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3 years ago

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In the figure the triangle is right triangle at Y and N is the foot of the perpendicular from Y to XZ. Given that XY=6 cm

and XZ=10 cm. What is the length of XN

2 answers:

evablogger [386]3 years ago

6 0

Answer:

$\underline{ \underline{ \red{ \large {\tt{✧ G \: I \: V \: E \: N}}} }}:$

XY = 6 cm , XZ = 10 cm

$\underline{ \underline{ \purple{ \large{ \tt{✧ T \: O \: \: F\: I\: N\: D}}}}} :$

Length of XN

$\underline{ \underline{ \pink{\large {\tt{✧ S \: O \: L \: U\: T\: I \: O\: N}}}} }:$

Let the length of XN be ' x ' & that of NZ be ' 10 - x '

In rt. XYZ :

Hypotenuse (h ) = 10 , Perpendicular ( p ) = YZ & Base ( b ) = 6

Using Pythagoras theorem:

$\large{ \sf{p = \sqrt{ {h}^{2} - {b}^{2} } }}$

Plug the values & then simplify!

⇢ $\large{ \sf{YZ = \sqrt{( {10)}^{2} - {(6)}^{2} } }}$

⇢ $\large{ \sf{YZ = \sqrt{100 - 36}}}$

⇢ $\large{ \sf{YZ = \sqrt{64}}}$

⇢ $\large{ \sf{YZ = 8} \: cm} \:$

In rt. YNZ ,

Hypotenuse = 8 cm , Perpendicular = YN , base = 10 - x

Using Pythagoras theorem :

⇾ $\large{ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

⇾ $\large{ \sf{ {YN}^{2} = {8}^{2} - {(10 - x)}}}^{2}$

⇾ $\large{ \sf{ {YN}^{2} = 64 - \{ {(10)}^{2} - 2 \times 10 \times x + {(x)}^{2} \} }}$

⇾ $\large{ \sf{ {YN}^{2} = 64 - \{100 - 20x + {x}^{2} \}}}$

⇾ $\large{ \sf{ {YN}^{2} = 64 - 100 + 20x - {x}^{2} }}$

⇾ $\large{ \sf{ {YN}^{2} = 20x - {x}^{2} - 36}} \longrightarrow \: eq (i)$

In rt. XYN ,

Hypotenuse = 6 , Perpendicular = YN , base = x

Using Pythagoras theorem :

$\large{ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

⇾ $\large{ \sf{ {YN}^{2} = {(6)}^{2} - {(x)}}}^{2}$

⇾ $\large{ \sf{ {YN}^{2} = 36 - {x}^{2} \longrightarrow \: eq(ii) }}$

Now , From Equation ( i ) & Equation ( ii )

⤑ $\large{ \sf{20x - {x}^{2} - 36 = 36 - {x}^{2} }}$

⤑ $\large{ \sf{20x - \cancel{ {x}^{2} } - 36 = 36 - \cancel{ {x}^{2} }}}$

⤑ $\large{ \sf{20x = 36 + 36}}$

⤑ $\large{ \sf{20x = 72}}$

⤑ $\large{ \sf{x = 3.6}}$ cm

Hence, The length of XN is $\large{ \boxed{\red{ \bold{ \tt{3.6 \: cm}}}}}$

[ Correct me if I am wrong ]

♨ Hope I helped! ♡

♪ Have a wonderful day / night ! ☃

☥ $\underbrace{ \overbrace{{ \mathfrak{Carry \: On \: Learning}}}}$ ✎

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IrinaVladis [17]3 years ago

4 0

Answer:

3.6cm

Step-by-step explanation:

YZ = 10² - 6² = 8

sinX = 8/10 = 53.13º

cos 53.13º = XN/6

XN = 3.6cm

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3 years ago

Given u = 3i − 8j and v = −4i + 8j, what is u • v? 73 8 −76 −56

Ivan

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2 years ago

Weights and heights of turkeys tend to be correlated. For a population of turkeys at a farm, this correlation is found to be 0.6

LenaWriter [7]

Answer:

a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The average height for turkeys at the 90th percentile for weight is 34.554

Of the turkeys at the 90th percentile for weight, roughly the percentage that would be taller than 28 inches 79.37%

Step-by-step explanation:

Given that:

For a population of turkeys at a farm, the correlation found between the weights and heights of turkeys is r = 0.64

the average weight in pounds $\overline x$ = 17

the standard deviation of the weight in pounds $S_x$ = 5

the average height in inches $\overline y$ = 28

the standard deviation of the height in inches $S_y$ = 8

Also, given that the weight and height both roughly follow the normal curve

For this study , the slope of the regression line can be expressed as :

$\beta_1 = r \times ( \dfrac{S_y}{S_x})$

$\beta_1 = 0.64 \times ( \dfrac{8}{5})$

$\beta_1 = 0.64 \times 1.6$

$\beta_1 = 1.024$

To the intercept of the regression line, we have the following equation

$\beta_o = \overline y - \beta_1 \overline x$

replacing the values:

$\beta_o = 28 -(1.024)(17)$

$\beta_o = 28 -17.408$

$\beta_o = 10.592$

However, the regression line needed for this study can be computed as:

$\hat Y = \beta_o + \beta_1 X$

$\hat Y = 10.592 + 1.024 X$

Recall that;

both the weight and height roughly follow the normal curve

As such, the weight related to 90th percentile can be determined as shown below.

Using the Excel Function at 90th percentile, which can be computed as:

(=Normsinv (0.90) ; we have the desired value of 1.28

∴

$\dfrac{X - \overline x}{s_x } = 1.28$

$\dfrac{X - 17}{5} = 1.28$

$X - 17 = 6.4$

X = 6.4 + 17

X = 23.4

The predicted height $\hat Y = 10.592 + 1.024 X$

where; X = 23.4

$\hat Y = 10.592 + 1.024 (23.4)$

$\hat Y = 10.592 + 23.9616$

$\hat Y = 34.5536$

Now; the probability of predicted height less than 34.5536 can be computed as:

$P(Y < 34.5536) = P( \dfrac{Y - \overline y }{S_y} < \dfrac{34.5536-28}{8})$

$P(Y < 34.5536) = P(Z< \dfrac{6.5536}{8})$

$P(Y < 34.5536) = P(Z< 0.8192)$

From the Z tables;

P(Y < 34.5536) =0.7937

Hence, a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The average height for turkeys at the 90th percentile for weight is :

$\hat Y = 10.592 + 1.024 X$

where; X = 23.4

$\hat Y = 10.592 + 1.024 (23.4)$

$\hat Y = 10.592 + 23.962$

$\mathbf{\hat Y = 34.554}$

Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

i.e

P(Y >28) = 1 - P (Y< 28)

$P(Y >28) = 1 - P( Z < \dfrac{28 - 34.554}{8})$

$P(Y >28) = 1 - P( Z < \dfrac{-6.554}{8})$

$P(Y >28) = 1 - P( Z < -0.8193)$

From the Z tables,

$P(Y >28) = 1 - 0.2063$

$\mathbf{P(Y >28) = 0.7937}$

= 79.37%

7 0

4 years ago