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Pertanyaan

Tabel permintaan, penawaran dan harga

Berdasarkan tabel diatas, maka titik keseimbangan yang terjadi adalah....space space 
 

A. begin mathsize 14px style left parenthesis 10 bevelled 1 half comma space 9.000 right parenthesis end style

B. (30, 6.000)

C. (30, 9.000)

D. begin mathsize 14px style left parenthesis 40 bevelled 1 half comma 9.000 right parenthesis end style

E. (60, 6.000)

  1. A

  2. B

  3. C

  4. D

  5. E

D. Setiani

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style Fungsi space Permintaan space colon fraction numerator straight P minus straight P subscript 1 over denominator straight P subscript 2 minus straight P subscript 1 end fraction equals fraction numerator straight Q minus straight Q subscript 1 over denominator straight Q subscript 2 minus straight Q subscript 1 end fraction fraction numerator straight P minus 10.000 over denominator 8.000 minus 10.000 end fraction equals fraction numerator straight Q minus 10 over denominator 20 minus 10 end fraction fraction numerator straight P minus 10.000 over denominator negative 2.000 end fraction equals fraction numerator straight Q minus 10 over denominator 10 end fraction 10 left parenthesis straight P minus 10.000 right parenthesis equals negative 2.000 left parenthesis straight Q minus 10 right parenthesis 10 straight P minus 100.000 equals negative 2.000 straight Q plus 20.000 10 straight P equals negative 2.000 straight Q plus 20.000 plus 100.000 10 straight P equals negative 2000 straight Q plus 120.000 straight P subscript straight d equals negative 200 straight Q plus 12.000  Fungsi space Penawaran space colon fraction numerator straight P minus straight P subscript 1 over denominator straight P subscript 2 minus straight P subscript 1 end fraction equals fraction numerator straight Q minus straight Q subscript 1 over denominator straight Q subscript 2 minus straight Q subscript 1 end fraction fraction numerator straight P minus 10.000 over denominator 8.000 minus 10.000 end fraction equals fraction numerator straight Q minus 50 over denominator 40 minus 50 end fraction fraction numerator straight P minus 10.000 over denominator negative 2.000 end fraction equals fraction numerator straight Q minus 50 over denominator negative 10 end fraction minus 10 left parenthesis straight P minus 10.000 right parenthesis equals negative 2.000 left parenthesis straight Q minus 50 right parenthesis minus 10 straight P plus 100.000 equals negative 2.000 straight Q plus 100.000 minus 10 straight P equals negative 2.000 straight Q plus 100.000 minus 100.000 minus 10 straight P equals negative 2.000 straight Q 10 straight P equals 2.000 straight Q straight P subscript straight s equals 200 straight Q  Keseimbangan space Pasar straight P subscript straight d equals straight P subscript straight s minus 200 straight Q plus 12.000 equals 200 straight Q 12.000 equals 200 straight Q plus 200 straight Q 400 straight Q equals 12.000 straight Q equals 30 comma space maka space colon straight P subscript straight s equals 200 straight Q straight P subscript straight s equals 200 left parenthesis 30 right parenthesis straight P equals 6.000  Titik space Keseimbangan space left parenthesis straight Q comma straight P right parenthesis equals left parenthesis 30 comma 6.000 right parenthesis  end style


space spaceJadi, jawaban yang sesuai adalah B.space space 

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