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artcher
8 days ago
13

If the farmer has 234 feet of fencing, what are the dimensions of the region which enclose the maximal area?

Mathematics
2 answers:
lawyer [11.3K]8 days ago
8 0
The dimensions are 58 ft × 58 ft. Step-by-step explanation: Let the length of the region be represented as x feet, and the width as y feet. Given a perimeter of 234 feet, the area A can be represented as xy. By differentiating the equation with regards to x, we can determine the point of maximum area, revealing that for x = 58.5 feet, the area's maximum occurs when both dimensions are 58.5 ft.
lawyer [11.3K]8 days ago
7 0
The answer is: Let the area to be enclosed be a square with side length s. The perimeter equals 234 feet. Since the perimeter of a square is 4×side, solving gives s = 58.5 feet. The area of the square is side × side, resulting in a maximum area that can be enclosed equal to square feet. Thus, the dimensions will be 58.5 feet by 58.5 feet.
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Graph the line that represents a proportional relationship between ddd and ttt with the property that an increase of 0.20.20, po
Leona [11225]

Answer:

Part a) The unit rate of change of d with respect to t is equal to 9

Part b) The graph of the line corresponding to the attached diagram

Step-by-step explanation:

we know that

The relationship between the variables t and d depicts a proportional change if described in the form d/t=k or d=kt

step 1

Determine the value of k

we have

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so

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28 days ago
An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the
Zina [11254]

Answer:

The chance that there is one or more defective integrated circuits is 33.10%.

Detailed solution:

Each integrated circuit can be either defective or not, which provides only two possible outcomes. As a result, this scenario is well-suited to be analyzed using the binomial probability distribution.

About the binomial distribution

This distribution calculates the likelihood of exactly x successes in n repeated trials where each trial has two possible results.

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

Here, C_{n,x} represents the count of combinations of x items selected from n elements, described by the formula:

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

And \pi is the probability of the event X occurring.

Applying it to the problem

The product contains 40 integrated circuits, so n = 40.

The probability that an individual integrated circuit is defective is 0.01, so \pi = 0.01.

Finding the probability of at least one defective circuit

There are two cases: either at least one integrated circuit is defective (probability P(X > 0)) or none are defective (probability P(X = 0)). Since probabilities sum to 1, we want to determine P(X>0).

P(X > 0) + P(X = 0) = 1

P(X > 0) = 1 - P(X = 0)

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

P(X = 0) = C_{40,0}.(0.01)^{0}.(0.99)^{40} = 0.6690

P(X > 0) = 1 - P(X = 0) = 1 - 0.6690 = 0.3310

Hence, the probability of having one or more defective integrated circuits is 33.10%.

3 0
1 month ago
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If your grocery bills over the last four weeks were $203, $195, $180 and $238, what was the average bill
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To solve this problem, simply sum all of the grocery expenses and then divide that total by 4...

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Joshua Freeman has a family membership in his company’s health insurance plan with the benefits shown. His recent non-network he
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17 days ago
The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm
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Answer:

(a) P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To address this problem, one needs to grasp the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems involving normally distributed samples are resolved through the application of the z-score formula.

In a data set with a mean \mu and standard deviation \sigma, the z-score for a measurement X is defined as:

Z = \frac{X - \mu}{\sigma}

The Z-score indicates how many standard deviations the measurement deviates from the mean. After determining the Z-score, one can consult the z-score table to find the relevant p-value for that score. This p-value represents the probability that the measure's value is less than X, which indicates the percentile of X. By subtracting this p-value from 1, we obtain the likelihood that the measure's value exceeds X.

Central Limit Theorem

According to the Central Limit Theorem, for a normally distributed random variable X with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated by a normal distribution, characterized by mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

The Central Limit Theorem can also be applicable to a skewed variable, provided that n is 30 or more.

In this case, we are given that:

\mu = 13, \sigma = 0.08

(a) Compute P(12.99 ≤ X ≤ 13.01) when n = 16.

In this scenario, we find n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02

This probability equates to the p-value of Z at X = 13.01 minus the p-value of Z at X = 12.99.

X = 13.01

Z = \frac{X - \mu}{\sigma}

According to the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

has a p-value of 0.6915

Z = 0.5

X = 12.99

Z = \frac{X - \mu}{s}

Z = \frac{12.99 - 13}{0.02}

Z = -0.5

Z = -0.5 yields a p-value of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) What is the probability that the sample mean diameter is greater than 13.01 when n = 25?

P(X ≥ 13.01) =

This is obtained by subtracting the p-value of Z when X = 13.01 from 1. Hence

Z = \frac{X - \mu}{s}

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a p-value of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

7 0
20 days ago
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