What is 73/50 as a percent

Bob sees a very tall white pine tree. He estimates that the angle of elevation from him to the top of the tree is 77°

vlabodo [156]

Answer:

86.63 ft

Step-by-step explanation:

The tangent relation applies.

Tan = Opposite/Adjacent

tan(77°) = (tree height)/(20 ft)

tree height = (20 ft)·tan(77°) ≈ 86.63 ft

_____

<em>Additional comment</em>

It is somewhat problematic for Bob to see the top of a white pine tree at that angle of elevation. His view would likely be obscured by the branches of the tree.

8 0

2 years ago

What is the Mean Average deviation from the Mean?<br> Numbers: 6, 2, 3, 3, 2, 2

Irina-Kira [14]

Answer:

2.42

Step-by-step explanation:

standard deviation= sqrt(variance)

Variance=E[X2] - E[X]2

E[X] = sum(X)/n=35/9, so E[X]2 = 1225/81

E[X2] =sum(X2)/n = 189/9 = 170/81

Variance= 476/81

Standard deviation = sqrt(476/81) ~2.42

6 0

3 years ago

You choose two different songs on a music playlist at random. Out of 80 songs on the playlist, 36 are hip hop songs. The first s

dexar [7]

Answer:

35/79

Step-by-step explanation:

Assuming you can't choose the same song a second time, the probability is 35/79.

If you can choose the same song, the probability is 36/80, which reduces to 9/20.

7 0

4 years ago

Filipo is building a rectangular sandbox . The length of the sandbox is 1 foot longer than twice the width of the sandbox .the p

Alla [95]

Answer:

Part A : D.) $2w + 2(2w+ 1) = 29$

Part B : Length of the sandbox is 10 feet.

Step-by-step explanation:

Given,

Perimeter = 29 ft

We need to find the equation for the perimeter and also the length of the sandbox.

Solution,

Let the width of the sandbox be 'w'.

Now as per question said;

The length of the sandbox is 1 foot longer than twice the width of the sandbox.

So we can say that;

Length = $2w+1$

Now we know that the perimeter is equal to the sum of twice of length and width.

framing in equation form, we get;

Perimeter = $2(2w+1)+2w$

we have given the perimeter, so on substituting the value, we get;

$29=2(2w+1)+2w$

Hence The equation used to find the width is $29=2(2w+1)+2w$ .

Now we solve for 'w'.

Applying distributive property, we get;

$4w+2+2w=29\\\\6w+2=29$

Subtracting both side by '2' we get

$6w+2-2=29-2\\\\6w=27$

Dividing both side by 6 we get;

$\frac{6w}{6}=\frac{27}{6}\\\\w=4.5\ ft$

Width of the sandbox = 4.5 ft

Length of the sandbox = $2w+1=2\times4.5 +1 = 9+1=10\ ft$

Hence Length of the sandbox is 10 feet.

3 0

3 years ago

A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o

vivado [14]

Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

7 0

3 years ago