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3 years ago

7

△JKL has vertices at J(−2, 4), K(1, 6), and L(4, 4). Determine whether △JKL is a right triangle. Explain. please with full calcu

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1 answer:

KATRIN_1 [288]3 years ago

3 0

Answer:

below

Step-by-step explanation:

basically find the line that connects JK and KL, and since its 2/3 and -2/3 which are not opposite reciprocals, the points dont form a right triangle

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Leya [2.2K]

Answer:

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Step-by-step explanation:

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Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean score = 130

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<em>Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.0225 < $t_3_9$ < 2.0225) = 0.95 {As the critical value of t at 39 degree of

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P(-2.0225 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.0225) = 0.95

P( $-2.0225 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u><em /></u>

<u><em>95% confidence interval for</em></u> $\mu$ = [ $\bar X-2.0225 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.0225 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $130-2.0225 \times {\frac{10}{\sqrt{40} } }$ , $130+2.0225 \times {\frac{10}{\sqrt{40} } }$ ]

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Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

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