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geniusboy [140]

2 years ago

15

Point K is on line segment JL. Given KL = 2x – 2, JL = 4x + 9, and

1 answer:

sergeinik [125]2 years ago

7 0

Answer:

JL = 21

Step-by-step explanation:

Given that K is on line segment JL, therefore:

KL + JK = JL (according to segment addition postulate)

KL = 2x - 2

JK = 5x + 2

JL = 4x + 9

Thus:

$(2x - 2) + (5x + 2) = (4x + 9)$

Solve for x

$2x - 2 + 5x + 2 = 4x + 9$

$2x +5x - 2 + 2 = 4x + 9$

$7x = 4x + 9$

Subtract 4x from both sides

$7x - 4x = 4x + 9 - 4x$

$3x = 9$

Divide both sides by 3

$\frac{3x}{3} = \frac{9}{3}$

$x = 3$

Find the numerical length of JL

$JL = 4x + 9$

Plug in the value of x

$JL = 4(3) + 9 = 12 + 9 = 21$

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

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a

$\= x = 18.5$ , $\sigma = 5.15$

b

$15.505 < \mu < 21.495$

c

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

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

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

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

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

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

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