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Varvara68 [4.7K]
3 years ago
5

2. (10.03)

Mathematics
1 answer:
faust18 [17]3 years ago
3 0

This question is incomplete because it was not written properly

Complete Question

A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)

a) 20%

b) 40%

c) 60%

d) 75%

Answer:

d) 75%

Step-by-step explanation:

We would be solving this question using conditional probability.

Let us represent the percentage of those who passed the first quiz as A = 80%

and

Those who passed the first quiz as B = unknown

Those who passed the first and second quiz as A and B = 60%

The formula for conditional probability is given as

P(B|A) = P(A and B) / P(A)

Where,

P(B|A) = the percent of those who passed the first one passed the second

Hence,

P(B|A) = 60/80

= 0.75

In percent form, 0.75 × 100 = 75%

Therefore, from the calculations above, 75% of those who passed the first quiz to also passed the second quiz.

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A population of bacteria is initially 6000. After three hours the population is 3000. If this rate of decay continues, find the
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Answer:

The equation is A=6000e^{-0.23t}  at a rate of -23%.

Step-by-step explanation:

Decay can be represented by the equation A=A_0e^{rt}. We can find the rate at which it decays by using t=3 hours and A=3000. This means A_0=6000 in this context.

A=A_0e^{rt}\\3000=6000e^{r(3)}

0.5=e^{(24.5)r}

After substituting, we divided by 6000 to each side to get 0.5 on the left. Now to solve for r, we will take the natural log of both sides and use log rules to isolate r.

ln 0.5=ln e^{(3)r}\\ln 0.5=3r (ln e)\\\frac{ln0.5}{3} =r

We know lne=1 so we were able to cancel it out and divide both sides by 3.

We solve with a calculator \frac{ln0.5}{3} =r\\-0.23=r

We change -0.23 into a percent by multiplying by 100 to get -23% as the rate.

The equation is A=6000e^{-0.23t}


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