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Marianna [84]
3 years ago
6

A dog kennel owner has 147ft. of fencing to enclose a rectangular dog run. She wants it to be 6times as long as it is wide. Find

the dimensions
Mathematics
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

63 feet long and 10.5 feet wide

Step-by-step explanation:

147 is the maximum amount of fencing, which means that this is dealing with perimeter.

So we set up our equation:

6x + 6x + x + x ≤ 147

(x being the width, and 6x being the length, which is 6 times the width)

this simplifies to:

14x ≤ 147

simplify:

x ≤ 10.5

So the width of the fencing is 10.5 feet long, meaning that the length is 63 feet long.

Check answers:

6(10.5) + 6(10.5) + 10.5 + 10.5 = 147

63 + 63 + 21 = 147

126 + 21 = 147

147 = 147

It works, so that is the solution.

Hope this helped :)

igomit [66]3 years ago
7 0

Answer:

length = 63 ft; width = 10.5 ft

Step-by-step explanation:

Let the width be x.

The length is 6 times the width, so the length is 6x.

The perimeter is the sum of 2 lengths and 2 widths.

6x + 6x + x + x

The perimeter equals 147 ft.

6x + 6x + x + x = 147

14x = 147

x = 10.5

The width is 10.5 ft.

The length is 6x, so the length is 6 * 10.5 ft = 63 ft

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a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

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P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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