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Ivahew [28]
3 years ago
9

The bacteria has an initial population of 2000 cells and increases at a rate of 40% per hour. Use the exponential function to mo

del the growth/decay of bacteria.
Mathematics
2 answers:
Fynjy0 [20]3 years ago
5 0
2000% x .4 will give you your answer! 
BaLLatris [955]3 years ago
4 0

Here given, initial population of bacteria = 2000 cells

Also given, the population increases at a rate of 40% per hour.

As the population is increasing so we will use the formula for exponential growth. The formula is A = Pe^{(rt)}

Where, P = initial amount, r = growth rate, t = time, A = final amount.

So here, P = 2000, r = 40% = \frac{40}{100} = 0.4, t = time in hours

We will plug in the values in the formula. we will get,

A = 2000e^{(0.4t)}

We have got the required function to model the growth of the bacteria.

The required function is A = 2000e^{(0.4t)}

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F(x) = 2x + 5
Finger [1]

9514 1404 393

Answer:

  15. f(x) +g(x) = 7x +2

  16. f(x) -g(x) = -3x +8

  17. h(x) · g(x) = 35x -21

Step-by-step explanation:

Substitute the function definition for each function expression and simplify.

__

15. f(x) + g(x) = (2x +5) +(5x -3)

  f(x) +g(x) = 7x +2

__

16. f(x) - g(x) = (2x +5) -(5x -3)

  f(x) -g(x) = -3x +8

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17. h(x) · g(x) = (7) · (5x -3)

  h(x) · g(x) = 35x -21

4 0
3 years ago
Plot the flowing points on the number line without measuring.​
N76 [4]

Answer:

1/4, 3/5, 8/10

Step-by-step explanation:

3/5 is greater than 1/4

(2/4 is equal to 2.5/5---that should give you a visual picture why)  

8/10 is greater than 3/5

3/5 multipled by 2 is 6/10

so, 6/10<8/10

3 0
3 years ago
Read 2 more answers
PLZ HELP! I will give brainliest!!!
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Answer: x axis is time in hours, y axis is water level in centimeters

I can't see the rest of the problem since you didnt show it

Step-by-step explanation:

7 0
3 years ago
Jana baked some muffins. After she served 13 of them to her friends for brunch, she had 17 left. How many muffins did she bake?
aleksley [76]

Answer:

30

Step-by-step explanation:

Add 13 served to 17 left over to get how many muffins she originally baked.

6 0
1 year ago
Read 2 more answers
Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l
laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

<u><em>Let </em></u>\mu<u><em> = mean lead concentration for all such medicines.</em></u>

So, Null Hypothesis, H_0 : \mu = 17 mu      {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, H_A : \mu < 17 mu       {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here <u>One-sample t test</u> <u>statistics</u> as we don't know about population standard deviation;

                      T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample mean lead concentration = \frac{\sum X}{n} = 12.25 mu

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 6.96 mu

             n = sample size = 10

So, <u><em>test statistics</em></u>  =  \frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}  ~ t_9

                              =  -2.16

The value of t test statistics is -2.16.

<u>Now, the P-value of the test statistics is given by the following formula;</u>

                P-value = P( t_9 < -2.16) = <u>0.031</u>

<u><em>Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test.</em></u><em> Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0
3 years ago
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