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ElenaW [278]
3 years ago
9

The perimeter of a rectangle is 16 inches. The equation that represents the perimeter of the rectangle is , where l represents t

he length of the rectangle and w represents the width of the rectangle. Which value is possible for the length of the rectangle?
7 in.
8 in.
9 in.
10 in.
Mathematics
1 answer:
Gekata [30.6K]3 years ago
5 0

Answer:

A. 7 in.

Step-by-step explanation:

We have been given that the perimeter of a rectangle is 16 inches. The equation that represents the perimeter of the rectangle is , where l represents the length of the rectangle and w represents the width of the rectangle.

We know that perimeter of rectangle is 2 times the sum of width and length of rectangle.

\text{Perimeter}=2(l+w)

\text{16 in}=2(l+w)

\frac{\text{16 in}}{2}=\frac{2(l+w)}{2}

\text{8 in}=l+w

To be a rectangle length cannot be 8 as length and width of the rectangle is 8 inches.

Therefore, 7 inches the possible value for the length of the rectangle.

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3 0
4 years ago
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A.f(x)=-3/4x+10<br> B.f(x)=3/4x-10<br> C.f(x)=4/3x+10<br> D.f(x)=-4/3x-10
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7 0
3 years ago
As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probabi
nordsb [41]

Answer: N(20, 4) distribution.

Step-by-step explanation:

Normal approximation to Binomial :

The normal approximation is used for binomial distribution having parameters n and p as

\mu=np\\\\ \sigma=\sqrt{np(1-p)}

if x is the random variable then x has N(\mu, \sigma).

Given : As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket.

The probability that a shopper will buy a packet of crackers after tasting the free sample : p=0.20.

Different shoppers can be regarded as independent trials.

if X is the number among the next 100 shoppers who buy a packet of crackers after tasting a free sample.

Then, Mean and standard deviation for x will be :

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7 0
3 years ago
I'm stuck on this question, any ideas? thanks
True [87]

Answer:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}=-1/36

Step-by-step explanation:

So we have the limit:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}

Let's remove the fractions in the denominator by multiplying both layers by (6+x)(6). So:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}\cdot (\frac{(6+x)(6)}{(6+x)(6)})

Distribute:

=\lim_{x \to 0} \frac{(6)-(6+x)}{x(6+x)(6)}

Simplify the numerator:

=\lim_{x \to 0} \frac{6-6-x}{x(6+x)(6)}\\=\lim_{x \to 0} \frac{-x}{x(6+x)(6)}

Both the numerator and the denominator have an x. Cancel:

=\lim_{x \to 0} \frac{-1}{(6+x)(6)}

Direct substitution:

= \frac{-1}{(6+0)(6)}

Simplify:

=-1/36

And that's our answer.

And we're done!

8 0
3 years ago
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