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Vlada [557]
3 years ago
5

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 40' Building Als 120 feet away, and b

uilding B is 60 feet away which building is taller and by approximately how many feet?
Building A It is around 100 69 feet taller than building B
Building A it is around 50 34 feet taller than building B
O Building B, it is around 100.69 feet taller than building A
Building B, it is around 50.34 feet taller than building A
Mathematics
2 answers:
-BARSIC- [3]3 years ago
5 0

Answer:

A. Building A, it is around 100.69 feet taller than building B.

Step-by-step explanation:

Since building A is 120 feet away at and elevation of 40^{0}, then:

Tan 40^{0} = \frac{x}{120}

Where x is the height of the building at the horizontal plane.

x = Tan  40^{0} × 120

   = 100.692 feet

For building B at the same elevation,

Tan  40^{0} = \frac{x}{60}

 x = Tan  40^{0} × 60

    = 50.346 feet

Since the angle of elevation are the same, building A is around 100.69 feet taller than building B.

OlgaM077 [116]3 years ago
3 0

Answer:

It is Building A is 50.34 feet taller than Building B. You are supposed to subtract both values.

Step-by-step explanation:

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And replacing we got:

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Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

P(X=3) = 0.1

Part b

We want this probability:

P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

Part c

For this case we want this probability:

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Part e

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E(X)= \sum_{i=1}^n X_i P(X_i)

And replacing we got:

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Part f

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E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

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And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

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