1. The abnormal rise in total count of RBC is called?

Answer: Polycythemia.

Reply

Type in
(Press Ctrl+g to toggle between English and the chosen language)

Comments

Tags
Show Similar Question And Answers
QA->The abnormal rise in total count of RBC is called?....
QA->Abnormal decrease in total count of WBC is known as ?....
QA->Abnormal increase in WBC count is known as ?....
QA->Normal count level of RBC in Blood ?....
QA->The Red blood cells (RBC) is also called as....
MCQ->Which of the following statements are correct about the C#.NET code snippet given below? namespace IndiabixConsoleApplication { class index { protected int count; public index() { count = 0; } } class index1: index { public void increment() { count = count +1; } } class MyProgram { static void Main(string[] args) { index1 i = new index1(); i.increment(); } } } count should be declared as public if it is to become available in the inheritance chain. count should be declared as protected if it is to become available in the inheritance chain. While constructing an object referred to by i firstly constructor of index class will be called followed by constructor of index1 class. Constructor of index class does not get inherited in index1 class. count should be declared as Friend if it is to become available in the inheritance chain....
MCQ->What will be the output of the program? class BoolArray { boolean [] b = new boolean[3]; int count = 0; void set(boolean [] x, int i) { x[i] = true; ++count; } public static void main(String [] args) { BoolArray ba = new BoolArray(); ba.set(ba.b, 0); ba.set(ba.b, 2); ba.test(); } void test() { if ( b[0] && b[1] | b[2] ) count++; if ( b[1] && b[(++count - 2)] ) count += 7; System.out.println("count = " + count); } }...
MCQ-> Directions : Study the following information carefully and answer these questions. A word and number arrangement machine when given an input line of words and numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement. (All the numbers are two digits numbers) Input : tall 48 13 rise alt 99 76 32 wise jar high 28 56 barn Step I : 13 tall 48 rise 99 76 32 wise jar high 28 56 barn alt Step II : 28 13 tall 48 rise 99 76 32 wise jar high 56 alt barn Step III : 32 28 13 tall 48 rise 99 76 wise jar 56 alt barn high Step IV : 48 32 28 13 tall rise 99 76 wise 56 alt barn high jar Step V : 56 48 32 28 13 tall 99 76 wise alt barn high jar rise Step VI : 76 56 48 32 28 13 99 wise alt barn high jar rise tall Step VII : 99 76 56 48 32 28 13 alt barn high jar rise tall wise and Step VII is the last step of the above input, as the desired arrangement is obtained. As per the rules followed in the above steps, find out in each of the following questions the appropriate step for the given input. Input : 84 why sit 14 32 not best ink feet 51 27 vain 68 92 (All the numbers are two digits numbers)Which step number is the following output? 32 27 14 84 why sit not 51 vain 92 68 feet best ink...
MCQ->What will be the output of the following program? #include<iostream.h> class IndiaBix { static int count; public: static void First(void) { count = 10; } static void Second(int x) { count = count + x; } static void Display(void) { cout<< count << endl; } }; int IndiaBix::count = 0; int main() { IndiaBix :: First(); IndiaBix :: Second(5); IndiaBix :: Display(); return 0; }...
MCQ->What will be the output of the following program? #include<iostream.h> class BaseCounter { protected: long int count; public: void CountIt(int x, int y = 10, int z = 20) { count = 0; cout<< x << " " << y << " " << z << endl; } BaseCounter() { count = 0; } BaseCounter(int x) { count = x ; } }; class DerivedCounter: public BaseCounter { public: DerivedCounter() { } DerivedCounter(int x): BaseCounter(x) { } }; int main() { DerivedCounter objDC(30); objDC.CountIt(40, 50); return 0; }...
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
DMCA.com Protection Status Powered By:Omega Web Solutions
© 2002-2017 Omega Education PVT LTD...Privacy | Terms And Conditions