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sergejj [24]
2 years ago
13

A rectangle has a length of (3x - 1) and a width of (x + 1). part a determine the perimeter and area of the rectangle as polynom

ials. part b determine the perimeter and area of the rectangle if x equals 5. part c determine the perimeter and area of the rectangle if x equals 8.
Mathematics
1 answer:
liubo4ka [24]2 years ago
6 0

Perimeter of rectangle is 8x and Area is 3x^2 + 2x -1..

Lets try to solve out the problem,

Length = 3x-1

Width = x+1

Perimeter = 2(l+b)

Area = l*b

Perimeter= 2 (3x-1 + x+1)

P= 8x

Area = (3x-1) (x+1)

=> 3x(x+1) -1 (x+1)

=> 3x^2 + 3x -x -1

=> 3x^2 + 2x -1.

Hence Perimeter of rectangle is 8x and Area is 3x^2 + 2x -1..

Learn more about Perimeter on:

brainly.com/question/397857

#SPJ4

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According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e
slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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