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makkiz [27]
1 year ago
10

What number could be added to 0.40 ml for the level of precision to be 0.01 ml? check all that apply 0.2 ml 0.154 ml 6 ml 2.02 m

l 8.8331 ml
Mathematics
1 answer:
Svetach [21]1 year ago
5 0

The following volumes of fluid can be added to 0.40 mL of liquid for the level of precision of 0.01 mL: a) 0.2 mL, b) 6 mL, c) 2.02 mL. The amounts 0.154 mL and 8.8331 mL are not possible due to given level of precision.

<h3>What is the amount of liquid to be added to sample according to a given precision?</h3>

If the measuring has a level of precision of 0.01 ml, this means that the <em>measured</em> quantities are only sensible to the <em>smallest</em> hundreths. Any change less than 0.01 ml and any decimal less than a hundreths are "invisible" for measuring processes.

Hence, the following volumes of fluid can be added to 0.40 mL of liquid for the level of precision of 0.01 mL: a) 0.2 mL, b) 6 mL, c) 2.02 mL. The amounts 0.154 mL and 8.8331 mL are not possible due to given level of precision.

To learn more on precision: brainly.com/question/1311561

#SPJ1

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Answer:

a) On January 1, 2017 the fish population will be the same as the initial population.

b) On September 18th, 2018 the fish population will be zero.

Step-by-step explanation:

Hi there!

a) First, let´s write the function:

F(t) = 3000(23 + 11t − t²)

The population on January 1, 2006 is the population at t = 0. Then:

F(0) = 3000(23 + 11· 0 - 0²)

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This will be the population every time at which t² - 11t = 0. Then let´s find the other value of t (besides t = 0) that makes that expression to be zero:

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On January 1, 2017 (2006 + 11), the fish population will be the same as the initial population.

b) We have to obtain the value of t at which F(t) = 0

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0 = 3000(23 + 11t − t²)

divide both sides of the equation by 3000

0 = 23 + 11t − t²

Let´s solve this quadratic equation using the quadratic formula:

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x = [-b ± √(b² - 4ac)] / 2a

x = 12.8  ( the other value of x is negative and therefore discarded).

After 12.8 years all the fish in the lake will have died.

If 1 year is 12 months, 0.8 years will be:

0.8 years · 12 months/year = 9.6 months

If 1 month is 30 days, 0.6 month will be:

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All the fish will have died after 12 years, 9 months and 18 days from January 1, 2006. That is, on September 18th, 2018.

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