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Solution:
- 伝感器Aの振幅は ( A = 1.5B )。
- 伝感器Bの振幅は ( B = \frac{2}{3}A )。
( A = 1.5B ) を ( B = \frac{2}{3}A ) に代入します:
[ B = \frac{2}{3}(1.5B) ]
[ B = B ]
これは関係性が確認されたことを示していますが、具体的な値を求めるために:
( B = 2 ) と仮定します。すると ( A = 1.5 \ imes 2 = 3 )。
したがって、( A = 3 ) かつ ( B = 2 )。
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( A = 3, B = 2 )
Question 2
博物館の学芸員がísticosの展示のカタログを作成しています。いてんしゃくの数は機器の数の2倍で、コレクション全体のアイテム数は90である場合、てんしゃくと機器の数をそれぞれ求めてください。
Solution:
-機器の数を ( x ) とします。 - いてんしゃくの数は ( 2x ) です。 - 式: ( x + 2x = 90 )。
[ 3x = 90 ]
[ x = 30 ]
したがって、機器は30個いて、てんしゃくは ( 2 \ imes 30 = 60 ) 個です。
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機器 = 30,てんしゃく = 60
Question 3
科学哲学者が2つの理論を研究しています。理論Xは理論Yの3倍の証拠によって支持されています。両方の理論の証拠の合計が64の場合、理論Xと理論Yをそれぞれ支持する証拠の数を求めてください。
Solution:
- 理論Yの証拠を ( y ) とします。
- 理論Xの証拠は ( 3y ) です。
- 式: ( y + 3y = 64 )。
[ 4y = 64 ]
[ y = 16 ]
したがって、理論Yは16の証拠、理論Xは ( 3 \ imes 16 = 48 ) の証拠によって支持されています。
理論X = 48, 理論Y = 16
Question 4
地震学者が、マグニチュードMの地震を予測するAIモデルを開発しています。このモデルはM = (\log_{10}(E) + 4.8) の式を使用します。ここで( E ) はジュール単位のエネルギーです。マグニチュードが5の地震を予測する場合、推定エネルギー ( E ) を求めてください。
Solution:
- ( M = 5 ) が与えられています。
[ 5 = \log_{10}(E) + 4.8 ]
[ \log_{10}(E) = 5 - 4.8 ]
[ \log_{10}(E) = 0.2 ]
( E ) を求めるために:
[ E = 10^{0.2} \approx 1.5849 ]
推定エネルギー ( E ) は約1.5849ジュールです。
( E \approx 1.5849 ) ジュール
Question 5
博物館の学芸員が、歴史的な楽器のデジタルカタログを管理しています。デジタル記録の数は拡 Tablesquareの数の40%多く、記録の総数は (\frac{15}{4}) の拡 Tablesquareに相当する場合、拡 Tablesquareの数を求めてください。
Solution:
- 拡 Tablesquareの数を ( x ) とします。
- デジタル記録: ( 1.4x )。
- 式: ( x + 1.4x = \frac{15}{4}x )。
[ 2.4x = \frac{15}{4} ]
[ x = \frac{15}{4 \ imes 2.4} = \frac{15}{9.6} = \frac{25}{16} ]
したがって、( x = 1.5625 ) 、約1.56の拡 Tablesquare です。
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拡 Tablesquare ≈ 1.56
Question 6
科学哲学者が2つのパラダイムを比較しています。パラダイムAはパラダイムBの2.5倍の支持thersによって支持されています。両方のパラダイムを支持する科学者の総数が84の場合、パラダイムAとBを支持する科学者の数を求めてください。
Solution:
- パラダイムBを支持する科学者の数を ( b ) とします。
- パラダイムAを支持: ( 2.5b )。
- 式: ( b + 2.5b = 84 )。
[ 3.5b = 84 ]
[ b = 24 ]
したがって、パラダイムBは24人の支持者、パラダイムAは ( 2.5 \ imes 24 = 60 ) の支持者によって支持されています。
パラダイムA = 60, パラダイムB = 24
Question 7
地震学者が、2つの異なる地震の期待されるリヒター規模を評価しています。初期振幅の比が ( R_1 : R_2 = 100 : 1 ) の場合、規模の差 ( M_1 - M_2 ) を計算してください。
Solution:
- スケール公式 ( M = \log_{10}(R) ) を使用します。
- ( R_1 = 100 ), ( R_2 = 1 )。
[ M_1 = \log_{10}(100) = 2 ]
[ M_2 = \log_{10}(1) = 0 ]
[ M_1 - M_2 = 2 - 0 = 2 ]
規模の差は2です。
( M_1 - M_2 = 2 )
Question 8
博物館の学芸員が、8行6列のグリッドで文明のタイムラインを配置しています。文明Aのエ entriesが30であり、文明Bが文明Aのエ entriesの2倍の場合、合計のエ entries はいくつありますりますか？
Solution:
- 文明A: 30 entries。
- 文明B: ( 2 \ imes 30 = 60 ) entries。
- 合計エ entries: ( 30 + 60 = 90 )。
合計エ entries = 90
Question 9
科学哲学者が2つの探究分野の影響を研究しています。分野Xの影響スコアは分野Yの4倍であり、両方の合計スコアが500の場合、分野XとYのスコアを求めてください。
Solution:
- 分野Yのスコアを ( y ) とします。
- 分野Xのスコア: ( 4y )。
- 式: ( y + 4y = 500 )。
[ 5y = 500 ]
[ y = 100 ]
分野Xは ( 4 \ imes 100 = 400 ) のスコアを持ちます。
分野X = 400, 分野Y = 100
Question 10
地震学者が、初期振幅に基づく関数 ( A(t) = A_0 e^{kt} ) を使用して地震の規模を予測しています。初期振幅 ( A_0 = 1 ) で、スケール係数 ( k = 0.5 ) の場合、3単位の時間後の振幅 ( A(3) ) を求めてください。
Solution:
- ( A(t) = e^{kt} )。
- ( A(3) = e^{0.5 \ imes 3} = e^{1.5} )。
[ e^{1.5} \approx 4.4817 ]
3単位の時間後の振幅は約4.4817です。