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vova2212 [387]
3 years ago
5

A rectangle has a length of 45 inches less than 4 times its width. if the area of the rectangle is 3325 square inches, find the

length of the rectangle.
Mathematics
1 answer:
Maru [420]3 years ago
8 0

The length of the rectangle is based on its width, so let's call width w and put the length in terms of the width. We are told that the width is 45 less than 4 times the width, so the length is 4w - 45. Area is found by multiplying length times width and we are given the area as 3325. So we will set up length times width and solve for w, which we will then use to solve for l. 3325 = (4w - 45)(w). Multiplying out we have 3325=4w^2-45w. Move the constant over by subtraction and then we will have a quadratic that can be factored to solve for w. 4w^2-45w-3325=0. We would put that through the quadratic formula to solve for w. When we do that we get that w = 35 and w = -23.75. The 2 things in math that will never EVER be negative is time and distance/length, so -23.75 is out. That means that the width is 35. The length is 4(35) - 45 which is 95. The dimensions of your rectangle are length is 95 and width is 35. There you go!

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Answer:

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Step-by-step explanation:

It can be useful to start with the point-slope form when you are given a point and a slope.

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  y = -8x -8 -7 . . . . eliminate parentheses, subtract 7

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Explain why it is not reasonable to say that 4.23is less than 4.13
Alina [70]
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7 0
3 years ago
Read 2 more answers
For i≥1 , let Xi∼G1/2 be distributed Geometrically with parameter 1/2 . Define Yn=1n−−√∑i=1n(Xi−2) Approximate P(−1≤Yn≤2) with l
murzikaleks [220]

Answer:

The answer is "0.68".

Step-by-step explanation:

Given value:

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E(X_i)=2 \\

Var (X_i)= \frac{1- \frac{1}{2}}{(\frac{1}{2})^2}\\

             = \frac{ \frac{2-1}{2}}{\frac{1}{4}}\\\\= \frac{ \frac{1}{2}}{\frac{1}{4}}\\\\= \frac{1}{2} \times \frac{4}{1}\\\\= \frac{4}{2}\\\\=2

Now we calculate the \bar X \sim N(2, \sqrt{\frac{2}{n}})\\

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\to \sum^n_{i=1}  \frac{X_i - 2}{n}  \times\sqrt{\frac{n}{2}}}  \sim  N(0, 1)\\\\\to  \sum^n_{i=1}  \frac{X_i - 2}{\sqrt{2n}}  \sim  N(0, 1)\\

\to Z_n = \frac{1}{\sqrt{n}} \sum^n_{i=1} (X_i -2) \sim N(0, 2)\\

\to P(-1 \leq X_n \leq 2)  = P(Z_n \leq Z) -P(Z_n \leq -1) \\\\

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3 years ago
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Answer:

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Step-by-step explanation:

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