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PolarNik [594]
3 years ago
14

A biologist tracked the growth of a strain of bacteria, as shown in the graph.

Mathematics
1 answer:
ozzi3 years ago
7 0

Answer:

Part A -> Function

Part B -> Function

Step-by-step explanation:

<u>Concept</u>

A relationship is a function only if there's a single y value corresponding to an x value. It can have the same y value for different x values but it <em>must not have multiply y values for a single x value</em>.

Below is an example

<u>Function</u>                                        <u>Not a Function</u>

<u>x</u>           <u>y</u>                                        <u>x</u>           <u>y</u>

1           5                                        1            5

2          6                                        2           7

3          5                                         1           6

<u>Part A</u>

Looking at the given graph, we see that the relationship has a single y value corresponding to an x value and so it is a function.

<u>Part B</u>

If there was the same number of bacteria for two consecutive hours it would mean the same y value for different x values. So, it would still remain a function.

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Read 2 more answers
Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c
steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}

Compute the sample proportions as follows:

\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23

The critical value of <em>z</em> for 90% confidence interval is:

z_{0.10/2}=z_{0.05}=1.645

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}

     =(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0
3 years ago
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