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Annette [7]
3 years ago
13

A rectangle has an area of 72 in². The length and the width of the rectangle are changed by a scale factor of 3.5.

Mathematics
2 answers:
yuradex [85]3 years ago
5 0

Answer:

Area of the rectangle(A) is given by:

where, l is the length and w is the width of the rectangle.

As per the statement:

A rectangle has an area of 72 in²

⇒A = 72 in²

then;

It is given that:

The length and the width of the rectangle are changed by a scale factor of 3.5.

Length of new rectangle become = 3.5l

Width of new rectangle become = 3.5 w

then;

Area of new rectangle  = (Length of new rectangle)(Width of new rectangle)

⇒Area of new rectangle = (3.5l)(3.5w) =12.25(lw)

Substitute the given values we have;

Therefore, the area of the new rectangle is, 882 in²

nydimaria [60]3 years ago
4 0

Answer:

the area of the new rectangle is 882 in^2

Step-by-step explanation:

Area of new rectangle = (3.5l)(3.5w) = 12.25

12.25 x 72 = 882

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<h3>How to evaluate the probability of a random variable getting at least some fixed value?</h3>

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Then, the probability of X attaining at least 'a' is written as:

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It is evaluated as:

P(X \geq a) = \sum_{\forall \: x_i \geq a} P(X = x_i)

The probability distribution of X is:

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        6                               0.02
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Worker working at least 8 hours means X attaining at least 8 as its values.

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P(X ≥ 8) = P(X = 8) + P(X = 9) +P(X = 10) = 0.85 ≈ 0.84 approx.

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Learn more about probability distributions here:

brainly.com/question/14882721

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Answer:

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Step-by-step explanation:

We know that area of a rectangle = length x width

We also know  that perimeter = (2 x Length) + (2 x Width)

the goal is to to find Length and Width such that:

Condition 1: Length x width = 2020

Condition 2: (2 x Length) + (2 x Width)= maximum possible

We are also given that both Length and Width are whole numbers, hence we can start by finding the factors of 2020 into prime factors

Factor 2020: 2 x 2 x 5 x 101

By observation, we can see that the following combinations of the factors make up the required area of 2020:

Case 1: (1) x (2)(2)(5)(101) = 1 x 2020 = 2020

Perimeter = 2(1 + 2020) = 4042

Case 2: (2) x (2)(5)(101) = 2 x 1010 = 2020

Perimeter = 2(2+ 1010) = 2024

Case 3: (2)(2) x (5)(101) = 4 x 505 = 2020

Perimeter = 2(4 + 505) = 1018

Case 4: (2)(2)(5) x (101) = 20 x 101 = 2020

Perimeter = 2(20 + 101) = 242

Case 5: (5) x (2)(2)(101) = 5 x 404 = 2020

Perimeter = 2(5 + 404) = 818

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Perimeter = 2(10 + 202) = 424

From the above, it is clear that case 1 yields the largest perimeter

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