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Liula [17]
3 years ago
3

A rectangle has a length of (3.2a 1 0.18b) centimeters. The width is half the length. Sasha writes the expression (12.8a 1 0.72b

) to represent the perimeter of the rectangle in centimeters. Is Sasha’s reasoning correct? Explain.
Mathematics
2 answers:
MAVERICK [17]3 years ago
6 0
Perimeter of the rectangle will be given by:
P=2(L+W)
L=(3.2a 10.18b)
W=1/2(3.2a 10.18b)=(1.6a 5.9b)
Thus the perimeter will be:
P=2(3.2a 10.18b+1.6a 5.9b)
P=2(4.8a 16.08b)
P=(9.6a+ 32.16b)

Sasha was not correct
vfiekz [6]3 years ago
3 0

Answer:

No, her reasoning is incorrect.

Step-by-step explanation:

Given,

The length of the rectangle,

l = ( 3.2a + 0.18b ) \text{ cm},

Here, the width is half the length.

⇒ Width of the rectangle,

w = \frac{l}{2}=\frac{1}{2}(3.2a + 0.18b)=\frac{3.2a}{2}+\frac{0.18b}{2} = (1.6a + 0.09b)\text{ cm}

We know that,

The perimeter of a rectangle is,

P=2(l+w)

=2(3.2a + 0.18b+1.6a + 0.09b)

=2(4.8a+0.27b)

=(9.6a+0.54b)\text{ cm}

Since, 9.6a + 0.54b ≠ 12.8a + 0.72b

Hence, her reasoning is incorrect.

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Mr. Smith's wife works at an animal hospital. There is a equivalent fraction relationship between the number of dogs and cats in
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Answer:

There are 10 dogs in the kennels

Step-by-step explanation:

The equivalent fraction relationship between the number of dogs and cats in the kennels is 1/2.

Which means, if n_d the number of dods and n_c is the number of cats, then,

\frac {n_d}{n_c}=\frac{1}{2}\cdots(i)

Given that there are 20 cats, so n_c=20.

Now, from equation (i) we have,

\frac {n_d}{20}=\frac{1}{2}

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3 years ago
If the angles of a triangle are 10 45 and 23 what is the perimerter of the triangle
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Answer:

No Solutions

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3 0
3 years ago
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Read 2 more answers
suppose that incomes of families in Newport Harbor are normally distributed with a mean of $750,000 and a standard deviation of
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Answer: 0.345

Step-by-step explanation:

Given : The incomes of families in Newport Harbor are normally distributed with Mean : \mu=\$ 750,000 and Standard deviation : \sigma= $250,000

Samples size : n=4

Let x be the random variable that represents the incomes of families in Newport Harbor.

The z-statistic :-

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x= $800,000

z=\dfrac{800000-750000}{\dfrac{250000}{\sqrt{4}}}=0.4

By using the standard normal distribution table , we have

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Hence, the probability that the average income of these 4 families exceeds $800,000 =0.345

6 0
4 years ago
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