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Klio2033 [76]
3 years ago
9

A medical researcher has two petri dishes containing viruses. Dish A has a population density of 1.2 viruses per square millimet

er. Dish A has an area of about 2,826 square millimeters. If both petri dishes have the same population density, approximately how many viruses are in Dish A?
Mathematics
2 answers:
Lyrx [107]3 years ago
7 0
<h3><u>Answer;</u></h3>

= 3,391 viruses

<h3><u>Explanation;</u></h3>
  • The population density is 1.2 viruses per square millimeter.
  • This means, each square millimeter is occupied by a population of 1.2 viruses.

Therefore; for an area of 2,826 square millimeters, there will be;

  = 2,826 multiplied by 1.2 viruses

 = 2,826 × 1.2

 = 3391.2 viruses,

and since we can have a fraction or decimal virus then the population would be <u>3391 viruses</u>

Harlamova29_29 [7]3 years ago
7 0

Answer: There are 3391.2 viruses in petri dish A.

Step-by-step explanation:

Given statements : The population density is 1.2 viruses per square millimeter.

each square millimeter is representing  population of 1.2 viruses.

There is a directly proportional relation between the area and the population  .

Let x be the population of virus in perti dish A.

Therefore,

x=1.2\times2826=3391.2\ vruses

Hence, there are 3391.2 viruses in petri dish A.

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Answer:

20.9-2.01\frac{7.65}{\sqrt{50}}=18.73  

20.9+2.01\frac{7.65}{\sqrt{50}}=23.07  

The 95% confidence interval would be given by (18.73;23.07)  

Step-by-step explanation:

Assuming these data

20 40 22 22 21 21 20 10 20 20

20 13 18 50 20 18 15 8 22 25

22 10 20 22 22 21 15 23 30 12

9 20 40 22 29 19 15 20 20 20

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Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

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

First we need to find the sample mean with the following formula:

\bar X= \frac{\sum_{i=1}^n X_i}{50}=20.9

And in order to find the sample standard deviation we can use the following formula:

s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}}=7.65

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=50-1=49  

Since the Confidence is 0.95or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01  

Now we have everything in order to replace into formula (1):  

20.9-2.01\frac{7.65}{\sqrt{50}}=18.73  

20.9+2.01\frac{7.65}{\sqrt{50}}=23.07  

So on this case the 95% confidence interval would be given by (18.73;23.07)  

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