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liberstina [14]
3 years ago
8

The graph shows a bacteria population as a function of the number of days since an antibiotic was introduced.

Mathematics
2 answers:
spayn [35]3 years ago
6 0

Answer:

14 million bacteria


Step-by-step explanation:

At time=0, the bacteria was introduced.

<em>The first point on the graph, the y-point corresponding to t=0, is the initial amount of bacteria</em>. This point is 14.

Since y-axis is given in millions, 14 corresponds to 14 million.


Hence, 14 million bacteria were present when the antibiotic was first introduced.

miss Akunina [59]3 years ago
4 0
14 million bacteria were present when the antibiotic was first introduced.
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<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more number and variables.

Let x be the number of cheese squares that she eats and y is the number of turkey slices that she eats. If nina can eat 12 or more grams of protein, hence:

2x + 3y ≥ 12

The inequality that represents the possible ways nina can eat 12 or more grams of protein is 2x + 3y ≥ 12

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You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w
Margaret [11]

Answer:

a

  The null hypothesis is  

         H_o  : \mu  =  21

The Alternative  hypothesis is  

           H_a  :  \mu<   21

b

     \sigma_{\= x} =   0.8944

c

   t = -2.236

d

  Yes the  mean population is  significantly less than 21.

Step-by-step explanation:

From the question we are given

           a set of  data  

                               20  18  17  22  18

       The confidence level is 90%

       The  sample  size  is  n =  5  

Generally the mean of the sample  is  mathematically evaluated as

        \= x  =  \frac{20 + 18 +  17 +  22 +  18}{5}

       \= x  =  19

The standard deviation is evaluated as

        \sigma =  \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }

         \sigma =  \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }

         \sigma = 2

Now the confidence level is given as  90 %  hence the level of significance can be evaluated as

         \alpha = 100 - 90

        \alpha = 10%

         \alpha =0.10

Now the null hypothesis is  

         H_o  : \mu  =  21

the Alternative  hypothesis is  

           H_a  :  \mu<   21

The  standard error of mean is mathematically evaluated as

         \sigma_{\= x} =   \frac{\sigma}{ \sqrt{n} }

substituting values

         \sigma_{\= x} =   \frac{2}{ \sqrt{5 } }

        \sigma_{\= x} =   0.8944

The test statistic is  evaluated as  

              t =  \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }

substituting values

              t =  \frac{ 19  - 21 }{ 0.8944 }

              t = -2.236

The  critical value of the level of significance is  obtained from the critical value table for z values as  

                   z_{0.10} =  1.28

Looking at the obtained value we see that z_{0.10} is greater than the test statistics value so the null hypothesis is rejected

6 0
3 years ago
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