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3 years ago

A, B, and C are midpoints of ∆XYZ. What is the length of cy

Mathematics

1 answer:

iogann1982 [59]3 years ago

5 0

Answer:

$XY=36\ units$

Step-by-step explanation:

The correct question is

A, B, and C are midpoints of ∆XYZ. What is the length of XY

we know that

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

step 1

Find the length of YZ

$AB=\frac{1}{2}YZ$

we have

$AB=24\ units$

substitute

$24=\frac{1}{2}YZ$

solve for YZ

$YZ=48\ units$

step 2

Find the length of XY

Applying Pythagoras Theorem in the right triangle XYZ

$XZ^2=XY^2+YZ^2$

substitute the given values

$60^2=XY^2+48^2$

solve for XY

$3,600=XY^2+2,304$

$XY^2=3,600-2,304$

$XY^2=1,296$

$XY=36\ units$

Applying the Midpoint Theorem

$BC=\frac{1}{2}XY$ -----> $BC=\frac{1}{2}(36)=18\ units$

$AC=\frac{1}{2}XZ$ -----> $AC=\frac{1}{2}(60)=30\ units$

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Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

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P.Q. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean speed for the 25-mil film = 1.15

$\bar X_1$ = sample mean speed for the 20-mil film = 1.06

$s_1$ = sample standard deviation for the 25-mil film = 0.11

$s_2$ = sample standard deviation for the 20-mil film = 0.09

$n_1$ = sample of 25-mil film = 8

$n_2$ = sample of 20-mil film = 8

$\mu_1$ = population mean speed for the 25-mil film

$\mu_2$ = population mean speed for the 20-mil film

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }$ = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ( $\mu_1-\mu_2$ ) is;

P(-2.624 < $t_1_4$ < 2.624) = 0.98 {As the critical value of t at 14 degrees of

freedom are -2.624 & 2.624 with P = 1%}

P(-2.624 < $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 2.624) = 0.98

P( $-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < $2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ) = 0.98

P( $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.98

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= [ $(1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ , $(1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ]

= [-0.042, 0.222]

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