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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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Can someone please solve 5-2x&lt;7
Dovator [93]
Add 2x and subtract 7
   5 -2x +2x < 7 +2x . . . . . . showing 2x added
   5 < 7 +2x . . . . . . . . . . . .. simplified
   5 -7 < 7 -7 +2x . . . . . . . . showing 7 subtracted
  -2 < 2x . . . . . . . . . . . . . . . simplified
Now, divide by 2
   -1 < x

The solution is -1 < x.

4 0
3 years ago
How do you sove 673 times 86?
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x      86
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6 0
3 years ago
20POINTS <br> help me please
max2010maxim [7]

Answer:

a. 50 students are randomly selected from the school; 11 drive to school

b. 407

Step-by-step explanation:

a. to get an accurate viewing, use random people that share 1 large common denominator (from the same school) that also has to do with the question

b. 11/50 is only between 50 students. make the denominator the same (1850) and multiply the same to the numerator

1850/50 = 37

11/50 x 37 = 407/1850

8 0
2 years ago
Ten individuals are candidates for a committee. Two will be selected. How many different pairs of individuals can be selected?
Nina [5.8K]

Answer:

45

Step-by-step explanation:

The different pairs of individual can be computed using rule of combination.

The combination is denotes as nCr.

nCr=n!/(r!(n-r)!)

In the given problem n=total candidates=10 and r= Selected candidates=2.

10C2=10!/(2!(10-2)!)

10C2=10*9*8!/(2!8!)

10C2=90/2

10C2=45

So, 45 different pairs of individuals can be selected.

6 0
3 years ago
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If I'm understanding this correctly 20 rounded to the nearest ten would be 20
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3 years ago
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