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luda_lava [24]
3 years ago
12

In a lab experiment, 720 bacteria are placed in a Petri dish. The conditions are such that the number of bacteria is able to dou

ble every 26 hours. How long would it be, to the nearest 10th of an hour until there are 1310 bacteria present?
Mathematics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

22.4 hours

Step-by-step explanation:

The population of bacteria is modelled by the equation:

P=P_0e^{rt}

From the the question, the initial population of bacteria is 720.

So after 26 hours, we have:

P=2P_0

This implies that:

2P_0=P_0e^{26r}

2=e^{26r}

26r =  ln(2)

r =  \frac{ ln(2) }{26}

r = 0.0267

We want to find how long it will take for there to be 1310 bacteria present.

1310=720e^{0.0267t}

\frac{1310}{720}  =  {e}^{0.0267t}

\ln(\frac{1310}{720})  = {0.0267t}

0.59853= {0.0267t}  \\ t =  \frac{0.59853}{0.0267}

t = 22.417

To the nearest tenth , it will take 22.4 hours

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(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

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Step-by-step explanation:

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Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

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Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

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