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Sholpan [36]
3 years ago
5

A man, 1.5m tall, is on top of a building. He observes a car on the road at an angle of 75°. If the building is 30m, how far is

the car from the building?​
Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

The answer to this question is 8.44 m.

Step-by-step explanation:

This problem is best illustrated in the photo below.

First, we must know that the angle stated in the problem is the angle of depression. Angle of depression is the angle between the horizontal and the line of sight of the observer. On the other hand, if the observer is looking upward, then the angle between the horizontal and his line of sight is called the angle of elevation.

Since we have the given angle of depression, we must use its complementary angle to solve for the distance of the car from the building.

Let x = complementary angle of 75°

y = distance of the car from the building

To solve for x,

To solve for y, we need to use a trigonometric function that will relate the adjacent of 15° and the opposite of 15°. Let us look at the mnemonics

SOH: Sine =  

CAH: Cosine =  

TOA: Tangent =  

Therefore we must use the tangent function to solve for y.

 

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How do you graph -1/5x-1 and 4/5x-6
kakasveta [241]

Answer:

The points for the given to linear equations is (5 , - 2) and (5 , - 1)

The points is plotted on the graph shown .

Step-by-step explanation:

Given as :

The two linear equation are

y = \dfrac{-1}{5}x - 1                  ...........1

y = \dfrac{4}{5}x - 6                  ...........2

Now, Solving both the linear equations

Put the value of y from eq 2 into eq 1

I.e  \dfrac{4}{5}x - 6 = \dfrac{-1}{5}x - 1

Or, \dfrac{4}{5}x + \dfrac{1}{5}x  = 6 - 1

Or,  \dfrac{4 + 1}{5}x = 5

or, \dfrac{5}{5}x = 5

∴ x = 5

Now, Put the value of x in eq 1

So, y = \dfrac{-1}{5}x - 1      

Or, y = \dfrac{-1}{5}× 5 - 1              

or,  y = \dfrac{-5}{5} - 1

Or, y = - 1 - 1

I.e y = -2

So, For x = 5 , y = - 2

Point is (x_1 , y_1) = (5 , - 2)

Again , put the value of x in eq 2

So, y = \dfrac{4}{5}x - 6

Or, y = \dfrac{4}{5}× 5 - 6

Or, y = \frac{4\times 5}{5} - 6

Or, y = 4 - 6

I.e y = - 2

So, For x = 5 , y = - 2

Point is (x_2 , y_2) = (5 , - 2)

Hence, The points for the given to linear equations is (5 , - 2) and (5 , - 2)

The points is plotted on the graph shown . Answer

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Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund
lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

\sigma=\sqrt{\frac{p*(1-p)}{n} }

If n=1139 and p =0.96, the standard deviation is:

\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836

The minimum and maximum usual values are:

\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n

\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62

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