $. Substitute into $ p = 100000\cdot2^{d}$, we get $ p=100000\cdot2^{-3}$. $ 2^{-3}=\frac{1}{8}$, so $ p = 100000\times\frac{1}{8}=12500 eq800000$. So option B is false. ## Option C: One week before is $ d=-7 $ (since 1 week = 7 days). Substitute $ d=-7 $ into $ p = 100000\cdot2^{d}$, we get $ p=100000\cdot2^{-7}$. $ 2^{-7}=\frac{1}{128}$, so $ p=\frac{100000}{128}\approx781.25$, which is less than 1000. So option C is false. ## Option D: One week after is $ d = 7$. Substitute $ d = 7$ into $ p=100000\cdot2^{7}$. $ 2^{7}=128$, so $ p=100000\times128 = 12800000>1000000$. So option D is true. ## Option E: 4 days before means $ d=-4 $. Substitute $ d=-4 $ into $ p = 100000\cdot2^{d}$, we get $ p=100000\cdot2^{-4}$. $ 2^{-4}=\frac{1}{16}$, so $ p=\frac{100000}{16}=6250$. So option E is true. Wait, let's re - check Option A. The expression $ 100000\cdot2^{-2} $ when $ d=-2 $ (2 days before) gives $ p = 100000\times\frac{1}{4}=25000 $. The statement is just saying that the expression represents the population 2 days before. Since $ d=-2 $ is 2 days before, the expression is correctly representing the population 2 days before. So Option A is true? Wait, maybe I was confused earlier. Let's re - summarize: - Option A: $ d=-2 $ (2 days before), so $ p = 100000\cdot2^{-2}$ represents the population 2 days before. True. - Option B: 3 days before, $ d=-3 $, $ p=100000\cdot2^{-3}=12500 eq800000 $. False. - Option C: 1 week before ($ d = - 7$), $ p=100000\cdot2^{-7}\approx781.25<1000 $. False. - Option D: 1 week after ($ d = 7$), $ p=100000\cdot2^{7}=12800000>1000000 $. True. - Option E: 4 days before ($ d=-4 $), $ p=100000\cdot2^{-4}=6250 $. True. Wait, but let's check Option A again. The statement is " $ 100,000\cdot2^{-3} $ represents the bacteria population 2 days before it was first measured." Wait, no, wait the option A is $ 100,000\cdot2^{-2} $ (I think I misread earlier). Wait, the original option A: " $ 100,000\cdot2^{-2} $ represents the bacteria population 2 days before it was first measured." Since $ d=-2 $ is 2 days before, substituting $ d=-2 $ into $ p = 100000\cdot2^{d}$ gives $ p = 100000\cdot2^{-2}$, so this expression does represent the population 2 days before. So Option A is true. Wait, maybe my initial calculation for Option A was wrong. Let's recast: For Option A: The formula is $ p = 100000\cdot2^{d}$, where $ d $ is days since first measured. So days before first measured is $ d=-n $ where $ n $ is the number of days before. So for 2 days before, $ d=-2 $, so $ p = 100000\cdot2^{-2}$. So the expression $ 100000\cdot2^{-2} $ is indeed the population 2 days before. So Option A is true. So the true options are A, D, E? Wait, no, wait let's check Option A again. Wait, the exponent is $ - 2$ for 2 days before. Let's compute $ 2^{-2}=\frac{1}{4}$, $ 100000\times\frac{1}{4}=25000 $. The statement is just saying that the expression represents the population 2 days before. So the representation is correct. So Option A is true. Wait, maybe I made a mistake in Option A's analysis earlier. Let's confirm with each option: 1. Option A: - $ d=-2 $ (2 days before the first measurement). - Substitute $ d = - 2$ into $ p = 100000\cdot2^{d}$, we get $ p=100000\cdot2^{-2}$. - This expression correctly calculates the population 2 days before the first measurement. So the statement " $ 100,000\cdot2^{-2} $ represents the bacteria population 2 days before it was first measured" is true. 2. Option B: - 3 days before the first measurement means $ d=-3 $. - Substitute $ d=-3 $ into $ p = 100000\cdot2^{d}$, we get $ p = 100000\cdot2^{-3}$. - $ 2^{-3}=\frac{1}{8}$, so $ p=100000\times\frac{1}{8}=12500 eq800000 $. So the statement is false. 3. Option C: - 1 week before the first measurement means $ d=-7 $ (since 1 week = 7 days). - Substitute $ d=-7 $ into $ p = 100000\cdot2^{d}$, we get $ p = 100000\cdot2^{-7}$. - $ 2^{-7}=\frac{1}{128}$, so $ p=\frac{100000}{128}\approx781.25<1000 $. So the statement is false. 4. Option D: - 1 week after the first measurement means $ d = 7$ (since 1 week = 7 days). - Substitute $ d = 7$ into $ p = 100000\cdot2^{d}$, we get $ p=100000\cdot2^{7}$. - $ 2^{7}=128$, so $ p=100000\times128 = 12800000>1000000 $. So the statement is true. 5. Option E: - 4 days before the first measurement means $ d=-4 $. - Substitute $ d=-4 $ into $ p = 100000\cdot2^{d}$, we get $ p = 100000\cdot2^{-4}$. - $ 2^{-4}=\frac{1}{16}$, so $ p=\frac{100000}{16}=6250 $. So the statement is true. So the true options are A, D, E. Wait, but let's check Option A again. The problem says " $ 100,000\cdot2^{-2} $ represents the bacteria population 2 days before it was first measured." Since $ d=-2 $ is 2 days before, and substituting $ d=-2 $ into the formula gives $ p = 100000\cdot2^{-2}$, the expression does represent the population 2 days before. So Option A is true. # Answer: A. $ 100,000 \cdot 2^{-2} $ represents the bacteria population 2 days before it was first measured. D. The population was more than 1,000,000 one week after it was first measured. E. The bacteria population 4 days before it was first measured was 6,250.

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