Roboguru

10. Diketahui

Pertanyaan

10. Diketahui

 begin mathsize 14px style f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator 4 over denominator 3 x squared end fraction space comma space pada space open square brackets 1 comma p close square brackets end cell row cell 0 space comma space ditempat space lain end cell end table close. Tentukan space nilai space p space agar space f left parenthesis x right parenthesis space merupakan space fungsi space kerapatan space probabilitas. end style  

  1. ...space 

  2. ...undefined 

Pembahasan Soal:

begin mathsize 14px style Nilai space fungsi space kerapatan space probabilitas adalah space 1 integral subscript 1 superscript p space fraction numerator 4 over denominator 3 x squared end fraction space d x equals 1 integral subscript 1 superscript p space 4 over 3 space x to the power of negative 2 end exponent space d x equals 1 open square brackets 4 over 3 times fraction numerator 1 over denominator negative 1 end fraction x to the power of negative 1 end exponent close square brackets subscript 1 superscript p equals 1 open square brackets negative 4 over 3 x to the power of negative 1 end exponent close square brackets subscript 1 superscript p equals 1 minus 4 over 3 p to the power of negative 1 end exponent minus negative 4 over 3 open parentheses 1 close parentheses to the power of negative 1 end exponent equals 1 minus fraction numerator 4 over denominator 3 p end fraction plus 4 over 3 equals 1 minus fraction numerator 4 over denominator 3 p end fraction equals 1 minus 4 over 3 minus fraction numerator 4 over denominator 3 p end fraction equals negative 1 third p equals fraction numerator begin display style bevelled fraction numerator negative 4 over denominator 3 end fraction end style over denominator negative 1 divided by 3 end fraction p equals 4 end style 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 11 Februari 2021

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