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seropon [69]
3 years ago
13

On a standard reticulocyte preparation with new methylene blue, there are 100 cells counted with blue-stained granulofilamentous

material. The red blood cell count is 3.22 x 1012/L and the hematocrit is 30%. Calculate the absolute reticulocyte count.
Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

0.322 × 10¹² /L

Step-by-step explanation:

Data provided in the question:

Number of cells counted with blue-stained granulofilamentous material

i.e number of RET = 100

The red blood cells count = 3.22 × 10¹² /L

Hematocrit = 30%

Now,

RET% = [ [ Number of RET ] ÷ 1000 RBCs ] × 100%

= [ 100 ÷ 1000 ] × 100%

= 0.1 × 100%

= 10%

also,

Absolute reticulocyte count = ( %RET × RBC count ) ÷ 100

= [ 10 × 3.22 × 10¹² /L ] ÷ 100

= 0.322 × 10¹² /L

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-2\left(3x+2\right)\ge \:-6x-4

\mathrm{Expand\:}-2\left(3x+2\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-2,\:b=3x,\:c=2,\\-2\cdot \:3x+\left(-2\right)\cdot \:2,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-2\cdot \:3x-2\cdot \:2,\\\mathrm{Simplify}\:-2\cdot \:3x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6,\\-6x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4,\\-6x-4,\\-6x-4\ge \:-6x-4

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\mathrm{Add\:}6x\mathrm{\:to\:both\:sides},\\-6x+6x\ge \:-6x+6x

\mathrm{Simplify},\\0\ge \:0

\mathrm{Therefore,\:the\:final\:solution\:is},\\True\quad \forall \:x\in \mathbb{R}

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8 0
3 years ago
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