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Lana71 [14]
3 years ago
5

The length of a rectangle is 3 inches more that its width. The perimeter of the rectangle is at least 30 inches. Which inequalit

y shows the possible range of widths of the rectangle?
A. w ≥ 16.5 inches
B. w ≥ 13.5 inches
C. w ≥ 9 inches
D. w ≥ 6 inches
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0
<span>Perimeter of a rectangle: 2(l + w)
</span>
2(l + w) ≥ 30 in
<span>

Simplify- divide both sides by 2.

l + w ≥ 15 in


Substitute (w + 3) into l so only 1 variable is used.

(w + 3) + w ≥ 15 in


Simplify further- add variables and subtract 3 from both sides.

2w ≥ 12 in


Divide both sides by 2.

w ≥ 6 in


Answer: D</span>
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The probability distribution was used along with statistical software to simulate 25 repetitions of the experiment​ (25 games).
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Answer:

Hello here is the missing data to your question

x  P(x)

0  0.167

1  0.3289

2  0.2801

3  0.149

4  0.0382

5  0.0368

The theoretical values are lower than the approximate values and the property illustrated is the mean and standard deviation

A) 1.6729

B)  1.234061

C) 1.84

D)  1.462874

Step-by-step explanation:

A) calculate the theoretical mean

theoretical mean (u) = summation of ; x_{i}p(x_{i})

                 = 0(0.167)+1(0.3289)+2(0.2801)+3(0.149)+4(0.0382)+5(0.0369)

                 = 1.6729

B) calculate the theoretical standard deviation

     theoretical standard deviation = \sqrt{variance} = \sqrt{[summation of x_{i}^2 p(x_{i} ) ]-u^2}

= \sqrt{[0+0.3289+1.1204+1.341+0.6112+0.92]-(2.798594)}

= 1.234061

C) calculate the approximate mean

the generated data used for calculation is added below

mean (x) = \frac{summation of (x_{i}) }{n} = \frac{1+2...+2+1}{25} = 46/25 = 1.84

D) calculate the approximate standard deviation

std = \sqrt{variance}

     = \sqrt{\frac{summation of( x_{i}-x )^2}{n-1} }

    = \sqrt{2.14} = 1.462874

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Answer:

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Step-by-step explanation:

Let

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