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mr Goodwill [35]
3 years ago
6

The equation p(t) = 1.e represents a

Mathematics
1 answer:
zvonat [6]3 years ago
4 0

Answer:

(b), (d) and (e)

Step-by-step explanation:

Given

p(t) = 1 * e^t

See attachment for y = p(t)

Required

Select true statements from the given options

(a) \ln(30) = days the bacteria reaches 30000

We have:

p(t) = 1 * e^t

In this case:

t = \ln(30) and p(t) = 30000

So, we have:

30000 = 1 * e^{\ln(30)}

30000 = e^{\ln(30)}

Using a calculator, we have:

e^{\ln(30)} = 30

So:

30000 = 30

The above equation is false.

(a) is not true

(b) The graph shows that \ln(20) \approx 3

We have:

p(t) = 1 * e^t

Let t = 3

So;

p(3) = 1 * e^3

From the graph, p(3) = 20

So:

20 = 1 * e^3

20 = e^3

Take natural logarithm of both sides

\ln(20) = \ln(e^3)

This gives:

\ln(20) = 3

(b) is true

(c) \ln(t) = y is the logarithm form of y = e^t

We have:

y = e^t

Take natural logarithm of both sides

\ln(y) = \ln(e^t)

This gives:

\ln(y) = t

\ln(y) = t  \ne \ln(t) = y

(c) is false

(d) e^4 > 50 and  \ln(50) < 4

From the graph, we have:

e^4 = 54 --- rough readings

This implies that:

e^4 > 50 is true

Because 54 > 50

Take natural logarithm of both sides

\ln(54) > \ln(50)

Rewrite as:

\ln(50) < \ln(54)

We have:

e^4 = 54

Take natural logarithm of both sides

\ln(e^4) = \ln(54)

4 = \ln(54)

\ln(54) = 4

Substitute \ln(54) = 4 in \ln(50) < \ln(54)

\ln(50) < 4

(d) is true

(e) The graph shows that 10 \approx \ln(2.3)

We have:

p(t) = 1 * e^t

Let t = 2.3

So;

p(2.3) = 1 * e^{2.3}

From the graph,

p(2.3) = 10 ---- rough readings

So:

10 = 1 * e^{2.3}

10 = e^{2.3}

Take natural logarithm of both sides

\ln(10) = \ln(e^{2.3})

This gives:

\ln(10) = 2.3

(e) is true

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To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

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What is the probability that the sample mean would be greater than 101.63 WPM?

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Z = \frac{X - \mu}{\sigma}

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Z = \frac{X - \mu}{s}

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1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

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