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natulia [17]
3 years ago
5

Help!!!

Mathematics
1 answer:
sesenic [268]3 years ago
6 0

Answer:

Time it will take t = 23.1 (round to the nearest tenth)

Step-by-step explanation:

Given: A population of bacteria is growing exponentially;

According to the model;

P = 100e^{0.60t}   ......[1]; where P is the number of colonies and t be the time measured in hours.

After how many hours will 400 colonies be present.

Substitute value of P = 400 in [1];

400 = 100 e^{0.06t}

Divide both sides by 100 we get;

4 = e^{0.06t}

Taking ln both sides we get;

ln 4 = ln e^{0.06t}

Since; ln e^x = x

then;

ln 4 = 0.06 t

Divide both sides by 0.06 we get;

t = \frac{ln 4}{0.06}

or

t =\frac{1.38629436112}{0.06} =23.104906 hours

Therefore, the time it will take is, t = 23.1 (round to the nearest tenth)



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A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

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Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

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3 years ago
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Olegator [25]

Answer:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

Step-by-step explanation:

Data provided

\bar X=7.38 represent the sample mean for the fuel efficiencies

\mu population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size  

Confidence interval

The formula for the confidence interval of the true mean is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom for this case is given by:

df=n-1=601-1=600

The Confidence level for this case is 0.95 or 95%, and the significance level \alpha=0.05 and \alpha/2 =0.025 and the critical value is given by t_{\alpha/2}=1.96

Replcing into the formula for the confidence interval is given by:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

8 0
4 years ago
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