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exis [7]
3 years ago
13

A rectangular exercise mat has a perimeter of 36 feet. The length of the mat is twice the width. Write and solve an equation to

determine the length, in feet, of the mat. Then find the area, in square feet, of the mat.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0
<h2>Hello!</h2>

The answers are:

- Length

length=2width=12feet

- Width

width=6feet

- Area

Area=72feet^{2}

<h2>Why?</h2>

We are given the perimeter of a rectangular mat, and we are asked to find the area, and write an equation to determine the length, if feet, of the mat. To solve this problem we need to remember the equation of the perimeter of a rectangle, which is:

Perimeter=2length+2width

Using the given information we have,

36feet=2length+2width

Also, we know that the length is twice the width, so:

length=2width

Then, substituting the second equation into the first equation, we can find the width of the mat.

Substituting, we have:

36feet=2(2width)+2width

36feet=4width+2width

36feet=6width

[tex]\frac{36feet}{6} =width\\width=\frac{36feet}{6} =6feet[/tex]

Then, substituting the width into the second equation, we can find the length of the mat,

Substituting, we have:

length=2width

length=2(6feet)=12feet

Then, finding the area, we have:

A=length*width=12feet*6feet=72feet^{2}

Have a nice day!

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A map is being reduced to from a pocket sized version. What is the scale factor being used to reduce the map?
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8 0
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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408

#Define Y:

Y~(11.04,5.7408)

P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291

Hence, the probability of 12 or fewer is 0.8291

b. The  probability that 5 or more fish were caught.

#Using normal approximation:

P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above P(X\geq 5)=0.9968 and a ,P(X\leq 12)=0.7291

P(5\leq X\leq 120\approx P(4.5\leq Y\leq  12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259

Hence, the probability of between 5 and 12 is 0.7259

4 0
3 years ago
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