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Shalnov [3]
3 years ago
11

A typical adult human body contains approximately 2.500 L of blood plasma. How many grams of blood plasma are in the typical adu

lt human body? The density of blood plasma is 1.03 g/mL.
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer:

2,575 grams of blood plasma are in the typical adult human body

Step-by-step explanation:

Density is the property that matter, whether solid, liquid or gas, has to compress into a given space, so it relates the amount of mass per unit volume. So the density of a substance is the quotient between mass and volume:

density=\frac{mass}{volume}

In this case, you know:

  • density=1.03 \frac{g}{mL}
  • mass= ?
  • volume= 2.5 L= 2,500 mL (being 1 L= 1,000 mL)

Replacing:

1.03\frac{g}{mL} =\frac{mass}{2,500 mL}

and solving, you get:

mass= 1.03 \frac{g}{mL} *2,500 mL

mass= 2,575 g

<u><em>2,575 grams of blood plasma are in the typical adult human body</em></u>

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5/9

Step-by-step explanation:

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Of the Students that attended Roosevelt Elementary 6/8 of the school play a sport. Of the students are playing a sport 3/5 of th
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Answer:

\frac{9}{20} students.

Step-by-step explanation:

We have been given that of the students that attended Roosevelt Elementary 6/8 of the school play a sport. Of the students are playing a sport 3/5 of the students are also involved in the drama club.

To find the students are both play a sport and a part of the drama club, we need to find 3/5 part of 6/8 as:

\text{Students who play a sport and a part of the drama club}=\frac{6}{8}\times \frac{3}{5}

\text{Students who play a sport and a part of the drama club}=\frac{3}{4}\times \frac{3}{5}

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5 0
3 years ago
Given f(x) =17 -x2 what is the average rate of change in f(x) over the interval [1,5]
Artist 52 [7]
By definition <span>the average rate of change is given by:
 </span>AVR =  \frac{f(x2) - f(x1)}{x2-x1}
 <span>We have the following function:
 </span>f(x) =17 - x^2
 We evaluate the function for the given interval:
 For X = 1:
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 f(1) =17 - 1
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 For X = 5:
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 Then, replacing values we have:
 AVR = \frac{-8 - 16}{5-1}
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 Answer:
 <span>the average rate of change in f(x) over the interval [1,5] is:
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Answer:

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