Solve using the method of elimination , determine if the system has 1 solution, no solutions, or infinite solutions

2 answers:

8 0

$\stackrel{\textit{2nd equation }\times 2}{\begin{array}{rrrrr} -8x&+&2y&=&8\\ 2(4x&+&8y)&=&2(-4) \end{array}}\qquad \implies \qquad \begin{array}{rrrrr} -8x&+&2y&=&8\\ 8x&+&16y&=&-8\\\cline{1-5} 0&+&18y&=&0 \end{array} \\\\\\ 18y=0\implies y=\cfrac{0}{18}\implies \boxed{y=0} \\\\\\ \stackrel{\textit{we know that}}{4x+8y=-4}\implies 4x+8(0)=-4\implies 4x=-4\implies x\cfrac{4}{-4}\implies \boxed{x=-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{one solution}}{(-1~~,~~0)}~\hfill$

you'd want to often enough check their slopes first, if the slopes differ, they meet, if equal, they're parallel and either never meet or meet everywhere.

iren2701 [21]2 years ago

6 0

Here is the work, the answer is no sol

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Answer:

The sample size is $n = 600$

Step-by-step explanation:

From the question we are told that

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$\alpha = 100 - 95$

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$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the values is

$Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining critical value of $\frac{\alpha }{2}$ instead of $\alpha$ is because

$\alpha$ represents the area under the normal curve where the confidence level interval ( $1-\alpha$ ) did not cover which include both the left and right tail while