{"id":184,"date":"2024-03-12T10:00:14","date_gmt":"2024-03-12T01:00:14","guid":{"rendered":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=184"},"modified":"2024-06-05T16:37:45","modified_gmt":"2024-06-05T07:37:45","slug":"%e6%95%b0%e5%ad%a6ii_1q%e6%a8%a1%e6%93%ac","status":"publish","type":"post","link":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=184","title":{"rendered":"\u6570\u5b66II_1Q\u6a21\u64ec"},"content":{"rendered":"\n\n\n\n
\u518d\u751f\u30ea\u30b9\u30c8<\/span><\/a><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
4H1_\u524d\u534a<\/a><\/td>\n4H1_\u5f8c\u534a<\/a><\/td>\n2H1<\/a><\/td>\n4H2_\u524d\u534a<\/a><\/td>\n4H2_\u5f8c\u534a<\/a><\/td>\n2H2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
4H1_\u524d\u534a<\/span>4H1_\u5f8c\u534a<\/span>2H1<\/span>4H2_\u524d\u534a<\/span>4H2_\u5f8c\u534a<\/span>2H2<\/span><\/div>
\n

\u7b49\u5dee\u6570\u5217<\/span><\/p>\n

\uff11\uff0e\u521d\u9805$2$\u3001\u516c\u5dee$-3$\u306e\u7b49\u5dee\u6570\u5217\u306b\u3064\u3044\u3066\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1) \u4e00\u822c\u9805$a_n$\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$a_n=dn+(a-d)$\u3000\u306b\u304a\u3044\u3066\u3001$a=2, d=-3$\u3000\u3092\u4ee3\u5165\u3057\u3066
\n$a_n=-3n+5$ (\u7b54)<\/span><\/div><\/div>\n

(2) $-40$ \u306f\u3053\u306e\u6570\u5217\u306e\u7b2c\u4f55\u9805\u304b\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(1)\u3088\u308a$a_n=-3n+5$\u3060\u304b\u3089\u3001\u65b9\u7a0b\u5f0f$-40=-3n+5$\u3092\u89e3\u3044\u3066\u3001
\n$3n=40+5$ \u3088\u308a $n=15$ \u3060\u304b\u3089\u3001\u7b2c15\u9805\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3)\u7b2c1\u9805\u304b\u3089\u7b2c$n$\u9805\u307e\u3067\u306e\u548c$S_n$\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$S_n=\\dfrac{1}{2}n(a+a_n)$\u3060\u304b\u3089\u3001(1)\u3088\u308a$a_n=-3n+5$\u3001
\n\u307e\u305f $a=2$ \u306a\u306e\u3067\u3000$S_n=\\dfrac{1}{2}n(-3n+7)$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(4) \u7b2c6\u9805\u304b\u3089\u7b2c25\u9805\u307e\u3067\u306e\u548c$S$\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u548c\u306e\u6700\u521d\u306e\u9805\u3092$s$\u3001\u6700\u5f8c\u306e\u9805\u3092$t$\u3001\u9805\u6570\u3092$n$\u3068\u3059\u308b\u3068\u3001$S=\\dfrac{1}{2}n(s+t)$
\n$s=a_{6}=-3\\cdot6+5=-13$\u3001$t=a_{25}=-3\\cdot25+5=-70$\u3001
\n$n=25-6+1=20$ \u3088\u308a\u3000$S=\\dfrac{1}{2}\\cdot20(-13-70)=-830$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

\uff12\uff0e\u7b2c\uff13\u9805\u304c\uff15\u3001\u7b2c11\u9805\u304c29\u306e\u7b49\u5dee\u6570\u5217\u306e\u521d\u9805 $a$ \u3068\u516c\u5dee $d$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u9023\u7acb\u65b9\u7a0b\u5f0f $\\left\\{ \\begin{array}{l} a_3=a+2d=5 \\\\ a_{11}=a+10d=29 \\end{array} \\right.$ \u3092\u89e3\u3044\u3066\u3001$a=-1$\u3001$d=3$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u7b49\u6bd4\u6570\u5217<\/span><\/p>\n

\uff11\uff0e\u521d\u9805$3$\u3001\u516c\u6bd4$2$\u306e\u7b49\u6bd4\u6570\u5217\u306b\u3064\u3044\u3066\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1)\u4e00\u822c\u9805 $a_n$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$a_n=a\\cdot r^{n-1}$\u3000\u306b\u304a\u3044\u3066\u3001$a=3, r=2$\u3000\u3092\u4ee3\u5165\u3057\u3066
\n$a_n=3\\cdot2^{n-1}$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2)$192$ \u306f\u3053\u306e\u6570\u5217\u306e\u7b2c\u4f55\u9805\u304b\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$a_n=3\\cdot2^{n-1}=192$ \u3088\u308a $2^{n-1}=\\dfrac{192}{3}=64=2^6$
\n\u3088\u3063\u3066 $n-1=6$ \u3060\u304b\u3089 $n=7$\u3000\u3000\u7b2c\uff17\u9805\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3)\u7b2c1\u9805\u304b\u3089\u7b2c$n$\u9805\u307e\u3067\u306e\u548c$S_n$\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$S_n=a\\cdot\\dfrac{r^n-1}{r-1}=3\\cdot\\dfrac{2^n-1}{2-1}=3(2^n-1)$<\/span><\/div><\/div>\n

\uff12\uff0e\u7b2c\uff13\u9805\u304c\uff15\u3001\u7b2c8\u9805\u304c160\u306e\u7b49\u5dee\u6bd4\u5217\u306e\u521d\u9805 $a$ \u3068\u516c\u6bd4 $r$ (\u5b9f\u6570) \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u9023\u7acb\u65b9\u7a0b\u5f0f$\\left\\{ \\begin{array}{l} a_3=ar^2=5&\\cdots\\textstyle{\u2460} \\\\ a_{8}=ar^7=160&\\cdots\\textstyle{\u2461} \\end{array} \\right.$ \u3092\u89e3\u304f $\\dfrac{\\textstyle{\u2461}}{\\textstyle{\u2460}}$ \u3088\u308a
\n$r^5=\\dfrac{ar^7}{ar^2}=\\dfrac{160}{5}=32=2^5$ \u3088\u3063\u3066\u3000$r=2$\uff08\u7b54\uff09
\n\u3055\u3089\u306b ${\\textstyle{\u2460}}$ \u306b\u4ee3\u5165\u3057\u3066 $a=\\dfrac{5}{4}$\uff08\u7b54\uff09<\/span><\/div><\/div>\n<\/div>\n
\n

\u548c $\\Sigma$ \u306e\u8a18\u53f7<\/span><\/p>\n

\uff11\uff0e\u6b21\u306e\u548c\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle{\\sum_{k=1}^{n}(4k-1)}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $\\displaystyle{\\sum_{k=1}^{n}k}=\\dfrac{1}{2}n(n+1)$ \u304a\u3088\u3073 $\\displaystyle{\\sum_{k=1}^{n}c}=cn$ \u3092\u7528\u3044\u308b
\n$\\displaystyle{\\sum_{k=1}^{n}(4k-1)}=4\\displaystyle{\\sum_{k=1}^{n}k}-\\displaystyle{\\sum_{k=1}^{n}1}=4\\cdot\\dfrac{1}{2}n(n+1)-1\\cdot n$
\n$=2n(n+1)-n=n(2n+2-1)=n(2n+1)$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $\\displaystyle{\\sum_{k=1}^{n}k(3k+1)}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $\\displaystyle{\\sum_{k=1}^{n}k}=\\dfrac{1}{2}n(n+1)$ \u304a\u3088\u3073 $\\displaystyle{\\sum_{k=1}^{n}k^2}=\\dfrac{1}{6}n(n+1)(2n+1)$ \u3092\u7528\u3044\u308b
\n$\\displaystyle{\\sum_{k=1}^{n}k(3k+1)}=3\\displaystyle{\\sum_{k=1}^{n}k^2}+\\displaystyle{\\sum_{k=1}^{n}k}=3\\cdot\\dfrac{1}{6}n(n+1)(2n+1)+\\dfrac{1}{2}n(n+1)$
\n$=\\dfrac{1}{2}n(n+1)(2n+1+1)=\\dfrac{1}{2}n(n+1)2(n+1)=n(n+1)^2$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3) $\\displaystyle{\\sum_{k=1}^{n}\\dfrac{1}{(3k+1)(3k+4)}}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $a_{k+1}-a_k=d$ \u306e\u3068\u304d\u3000$\\displaystyle{\\sum_{k=1}^{n}}\\dfrac{1}{a_ka_{k+1}}=\\dfrac{1}{d}\\left(\\dfrac{1}{a_1}-\\dfrac{1}{a_{n+1}}\\right)$ \u3092\u7528\u3044\u308b
\n$a_k=3k+1$ \u3068\u304a\u304f\u3068 $a_{k+1}=3(k+1)+1=3k+4$ \u3088\u308a
\n$\\displaystyle{\\sum_{k=1}^{n}\\dfrac{1}{(3k+1)(3k+4)}}=\\displaystyle{\\sum_{k=1}^{n}}\\dfrac{1}{a_ka_{k+1}}=\\dfrac{1}{d}\\left(\\dfrac{1}{a_1}-\\dfrac{1}{a_{n+1}}\\right)$\u3000\u3060\u304b\u3089
\n\u3053\u306e\u5f0f\u306b\u3001$d=a_{k+1}-a_k=3k+4-(3k+1)=3$\u3001$a_1=3\\cdot1+1=4$
\n\u304a\u3088\u3073 $a_{n+1}=3n+4$ \u3092\u4ee3\u5165\u3057\u3066\u3001
\n$\\displaystyle{\\sum_{k=1}^{n}\\dfrac{1}{(3k+1)(3k+4)}}=\\dfrac{1}{3}\\left(\\dfrac{1}{4}-\\dfrac{1}{3n+4}\\right)=\\dfrac{1}{3}\\cdot\\dfrac{3n+4-4}{4(3n+4)}=\\dfrac{n}{4(3n+4)}$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(4) $\\displaystyle{\\sum_{k=1}^{n}\\dfrac{1}{\\sqrt{3k+4}+\\sqrt{3k+1}}}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{1}{\\sqrt{3k+4}+\\sqrt{3k+1}}=\\dfrac{1}{\\sqrt{3k+4}+\\sqrt{3k+1}}\\cdot\\dfrac{\\sqrt{3k+4}-\\sqrt{3k+1}}{\\sqrt{3k+4}-\\sqrt{3k+1}}$
\n$=\\dfrac{\\sqrt{3k+4}-\\sqrt{3k+1}}{(3k+4)-(3k+1)}=\\dfrac{\\sqrt{3k+4}-\\sqrt{3k+1}}{3}=\\dfrac{-1}{3}\\left(\\sqrt{3k+1}-\\sqrt{3k+4}\\right)$
\n$=\\dfrac{-1}{3}\\left(a_k-a_{k+1}\\right)$ \uff08$a_k=\\sqrt{3k+1}\u3000$\u3001$a_{k+1}=\\sqrt{3(k+1)+1}=\\sqrt{3k+4}$\u3000\uff09\u3088\u3063\u3066
\n$\\displaystyle{\\sum_{k=1}^{n}}\\dfrac{1}{\\sqrt{3k+4}+\\sqrt{3k+1}}=\\displaystyle{\\sum_{k=1}^{n}}\\dfrac{-1}{3}(a_k-a_{k+1})$
\n$=\\dfrac{-1}{3}\\{(a_1-a_2)+(a_2-a_3)+\\cdots+(a_n-a_{n+1})\\}=\\dfrac{-1}{3}(a_1-a_{n+1})$
\n$=\\dfrac{a_{n+1}-a_1}{3}=\\dfrac{\\sqrt{3n+4}-\\sqrt{4}}{3}=\\dfrac{\\sqrt{3n+4}-2}{3}$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u6f38\u5316\u5f0f\u3068\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5<\/span><\/p>\n

\uff11\uff0e\u6b21\u306e\u6f38\u5316\u5f0f\u3092\u89e3\u3051\u3002<\/p>\n

(1) $a_1=1,\\;a_{n+1}=a_n+2n-1$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $b_k=a_{k+1}-a_k$\u3000\u306e\u3068\u304d\u3000$a_n=a_1+\\displaystyle{\\sum_{k=1}^{n-1}}b_k$ \u3092\u7528\u3044\u308b
\n$a_{n+1}=a_n+2n-1$\u3000\u3088\u308a\u3000$a_{n+1}-a_n=2n-1$\u3000\u3053\u308c\u306b $n=k$ \u3092\u4ee3\u5165\u3057\u3066
\n$b_k=a_{k+1}-a_k=2k-1$\u3000\u3068\u306a\u308b\u306e\u3067\u3001
\n$a_n=a_1+\\displaystyle{\\sum_{k=1}^{n-1}}b_k=1+\\displaystyle{\\sum_{k=1}^{n-1}}(2k-1)=1+2\\displaystyle{\\sum_{k=1}^{n-1}}k-\\displaystyle{\\sum_{k=1}^{n-1}}1$
\n$=1+2\\cdot\\dfrac{1}{2}(n-1)n-1\\cdot(n-1)=1+(n-1)n-n+1=n^2-2n+2$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $a_1=2,\\;a_{n+1}=2a_n-1$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $a_{n+1}=ra_n+b\\;(r \\neq 1)$\u3000\u306e\u3068\u304d\u3000$a_n-\\lambda=\uff08a_1-\\lambda)r^{n-1}$ \u3092\u7528\u3044\u308b
\n\u305f\u3060\u3057\u3001$\\lambda=r\\lambda+b$
\n$a_{n+1}=2a_n-1$\u3000\u3088\u308a\u3000$\\lambda=2\\lambda-1$\u3000\u3092\u89e3\u304f\u3068\u3000$\\lambda=1$\u3000\u3088\u3063\u3066\u3001
\n$a_n-1=(a_1-1)2^{n-1}=(2-1)2^{n-1}=2^{n-1}$\u3000\u3060\u304b\u3089\u3000$a_n=2^{n-1}+1$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

\uff12\uff0e\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1) \u6f38\u5316\u5f0f $a_1=1,\\; a_{n+1}=\\dfrac{a_n}{3a_n+1}$ \u306e\u89e3\u306f
\n$a_n=\\dfrac{1}{3n-2}$ \u3067\u3042\u308b\u3053\u3068\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3067\u8a3c\u660e\u305b\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$a_n=\\dfrac{1}{3n-2}\\cdots(*)$\u3000\u3068\u304a\u304f
\n(1) $n=1$\u3000\u306e\u3068\u304d\u3000\u3000$((*)$\u306e\u5de6\u8fba$)=a_1=1$\u3000\u3000$((*)$\u306e\u53f3\u8fba$)=\\dfrac{1}{3\\cdot1-2}=\\dfrac{1}{1}=1$
\n$\\quad$\u3088\u3063\u3066\u3000(\u5de6\u8fba)$=$(\u53f3\u8fba)\u3000\u3060\u304b\u3089\u3000$n=1$\u3000\u306e\u3068\u304d $(*)$ \u304c\u6210\u308a\u7acb\u3064
\n(2) $n=k$ \u306e\u3068\u304d\u3001$a_k=\\dfrac{1}{3k-2}\\cdots$\u2460 \u304c\u6210\u308a\u7acb\u3064\u3068\u4eee\u5b9a\u3059\u308b\u3000\u3000\u6f38\u5316\u5f0f\u3068\u2460\u3088\u308a
\n$\\quad a_{k+1}=\\dfrac{a_k}{3a_k+1}=\\dfrac{\\dfrac{1}{3k-2}}{3\\cdot\\dfrac{1}{3k-2}+1}\\cdot\\dfrac{3k-2}{3k-2}=\\dfrac{1}{3+3k-2}=\\dfrac{1}{3k+1}=\\dfrac{1}{3(k+1)-2}$
\n$\\quad$\u3088\u3063\u3066\u3001$a_{k+1}=((*)$\u3067 $n=k+1$ \u3068\u304a\u3044\u305f\u5f0f\u306e\u53f3\u8fba$)$ \u304c\u6210\u308a\u7acb\u3064\u306e\u3067\u3001
\n$\\quad$$(*)$ \u306f $n=k+1$ \u306e\u3068\u304d\u3082\u6210\u308a\u7acb\u3064
\n(3) (1)(2)\u3000\u3088\u308a\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 $n$ \u306b\u3064\u3044\u3066 $(*)$ \u304c\u6210\u308a\u7acb\u3064\u3000(\u8a3c\u660e\u7d42)<\/span><\/div><\/div>\n

(2) $1+2+\\cdots+n=\\dfrac{1}{2}n(n+1)$\u3000\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3067\u8a3c\u660e\u305b\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$1+2+\\cdots+n=\\dfrac{1}{2}n(n+1)=\\dfrac{1}{2}n(n+1)\\cdots(*)$\u3000\u3068\u304a\u304f
\n(1) $n=1$\u3000\u306e\u3068\u304d\u3000\u3000$((*)$\u306e\u5de6\u8fba$)=1$\u3000\u3000$((*)$\u306e\u53f3\u8fba$)=\\dfrac{1}{2}\\cdot1\\cdot(1+1)=1$
\n$\\quad$\u3088\u3063\u3066\u3000(\u5de6\u8fba)$=$(\u53f3\u8fba)\u3000\u3060\u304b\u3089\u3000$n=1$\u3000\u306e\u3068\u304d $(*)$ \u304c\u6210\u308a\u7acb\u3064
\n(2) $n=k$ \u306e\u3068\u304d\u3001$1+2+\\cdots+k=\\dfrac{1}{2}k(k+1)\\cdots$\u2460 \u304c\u6210\u308a\u7acb\u3064\u3068\u4eee\u5b9a\u3059\u308b
\n$\\quad$\u2460\u306e\u4e21\u8fba\u306b $(k+1)$ \u3092\u305f\u3059\u3068
\n$1+2+\\cdots+k+(k+1)=\\dfrac{1}{2}k(k+1)+(k+1)=\\left(\\dfrac{1}{2}k+1\\right)(k+1)=\\dfrac{1}{2}(k+2)(k+1)$
\n$\\quad$\u3088\u3063\u3066\u3001$1+2+\\cdots+k+(k+1)=((*)$\u3067 $n=k+1$ \u3068\u304a\u3044\u305f\u5f0f\u306e\u53f3\u8fba$)$ \u304c\u6210\u308a\u7acb\u3064\u306e\u3067\u3001
\n$\\quad$$(*)$ \u306f $n=k+1$ \u306e\u3068\u304d\u3082\u6210\u308a\u7acb\u3064
\n(3) (1)(2)\u3000\u3088\u308a\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 $n$ \u306b\u3064\u3044\u3066 $(*)$ \u304c\u6210\u308a\u7acb\u3064\u3000(\u8a3c\u660e\u7d42)<\/span><\/div><\/div>\n<\/div>\n
\n

\u30c7\u30fc\u30bf\u306e\u6574\u7406<\/span><\/p>\n

\uff11\uff0e\u6570\u5024\u30c7\u30fc\u30bf $\\{1,1,1,2,3,4,4,4,5,5\\}$ \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1) \u5ea6\u6570\u5206\u5e03\u8868\u3092\u4f5c\u308a\u3001\u5e73\u5747$=\\mu$\u3001\u5206\u6563$=\\sigma^2$\u3001\u6a19\u6e96\u504f\u5dee$=\\sigma$ \u304a\u3088\u3073\u6700\u983b\u5024$=m$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u30c7\u30fc\u30bf\u30e9\u30d9\u30eb\u304c $x$ \u306e\u3068\u304d\u3000(\u516c\u5f0f) \u5206\u6563$=\\sigma^2=\\mu_{x^2}-(\\mu_x)^2$\u3000\u3092\u7528\u3044\u308b
\n\u30c7\u30fc\u30bf\u30e9\u30d9\u30eb\u3092 $x$\u3001\u5ea6\u6570\u3092 $f$ \u3068\u3057\u3066\u5ea6\u6570\u5206\u5e03\u8868\u3092\u4f5c\u308b \u8868\u3088\u308a \u6700\u983b\u5024$=m=1,4$\uff08\u7b54\uff09<\/span><\/p>\n\n\n\n\n\n\n
$x$<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n\u8a08<\/td>\n<\/tr>\n
$f$<\/td>\n3<\/td>\n1<\/td>\n1<\/td>\n3<\/td>\n2<\/td>\n10<\/td>\n<\/tr>\n
$xf$<\/td>\n3<\/td>\n2<\/td>\n3<\/td>\n12<\/td>\n10<\/td>\n30<\/td>\n<\/tr>\n
$x^2f$<\/td>\n3<\/td>\n4<\/td>\n9<\/td>\n48<\/td>\n50<\/td>\n114<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

$x$\u306e\u5e73\u5747$=\\mu=\\mu_x=\\dfrac{xf\u306e\u5408\u8a08}{f\u306e\u5408\u8a08}=\\dfrac{30}{10}=3$\uff08\u7b54\uff09\u3001$x^2$\u306e\u5e73\u5747$=\\mu_{x^2}=\\dfrac{x^2f\u306e\u5408\u8a08}{f\u306e\u5408\u8a08}=\\dfrac{114}{10}=11.4$
\n\u3088\u3063\u3066\u3001\u5206\u6563$=\\sigma^2=\\mu_{x^2}-(\\mu_x)^2=11.4-3^2=2.4$\uff08\u7b54\uff09\u3001\u6a19\u6e96\u504f\u5dee$=\\sigma=\\sqrt{\\sigma^2}=\\sqrt{2.4}$\uff08\u7b54\uff09
\n\u6ce8\u610f\uff1a\u5206\u6563\u306e\u5b9a\u7fa9\u306f\u300c(\u504f\u5dee\u306e\uff12\u4e57)$=(x-\\mu)^2$\u300d\u306e\u5e73\u5747\u306a\u306e\u3067\u3001\u4e0b\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u304b\u3089\u76f4\u63a5\u6c42\u3081\u3066\u3082\u826f\u3044
\n$\\quad \\quad \\sigma^2=\\dfrac{(x-\\mu)^2f\u306e\u5408\u8a08}{f\u306e\u5408\u8a08}=\\dfrac{24}{10}=2.4$<\/span><\/div><\/div>\n

(2) \u6700\u5927\u5024=Max\u30fb\u6700\u5c0f\u5024=Min\u304a\u3088\u3073\u4e2d\u592e\u5024$=Q_2$\u3001\u7b2c\uff11\u56db\u5206\u4f4d\u6570$=Q_1$\u3001\u7b2c3\u56db\u5206\u4f4d\u6570$=Q_3$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u6700\u5927\u5024=Max$=5$\u3001\u6700\u5c0f\u5024=Min$=1$ \uff08\u7b54\uff09\u3000\u30c7\u30fc\u30bf\u306e\u500b\u6570\u306f\uff11\uff10\u500b\u3068\u5076\u6570\u306a\u306e\u3067\u3001
\n\u4e2d\u592e\u5024\u306f5\u756a\u76ee\u30686\u756a\u76ee\u306e\u5e73\u5747\u3092\u3068\u3063\u3066\u3001$Q_2=\\dfrac{3+4}{2}=3.5$\uff08\u7b54\uff09
\n\u7b2c\uff11\u56db\u5206\u4f4d\u6570\u306f\u524d\u534a\uff15\u500b\u306e\u4e2d\u592e\u5024\u306a\u306e\u3067\u3001\uff13\u756a\u76ee\u306e\u5024\u3060\u304b\u3089 $Q_1=1$\uff08\u7b54\uff09
\n\u7b2c\uff14\u56db\u5206\u4f4d\u6570\u306f\u5f8c\u534a\uff15\u500b\u306e\u4e2d\u592e\u5024\u306a\u306e\u3067\u3001\uff18\u756a\u76ee\u306e\u5024\u3060\u304b\u3089 $Q_3=4$\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3)\u7bb1\u30d2\u30b2\u56f3\u3092\u66f8\u3051<\/p>\n

<\/span>\u89e3\u7b54<\/div>
Min$=Q_1=1$\u3001$\\mu=3$\u3001$Q_2=3.5$\u3001$Q_3=4$\u3001Max$=5$ \u306b\u6ce8\u610f\u3059\u308b<\/span>
\n\"\"<\/div><\/div>\n

2. $y=\\dfrac{x-30}{5}$ \u3068\u5909\u6570\u5909\u63db\u3057\u3066 $y$ \u306b\u95a2\u3059\u308b\u5e73\u5747 $E(y)$ \u3068\u5206\u6563 $V(y)$ \u3092\u6c42\u3081\u305f\u3089
\n$\\;\u3000E(y)=3, V(y)=2$ \u3067\u3042\u3063\u305f\u3002\u3000$x$ \u306b\u95a2\u3059\u308b\u5e73\u5747 $E(x)$ \u3068\u5206\u6563 $V(x)$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $E(ax+b)=aE(x)+b$\u3000\u304a\u3088\u3073 $V(ax+b)=a^2V(x)$ \u3092\u7528\u3044\u308b
\n$y=\\dfrac{x-30}{5}$ \u3088\u308a $x=5y+30$ \u3060\u304b\u3089\u3001\u516c\u5f0f\u3092\u9069\u7528\u3059\u308b\u3068
\n$E(x)=E(5y+30)=5E(y)+30=5\\cdot3+30=45$ (\u7b54)
\n$V(x)=V(5y+30)=5^2\\cdot V(y)=25\\cdot2=50$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u6570\u5217\u306e\u6975\u9650\u3068\u95a2\u6570\u306e\u6975\u9650<\/span><\/p>\n

\uff11\uff0e\u6b21\u306e\u6975\u9650\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{3n^2-5n+2}{2n^2+n+1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{3n^2-5n+2}{2n^2+n+1}=\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{(3n^2-5n+2)\\times\u3000\\dfrac{1}{n^2}}{(2n^2+n+1)\\times\u3000\\dfrac{1}{n^2}}=\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{3-\\dfrac{5}{n}+\\dfrac{2}{n^2}}{2+\\dfrac{1}{n}+\\dfrac{1}{n^2}}$
\n$=\\dfrac{3-0+0}{2+0+0}=\\dfrac{3}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{7^{n+3}-5^n}{7^{n+2}+5^{n+1}}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{7^{n+3}-5^n}{7^{n+2}+5^{n+1}}=\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{(7^{n+3}-5^n)\\times\\dfrac{1}{7^n}}{(7^{n+2}+5^{n+1})\\times\\dfrac{1}{7^n}}=\\displaystyle{\\lim_{n \\to \\infty}}\\dfrac{7^3-\\left(\\dfrac{5}{7}\\right)^n}{7^2+5\\cdot\\left(\\dfrac{5}{7}\\right)^n}$
\n$=\\dfrac{7^3-0}{7^2+5\\cdot0}=7$ (\u7b54)<\/span><\/div><\/div>\n

(3) $\\displaystyle{\\lim_{x \\to 4}}\\dfrac{x^2-x-12}{x-4}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{x \\to 4}}\\dfrac{x^2-x-12}{x-4}=\\displaystyle{\\lim_{x \\to 4}}\\dfrac{\\bcancel{(x-4)}(x+3)}{\\bcancel{x-4}}=\\displaystyle{\\lim_{x \\to 4}}(x+3)=4+3=7$ (\u7b54)<\/span><\/div><\/div>\n

(4) $\\displaystyle{\\lim_{x \\to \\infty}}\\left( \\sqrt{x^2+4x+3}-\\sqrt{x^2+x+1} \\right)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5206\u5b50\u3092\u6709\u7406\u5316\u3059\u308b
\n$\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{\\sqrt{x^2+4x+3}-\\sqrt{x^2+x+1}}{1}$
\n$=\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{\\sqrt{x^2+4x+3}-\\sqrt{x^2+x+1}}{1}\\times\\dfrac{\\sqrt{x^2+4x+3}+\\sqrt{x^2+x+1}}{\\sqrt{x^2+4x+3}+\\sqrt{x^2+x+1}}$
\n$=\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{(x^2+4x+3)-(x^2+x+1)}{\\sqrt{x^2+4x+3}+\\sqrt{x^2+x+1}}=\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{3x+2}{\\sqrt{x^2+4x+3}+\\sqrt{x^2+x+1}}$
\n$=\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{(3x+2)\\times\\dfrac{1}{x}}{\\left(\\sqrt{x^2+4x+3}+\\sqrt{x^2+x+1}\\right)\\times\\dfrac{1}{x}}$
\n$=\\displaystyle{\\lim_{x \\to \\infty}}\\dfrac{3+\\dfrac{2}{x}}{\\sqrt{1+\\dfrac{4}{x}+\\dfrac{3}{x^2}}+\\sqrt{1+\\dfrac{1}{x}+\\dfrac{1}{x^2}}}=\\dfrac{3+0}{\\sqrt{1+0+0}+\\sqrt{1+0+0}}=\\dfrac{3}{2}$ (\u7b54)<\/span><\/div><\/div>\n

\u7b49\u6bd4\u7d1a\u6570\u3068\u7121\u9650\u548c<\/span><\/p>\n

\uff11\uff0e\u6b21\u306e\u7b49\u6bd4\u7d1a\u6570\u306e\u53ce\u675f\u30fb\u767a\u6563\u3092\u8abf\u3079\u3001\u53ce\u675f\u3059\u308b\u5834\u5408\u306f\u305d\u306e\u5024\u3082\u6c42\u3081\u3088\u3002<\/p>\n

(1) $S=2+\\dfrac{3}{2}+\\dfrac{9}{8}+\\cdots$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$-1<r<1$\u306a\u3089\u3070\u3001$S=a+ar+ar^2+\\cdots=\\dfrac{a}{1-r}$ \u3092\u7528\u3044\u308b
\n$-1<r=\\dfrac{3}{4}<1$\u3001$a=2$\u3000\u3060\u304b\u3089\u3001
\n$S=2+\\dfrac{3}{2}+\\dfrac{9}{8}+\\cdots=\\dfrac{2}{1-\\dfrac{3}{4}}\\times\\dfrac{4}{4}=\\dfrac{8}{4-3}=8$ (\u7b54)<\/span><\/div><\/div>\n

(2) $S=2+\\dfrac{5}{2}+\\dfrac{25}{8}+\\cdots$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$-1<r<1$\u4ee5\u5916\u306e\u5834\u5408\u306f\u3001$S=a+ar+ar^2+\\cdots$ \u306f\u300c\u767a\u6563\u300d\u3059\u308b\u3053\u3068\u3092\u7528\u3044\u308b
\n$r=\\dfrac{5}{4}\\geq1$ \u3060\u304b\u3089\u3001$S=2+\\dfrac{5}{2}+\\dfrac{25}{8}+\\cdots$ \u306f\u300c\u767a\u6563\u300d\u3059\u308b (\u7b54)<\/span><\/div><\/div>\n

\uff12\uff0e\u6b21\u306e\u548c\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle{\\sum_{k=1}^{n}}\\left(\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}\\right)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$a_k=3k+1$ \u3068\u304a\u304f\u3068\u3001$a_{k+1}=3(k+1)+1=3k+4$ \u3060\u304b\u3089\u3001
\n$\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}=\\dfrac{1}{a_k}-\\dfrac{1}{a_{k+1}}$ \u304a\u3088\u3073
\n$\\dfrac{1}{a_1}=\\dfrac{1}{3\\cdot1+1}=\\dfrac{1}{4}$\u3001$\\dfrac{1}{a_{n+1}}=\\dfrac{1}{3n+4}$ \u306b\u6ce8\u610f\u3059\u308b\u3068
\n$\\displaystyle{\\sum_{k=1}^{n}}\\left(\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}\\right)=\\displaystyle{\\sum_{k=1}^{n}}\\left(\\dfrac{1}{a_k}-\\dfrac{1}{a_{k+1}}\\right)$
\n$=\\left(\\dfrac{1}{a_1}-\\bcancel{\\dfrac{1}{a_{2}}}\\right)+\\left(\\bcancel{\\dfrac{1}{a_2}}-\\bcancel{\\dfrac{1}{a_{3}}}\\right)+\\left(\\bcancel{\\dfrac{1}{a_3}}-\\bcancel{\\dfrac{1}{a_{4}}}\\right)+\\cdots+\\left(\\bcancel{\\dfrac{1}{a_n}}-\\dfrac{1}{a_{n+1}}\\right)$
\n$=\\left(\\dfrac{1}{a_1}-\\dfrac{1}{a_{n+1}}\\right)=\\dfrac{1}{4}-\\dfrac{1}{3n+4}\\left(=\\dfrac{3n}{4(3n+4)}\\right)$ (\u7b54)<\/span><\/div><\/div>\n

(2) $\\displaystyle{\\sum_{k=1}^{\\infty}}\\left(\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}\\right)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(1)\u306e\u7d50\u679c\u3092\u7528\u3044\u308b
\n$\\displaystyle{\\sum_{k=1}^{\\infty}}\\left(\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}\\right)=\\displaystyle{\\lim_{n \\to \\infty}}\\displaystyle{\\sum_{k=1}^{n}}\\left(\\dfrac{1}{3k+1}-\\dfrac{1}{3k+4}\\right)$
\n$=\\displaystyle{\\lim_{n \\to \\infty}}\\left(\\dfrac{1}{4}-\\dfrac{1}{3n+4}\\right)=\\dfrac{1}{4}-0=\\dfrac{1}{4}$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u5e73\u5747\u5909\u5316\u7387\u3068\u5fae\u5206\u306e\u5b9a\u7fa9<\/span><\/p>\n

\uff11\uff0e\u95a2\u6570 $y=f(x)=x^2-3x$ \u306b\u3064\u3044\u3066\u3001\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1)\u533a\u9593$[1,3]$\u306b\u304a\u3051\u308b\u5e73\u5747\u5909\u5316\u7387 $\\dfrac{\\Delta y}{\\Delta x}$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n$f(1)=1^2-3\\cdot1=-2$\u3001$f(3)=3^2-3\\cdot3=0$ \u3088\u308a
\n$\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{f(3)-f(1)}{3-1}=\\dfrac{0-(-2)}{2}=1$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2)\u5b9a\u7fa9\u306b\u5f93\u3063\u3066\u3001\u5c0e\u95a2\u6570 $y’=\\dfrac{dy}{dx}$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n$f(x+h)=(x+h)^2-3(x-h)=(x^2-3x)+h(2x-3+h)$ \u3088\u308a
\n$f(x+x)-(x^2-3x)=f(x+h)-f(x)=h(2x-3+h)$ \u3060\u304b\u3089
\n$y’=\\dfrac{dy}{dx}=\\displaystyle{\\lim_{h \\to 0}\\dfrac{f(x+h)-f(x)}{h}}=\\displaystyle{\\lim_{h \\to 0}\\dfrac{\\cancel{h}(2x-3+h)}{\\cancel{h}}}=2x-3$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u3079\u304d\u95a2\u6570\u3068\u6709\u7406\u95a2\u6570\u306e\u5fae\u5206<\/span><\/p>\n

\uff11\uff0e\u6b21\u306e\u95a2\u6570\u3092\u5fae\u5206\u305b\u3088\u3002<\/p>\n

(1) $y=3x^4+2x^3-7x+1$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\left( x^n \\right)^{\\prime}=nx^{n-1}$\u3001$(x)’=1$\u3001$(c)’=0$
\n\u304a\u3088\u3073\u5fae\u5206\u306e\u7dda\u5f62\u6027\u3000\u3000$(af+bg)’=af’+bg’$ \u3092\u7528\u3044\u308b
\n$\\dfrac{dy}{dx}=y’=3(x^4)’+2(x^3)’-7(x)’+(1)’=12x^3+6x^2-7$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $y=\\dfrac{1}{x^2}-2\\sqrt[4]{x^3}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\left( x^a \\right)^{\\prime}=ax^{a-1}$\u3001$\\dfrac{1}{x^n}=x^{-n}$\u3001$\\sqrt[n]{x^m}=x^{\\frac{m}{n}}$ \u3092\u7528\u3044\u308b
\n$y=x^{-2}-2x^{\\frac{3}{4}}$ \u3088\u308a
\n$\\dfrac{dy}{dx}=y’=-2x^{-3}-\\dfrac{3}{2}x^{-\\frac{1}{4}}=\\dfrac{-2}{x^3}-\\dfrac{3}{2\\sqrt[4]{x}}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3) $y=(x^2-x+1)^3$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\left( f^n \\right)^{\\prime}=nf’f^{n-1}$ \u3092\u7528\u3044\u308b
\n$y=(x^2-x+1)^3=f^3$ \u3068\u304a\u304f\u3068\u3001$f’=2x-1$ \u3088\u308a\u3000$\\dfrac{dy}{dx}=y’=3f’f^2=3(2x-1)(x^2-x+1)^2$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(4) $y=\\dfrac{3x+2}{x^2-x+1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\left( \\dfrac{f}{g^n} \\right)^{\\prime}=\\dfrac{f’g-nfg’}{g^{n+1}}$\u3001$\\left( \\dfrac{f}{g} \\right)^{\\prime}=\\dfrac{f’g-fg’}{g^{2}}$ \u3092\u7528\u3044\u308b
\n$y=\\dfrac{3x+2}{x^2-x+1}=\\dfrac{f}{g}$ \u3068\u304a\u304f\u3068 $f’=3$\u3001$g’=2x-1$ \u3088\u308a
\n$\\begin{array}{l} f’g &=& 3(x^2-x+1) &=& 3x^2-3x+3\\\\ fg’ &=& (3x+2)(2x-1) &=& 6x^2+x-2 \\\\ f’g-fg’ &=& &=&-3x^2-4x+5 \\end{array}$
\n$\\dfrac{dy}{dx}=y’=\\dfrac{-3x^2-4x+5}{(x^2-x+1)^2}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n<\/div>\n
\n

\u56de\u5e30\u5206\u6790<\/span><\/p>\n

\uff11\uff0e\u4e0b\u306e\u8868\u306f\u3001\u3042\u308b\u8a66\u9a13\u306e\u52c9\u5f37\u6642\u9593 $x$ \u3068\u5f97\u70b9 $y$ \u3092\u8868\u306b\u3057\u305f\u3082\u306e\u3067\u3042\u308b\u3002\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n\n\n\n\n
\u6642\u9593 $x$<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
\u5f97\u70b9 $y$<\/td>\n50<\/td>\n65<\/td>\n70<\/td>\n85<\/td>\n80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(1) \u52c9\u5f37\u6642\u9593 $x$ \u306e\u5e73\u5747 $\\mu_x$ \u3068\u5f97\u70b9 $y$ \u306e\u5e73\u5747 $\\mu_y$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\mu_x=\\dfrac{1+2+3+4+5}{5}=\\dfrac{15}{5}=3$\u3001$\\mu_y=\\dfrac{50+65+70+85+80}{5}=\\dfrac{350}{5}=70$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) \u52c9\u5f37\u6642\u9593 $x$ \u306e\u5206\u6563 $\\sigma_x^2$\u3001\u5f97\u70b9 $y$ \u306e\u5206\u6563 $\\sigma_y^2$\u3001$x$\u3068$y$\u306e\u5171\u5206\u6563$\\sigma_{xy}$ \u304a\u3088\u3073 \u76f8\u95a2\u4fc2\u6570 $r$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\sigma_x^2=\\overline{(x-\\mu_x)^2}=\\overline{(x-3)^2}$
\n$\\quad=\\dfrac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}$
\n$\\quad=\\dfrac{4+1+0+1+4}{5}=\\dfrac{10}{5}=2$ \uff08\u7b54\uff09
\n$\\sigma_y^2=\\overline{(y-\\mu_y)^2}=\\overline{(y-70)^2}$
\n$\\quad=\\dfrac{(50-70)^2+(65-70)^2+(70-70)^2+(85-70)^2+(80-70)^2}{5}$
\n$\\quad=\\dfrac{400+25+0+225+100}{5}=\\dfrac{750}{5}=150$ \uff08\u7b54\uff09
\n$\\sigma_{xy}=\\overline{(x-\\mu_x)(y-\\mu_y)}=\\overline{(x-3)(y-70)}$
\n$\\quad=\\dfrac{(-2)\\cdot(-20)+(-1)\\cdot(-5)+0\\cdot0+1\\cdot15+2\\cdot10}{5}=\\dfrac{80}{5}=16$ \uff08\u7b54\uff09
\n$r=\\dfrac{\\sigma_{xy}}{\\sqrt{\\sigma_x^2}\\sqrt{\\sigma_y^2}}=\\dfrac{16}{\\sqrt{2}\\sqrt{150}}=\\dfrac{16}{\\sqrt{300}}=\\dfrac{16}{10\\sqrt{3}}=\\dfrac{8\\sqrt{3}}{15}\\left(\\fallingdotseq0.92\\right)$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3) \u56de\u5e30\u76f4\u7dda\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$y=\\dfrac{\\sigma_{xy}}{\\sigma_x^2}(x-\\mu_x)+\\mu_y$\u3000\u306b(1)\u3001(2)\u306e\u7d50\u679c\u3092\u4ee3\u5165\u3057\u3066
\n$y=\\dfrac{16}{2}(x-3)+70$ \u3088\u308a $y=8x+46$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(\u81ea\u7531\u5ea61\u306e)\u72ec\u7acb\u6027\u306e\u30ab\u30a4\uff12\u4e57\u691c\u5b9a<\/span><\/p>\n

\uff11\uff0e\u98f2\u6599\u6c34A\u3068B\u306b\u95a2\u3057\u3066\u3001\u3069\u3061\u3089\u304c\u597d\u304d\u304b\u30a2\u30f3\u30b1\u30fc\u30c8\u3092\u53d6\u3063\u305f\u7d50\u679c\u304c\u4e0b\u306e\u8868\u3067\u3042\u308b\u3002\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n\n\n\n\n\n\n
<\/td>\n\u98f2\u6599\u6c34A<\/td>\n\u98f2\u6599\u6c34B<\/td>\n\u8a08<\/td>\n<\/tr>\n
\u7537<\/td>\n$a=$60<\/td>\n$b=$40<\/td>\n$C=$100<\/td>\n<\/tr>\n
\u5973<\/td>\n$c=$40<\/td>\n$d=$10<\/td>\n$D=$50<\/td>\n<\/tr>\n
\u8a08<\/td>\n$A=$100<\/td>\n$B=$50<\/td>\n$N=$150<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(1) \u691c\u5b9a\u7d71\u8a08\u91cf $T$ \u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u691c\u5b9a\u7d71\u8a08\u91cf\u3092\u6c42\u3081\u308b\u7c21\u6613\u516c\u5f0f\u306b\u4ee3\u5165\u3057\u3066
\n$T=\\dfrac{(ad-bc)^2N}{ABCD}=\\dfrac{(60\\cdot10-40\\cdot40)^2\\cdot150}{100\\cdot50\\cdot100\\cdot50}=6$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) \u5e30\u7121\u4eee\u8aac $H_0$: A\u3068B\u306e\u597d\u307f\u306b\u95a2\u3057\u3066\u7537\u5973\u5dee\u304c\u306a\u3044
\n$\\;\u3000$\u5bfe\u7acb\u4eee\u8aac $H_1$: A\u3068B\u306e\u597d\u307f\u306b\u95a2\u3057\u3066\u7537\u5973\u5dee\u304c\u3042\u308b
\n\u3068\u8a2d\u5b9a\u3057\u3066\u6709\u610f\u6c34\u6e965%\u3067\u30ab\u30a4\uff12\u4e57\u691c\u5b9a\u3092\u5b9f\u65bd\u3059\u308b\u3068\u304d\u3001
\n\u5e30\u7121\u4eee\u8aac $H_0$ \u304c\u68c4\u5374\u3055\u308c\u308b\u304b\u53d7\u5bb9\u3055\u308c\u308b\u304b\u3092\u5224\u5b9a\u305b\u3088
\n(\u305f\u3060\u3057\u3001\u81ea\u7531\u5ea6\uff11\u306e\u30ab\u30a4\uff12\u4e57\u5206\u5e03\u306e\u81e8\u754c\u5024\u306f\u3001\u6709\u610f\u6c34\u6e965%\u3067\u7d043.84\u3067\u3042\u308b\u3053\u3068\u3092\u4f7f\u3063\u3066\u826f\u3044)<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$T=6\\geq3.84$ \u306a\u306e\u3067\u3001\u5e30\u7121\u4eee\u8aac $H_0$ \u306f\u68c4\u5374\u3055\u308c\u308b \uff08\u7b54\uff09<\/span><\/div><\/div>\n<\/div><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

\u518d\u751f\u30ea\u30b9\u30c8 4H1_\u524d\u534a 4H1_\u5f8c\u534a 2H1 4H2_\u524d\u534a 4H2_\u5f8c\u534a 2H2<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[42,14,4,7],"tags":[],"class_list":["post-184","post","type-post","status-publish","format-standard","hentry","category-si","category-top","category-math2","category-math2-1q"],"_links":{"self":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/184","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=184"}],"version-history":[{"count":527,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/184\/revisions"}],"predecessor-version":[{"id":2000,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/184\/revisions\/2000"}],"wp:attachment":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=184"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=184"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=184"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}