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Talja [164]
2 years ago
7

the length of a rectangle is 2ft more than twice the width. The area is 180 ft^2. find the length and width of the rectangle

Mathematics
1 answer:
Mariana [72]2 years ago
3 0

Answer:

Area = length * width

 

112 = (2w+2)*w

 

112 = 2w^2 + 2w

 

0 =2w^2 + 2w - 112

 

0 = w^2 + w - 56

 

0 = (  w  +  8  )( w  -    7  )

 

w+8=0  results in negative measures

 

w-7=0 --> w = 7

 

width is 7

length is 16

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2. 5.4 + 3 = 3 + 5.4 illustrates the commutative property of addition.

<h3>What is the Commutative Property of Multiplication?</h3>

The commutative property of multiplication states that the arrangement of change of numbers that you want to multiply does not change what you would get as the product.

For example, a × b = b × a.

In the same vein, 8(-5/6) = (-5/6)8 illustrates the commutative property of multiplication.

<h3>What is the Commutative Property of Addition?</h3>

The commutative property of addition also states that the order or arrangement of addends will not change the sum you would get.

For example, 3 + 2 = 2 + 3.

Therefore, the statement, 5.4 + 3 = 3 + 5.4 illustrates the commutative property of addition.

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1 year ago
12/21 in its simplest form
UNO [17]
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12/21 in its simplest form is 4/7.

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5 0
3 years ago
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Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How
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In this question, the Poisson distribution is used.

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In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that \mu = 5.2r, in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

P(X \geq 1) = 1 - P(X = 0) = 0.99

Thus:

P(X = 0) = 1 - 0.99 = 0.01

We have that:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}

Then

e^{-5.2r} = 0.01

\ln{e^{-5.2r}} = \ln{0.01}

-5.2r = \ln{0.01}

r = -\frac{\ln{0.01}}{5.2}

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3 years ago
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kozerog [31]

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