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3 years ago

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The angle of elevation of a cliff top from a small boat on the sea is 36º. The boat is 100m from the foot of the vertical cliff.

Draw a diagram, and then calculate the height of the cliff.

1 answer:

storchak [24]3 years ago

3 0

Answer:

72.65m (2dp) is the height of the cliff

Step-by-step explanation:

The height of the cliff can be calculated using trigonometry

Since we know the angle of elevation and the adjacent = side of the angle, and we are looking for the opposite side of the angle, which say is h, we will use tan since we are not dealing with the hypotenuse

tan(36) = h/100

100tan(36) = h

h = 72.65m (2dp)

I have attached a diagram as reference

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Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.

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The Lagrangian for this function and the given constraints is

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)$

which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

Check the Hessian for $f(x,y,z)$ , given by

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite, since $\mathbf v^\top\mathbf{Hv}>0$ for any vector $\mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top$ , which means $f(x,y,z)=x^2+y^2+z^2$ attains a minimum value of $\dfrac{480}{11}$ at $\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ . There is no maximum over the given constraints.

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3 years ago

After reading The Lord of the Rings, Francesca signs up for archery lessons. At her first lesson, she sets up her target 5 feet

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Answer:

$3$ times

Step-by-step explanation:

$5x=15$

Divide both sides by 5

$x=3$

Hope this helps

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3 years ago

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⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

<u>95% confidence interval for p</u> = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

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