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Elena L [17]
4 years ago
11

The length of a rectangular sign is 5 times its width . If the sign's perimeter is 36 inches, what is the signs area in square i

nches
Mathematics
2 answers:
swat324 years ago
4 0

Answer: The area is 45 square inches.

Step-by-step explanation:

so if we know that the the length is 5 times the width then we could represent the length as  5x   and x for the width.  

And we  know the perimeter of the sign so we could set up equation to solve for the side lengths.

The perimeter of a rectangle is  2l + 2w  which means 2 times the length plus two times the width.

2(5x) + 2(x) = 36

10x + 2x = 36

12x = 36

x = 3

Now we know the value of x  so we will plot it into the length and with to find the actually side lengths.

length= 5(3) = 15

width=  3  

Check:  15 + 15 + 3 + 3 = 36  

Now we know the side lengths so to find the area we will have to multiply the length times the width.

3 * 15 = 45  

My name is Ann [436]4 years ago
3 0

Answer:

9 inches

Step-by-step explanation:

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The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo
muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

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The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

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