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vazorg [7]
3 years ago
5

A. David notices that the shape of the Ligersville toxic dumpsite is somewhat like a triangle. He claims that he can approximate

the perimeter of the dumpsite without doing any extra measurements. What assumption about side AB of the dumpsite would David need to make in order to justify his claim?
B. Based on the assumption in part A, determine the approximate length of side AB of the dumpsite. Explain your reasoning in at least two sentences.

C. Determine the approximate perimeter of the Ligersville toxic dumpsite. Explain your reasoning.

Mathematics
2 answers:
ziro4ka [17]3 years ago
6 0

Answer:

A. Side AB is approximately straight and forms the hypotenuse side of the right triangle

B. 40 yards

C. 96 yards

Step-by-step explanation:

A. The assumption about the curved side AB that David would need to make in order to justify his claim is that the side AB is straight and is the hypotenuse side of the right triangle.

B.  The length of AB is thus;

√(24² + 32²) = 40 yard

C. The approximate perimeter is therefore;

24 + 32 + 40 = 96 yards

allochka39001 [22]3 years ago
4 0

Answer:

A) AB is approximately straight line

B) 40 yards

C) 96 years

Step-by-step explanation:

A) For the toxic dump site to be traingular, AB must be a straight line to form hypotneuse of the traingle and that the traingle is right angle triangle

B) assuming a right angle triangle, using pythagoras theorem

H²=P²+B²

H²= 24²+ 32²

H= 40 yards

C) Parameter= 24+32+40=96 yards

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Read 2 more answers
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up
Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid \sum \limits ^{n}_{i=1} n^2 P(X=n)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

3 0
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