Archive
10th of March 2017
Mathematics
Q:

Past records show that 80% of the operations performed by a certain doctor were successful . If he performs 4 operations per day what is the probability that at least 3 operations will be successful?


Please give the solution to the following question posted on 8th March : The projection of a vector on the coordinate axes are 6,-3, 2 resp. Find its length and direction cosines.

Also you had stated to give the answer to the second part of the second question posted on 8th March : Also find if the plane thus obtained

contains the line x-2/3=y-1/1= z-2/5 Pls tell if you have answered this part of the question.

Also pls refer your answer to my question posted on 3rd March : Find a unit vector in the XY plane which makes an angle of 45degree with vector i+j and 60 degree with the vector 3i-4j . You have stated in your answer that there is no such vector in the XY Plane.

However there is such a vector as per answer given in the book and the vector is 1/14(13i +j). Hence I request you to please check the answer provided by you . Thanks


Ahuja8087 Ahuja - CBSE - Class XII Commerce

Friday, March 10, 2017 at 00:26:AM

A:

Dear student, the following are the solutions as we told you we shall provide. Hope this helps. 

begin mathsize 16px style Let space straight X space denote space the space number space of space successes space in space 4 space operations.
Then comma space straight X space follows space binomial space distribution space with
straight n equals 4
straight p equals successful space operation equals 80 over 100 equals 4 over 5
straight q equals failed space operation equals 20 over 100 equals 1 fifth
straight P left parenthesis at space least space 3 space operations space are space successful right parenthesis
equals straight P left parenthesis straight X greater or equal than 3 right parenthesis
equals straight P left parenthesis straight X equals 3 right parenthesis plus straight P left parenthesis straight x equals 4 right parenthesis
space equals straight C presuperscript 4 subscript 3 open parentheses 4 over 5 close parentheses cubed open parentheses 1 fifth close parentheses to the power of 1 plus straight C presuperscript 4 subscript 4 open parentheses 4 over 5 close parentheses to the power of 4 open parentheses 1 fifth close parentheses to the power of 0
equals 4 cross times open parentheses 4 over 5 close parentheses cubed cross times open parentheses 1 fifth close parentheses plus 1 cross times open parentheses 4 over 5 close parentheses to the power of 4 cross times 1
equals open parentheses 4 over 5 close parentheses cubed open square brackets 4 over 5 plus 4 over 5 close square brackets
equals 64 over 125 cross times 8 over 5
equals 512 over 625 end style


Error converting from MathML to accessible text.

Error converting from MathML to accessible text.

begin mathsize 16px style bold Solution bold space bold of bold space bold 3 bold rd bold space bold March bold colon
Let space vector space be space ai plus bj.
Now comma
straight a with rightwards harpoon with barb upwards on top. straight b with rightwards harpoon with barb upwards on top equals open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar cosθ
straight a.1 plus straight b.1 equals square root of straight a squared plus straight b squared end root left parenthesis square root of 2 right parenthesis fraction numerator 1 over denominator square root of 2 end fraction
straight a plus straight b equals square root of straight a squared plus straight b squared end root
Squaring comma
rightwards double arrow straight a squared plus straight b squared plus 2 ab equals straight a squared plus straight b squared
rightwards double arrow 2 ab equals 0

And comma space straight a squared plus straight b squared equals 1 space left parenthesis Since space unit space vector right parenthesis
and space since space straight a plus straight b equals square root of straight a squared plus straight b squared end root rightwards double arrow straight a plus straight b equals 1
rightwards double arrow straight a equals 1 space or space straight b equals 0 space OR space straight a equals 0 space or space straight b equals 1 space.... left parenthesis straight i right parenthesis

straight a.3 minus straight b.4 equals 5 cos 60
rightwards double arrow 3 straight a minus 4 straight b equals 5 over 2
rightwards double arrow 6 straight a minus 8 straight b equals 5 space..... left parenthesis ii right parenthesis
Now comma
straight a equals 1 space or space straight b equals 0 space does space not space satisfy space 6 straight a minus 8 straight b equals 5 space condition
Also comma space straight a equals 0 space or space straight b equals 1 space does space not space satisfy space 6 straight a minus 8 straight b equals 5 space condition

Both space the space conditions space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space have space to space be space satisfied. space That space is space why comma space we space said space there space is space no space vector space in space xy space plane space which space satisfies space given space conditions.

Now comma space if space straight I space consider space only space condition space left parenthesis ii right parenthesis space without space using space condition space left parenthesis straight i right parenthesis comma space straight I space can space get space the space vector space as space shown space below.
straight a.3 minus straight b.4 equals 5 cos 60 rightwards double arrow 3 straight a minus 4 straight b equals 5 over 2 open parentheses straight a plus straight b close parentheses rightwards double arrow straight a equals 13 straight b
So comma space if space straight b equals 1 comma space straight a equals 13
The space vector space would space be space 13 straight i plus straight i space and space hence space the space unit space vector space would space be space fraction numerator 13 straight i plus straight i over denominator square root of 13 squared plus 1 end root end fraction
Now comma space this space is space possible space since space by space the space first space condition comma space we space cannot space get space both space straight a space and space straight b space to space be space non minus zero space vectors.
And space according space to space the space definition space of space angle space between space two space vectors comma space straight a space and space straight b space space should space be space non minus zero.
So comma space the space vector space will space be space fraction numerator 13 straight i plus straight i over denominator square root of 13 squared plus 1 end root end fraction. space Please space check space dear space student. space Since space the space answers space are space still space not space matching. space end style

Friday, March 10, 2017 at 17:29:PM