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poizon [28]
3 years ago
12

A rectangular yard is fenced in using 319 feet of custom fence Your friends really like the fence and decide to fence in their y

ard using the same fence.
Their yard is similar but has a scale factor of 8/5 times the size of
yours, how much fence to the nearest foot, will they have to purchase?

A) about 321 feet
B) about 510 feet
C) about 542 feet
D) about 199 feet​
Mathematics
1 answer:
Vsevolod [243]3 years ago
5 0

We have been given that a rectangular yard is fenced in using 319 feet of custom fence Your friends really like the fence and decide to fence in their yard using the same fence.  Their yard is similar but has a scale factor of 8/5 times the size of  yours.

We are asked to find the length of fence that your friends should purchase.

Since the size of your friend's yard is \frac{8}{5} times your yard, so your friend should purchase \frac{8}{5} times the 319 feet.

\text{Length of fence your friend should purchase}=\frac{8}{5}\times 319\text{ ft}

\text{Length of fence your friend should purchase}=8\times 63.8\text{ ft}

\text{Length of fence your friend should purchase}=510.4\text{ ft}

Upon rounding our answer to nearest foot, we will get:

\text{Length of fence your friend should purchase}\approx 510\text{ ft}

Therefore, your friends should purchase 510 feet of fence and option C is the correct choice.

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Answer:

\mathbf{P(X=5) =0.0888}    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

\mathtt{P(X=x) =(^{n}_{x} )   \  \pi^x \  (1-\pi)^{n-x}}

\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} )   \  \pi^x \  (1-\pi)^{n-x}}

where;

n = 8 and π = 0.36

For x = 5

The probability \mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} )   \  0.36^5 \  (1-0.36)^{8-5}}

\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} )   \  0.36^5 \  (0.64)^{3}}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =({8 \times 7 } )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =0.0887645}

\mathbf{P(X=5) =0.0888}     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})

{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times  (0.36)^0  \times  (1-0.36)^8  \ )  +  \dfrac{8!}{1!(7!)} \times  (0.36)^1  \times  (1-0.36)^7  \ +\dfrac{8!}{2!(6!)} \times  (0.36)^2  \times  (1-0.36)^6  \ +  \dfrac{8!}{3!(5!)} \times  (0.36)^3  \times  (1-0.36)^5 +  \dfrac{8!}{4!(4!)} \times  (0.36)^4  \times  (1-0.36)^4  \  +  \dfrac{8!}{5!(3!)} \times  (0.36)^5  \times  (1-0.36)^3  \ )

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

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