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poizon [28]
2 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]2 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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Step-by-step explanation:

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Use the values on the number line to find the sampling error.
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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
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Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is [ {\rm{\$ 10,000 , \$ 15,000}], where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,  

The interval of the accepted bidding is [$10,000,$15,000],

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\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}

(d)  The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

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