A colony of bacteria decreases in population by 10 percent each hour. The initial bacteria culture contained 250 spores. What exponential equation can be used to find the terms in the geometric sequence?(1 point)

Question

Responses

b(x)=250−10(x−1)
𝑏
(
𝑥
)
=
250
−
10
(
𝑥
−
1
)
b left parenthesis x right parenthesis equals 250 minus 10 left parenthesis x minus 1 right parenthesis

b(x)=250(0.10)x−1
𝑏
(
𝑥
)
=
250
(
0.10
)
𝑥
−
1
b left parenthesis x right parenthesis equals 250 left parenthesis 0.10 right parenthesis superscript x minus 1 baseline

b(x)=250(0.99)x−1
𝑏
(
𝑥
)
=
250
(
0.99
)
𝑥
−
1
b left parenthesis x right parenthesis equals 250 left parenthesis 0.99 right parenthesis superscript x minus 1 baseline

b(x)=250(0.90)x−1

Answer 1

To find the exponential equation that represents the decrease in the bacterial population, we start with the information that the initial population is 250 and that it decreases by 10 percent each hour.

When we decrease a quantity by 10%, we are left with 90% of the original quantity. In terms of a decimal, 90% is represented as 0.90. Therefore, the population at any hour can be modeled using exponential decay.

The general form of the exponential decay function can be expressed as:

\[ b(x) = b_0 \cdot (r)^x \]

where:

• \( b_0 \) is the initial amount (250 spores),
• \( r \) is the decay factor (0.90), and
• \( x \) is the time in hours.

So, substituting the given values we have:

\[ b(x) = 250 \cdot (0.90)^x \]

None of the provided responses match this equation exactly, but based on the options presented, the correct option should be:

b(x) = 250(0.90)^(x−1)

This implies you've started the count from hour 1, rather than hour 0. Therefore, "x-1" accounts for the delay in considering the initial state, which can sometimes be common in certain contexts.

So the best fit would be:

b(x) = 250(0.90)^(x−1).