Question

For a t distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746
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Answer 1

$\huge{\underline{\bf\red{Questi {\mathbb{O}} n} :}}$

$\sf Fine \: the \: value \: of \: 'a' \: for \: which \\ \sf the \: system \: of \: linear \: equations \\ x + ay = 2 \: { \sf{and}} \: 2x - 5y = 3 \\ \sf has \: no \: solution.$

$\huge{\underline{\: \bf\green{Answ {\mathbb{E}} r}\ \: :}}$

CASE ( i ) :-

$\boxed{\qquad \qquad a = - \frac{5}{2} \qquad \qquad }$

CASE ( ii ) :-

$\boxed{a = \bigg\{ x :x \in {\mathbb{R}},x \neq - \frac{10}{3} \bigg\}}$

$\huge{\underline{\bf\blue{Soluti {\mathbb{O}} n}\ :}}$

The given system of linear equations is of the form :—

$a_1 x + b_1 y + c_1 = 0$

$a_2 x + b_2 y + c_2 = 0$

where,

$a_1 = 1, \ b_1 = a, \ c_1 = -2$

$a_2 = 2, \ b_2 = -5, \ c_2 = -3$

Now,

$\because$ the given system of linear equations has no solution.

$\therefore$ $\boxed{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} }$

CASE ( i ) :

$\begin{aligned} \\ \qquad \ \frac{a_{1}}{a_{2}} & = \frac{b_{1}}{b_{2}} \\ \\ \implies\frac{1}{2} & = \frac{a}{ - 5} \\ \\ \implies a & = - \frac{5}{2} & & & & & & & \end{aligned}$

CASE ( ii ) :

$\begin{aligned} \\ \qquad \ \frac{b_{1}}{b_{2}} & \neq \frac{c_{1}}{c_{2}} \\ \\ \implies\frac{a}{ - 5} & \neq \frac{ - 2}{ - 3} \\ \\ \implies \quad a & \neq \frac{ - 10}{3} & & & & & & & \end{aligned}$

$\red{\rule{5.5cm}{0.02cm}}$

$\purple{\rule{7.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$