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3 years ago

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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Dafna11 [192]

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

<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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2 years ago

K divided by 11 is 7

Pani-rosa [81]

K/11 = 7
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Answer
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3 years ago

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3 years ago

Increase £240 by 30%?

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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:

The null hypothesis is

$H_o : \mu = 21$

The Alternative hypothesis is

$H_a : \mu< 21$

$\sigma_{\= x} = 0.8944$

$t = -2.236$

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

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3 years ago