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Katyanochek1 [597]
3 years ago
5

What is the length of each side of a square that has the same area as a rectangle that is 14 mm long and 16 mm wide? Round to th

e nearest whole number.
Mathematics
1 answer:
swat323 years ago
7 0
1. You have the following information:

 - <span> The square has the same area as a rectangle.
 - The dimensions of the rectangle are: 14 mm long and 16 mm wide.

 2. You must use the formula for calculate the area of a square, which is shown below:

 A=S</span>²

 "A" is the area of the square.<span>
 "S" is the lenght of the side of the square. (It is important to remember that the sides of a squre have equals lenghts.

 3. The square and the rectangle have the same area. Therefore:

 Arectangle=(14 mm)(16 mm)
 Arectangle=224 mm</span>²
<span>
 A=224 mm</span>²
<span>
 4. Now, you must substitute the value A=224 mm</span>² into the formula A=S² and clear "S":
<span>
 A=S</span>²
<span> S=</span>√A
<span> S=</span>√(224 mm²)
<span> S=15 mm

 The answer is: 15 mm

 </span>
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Evaluate the expression t-6/4 for t =38
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Decimal Form:

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8 0
3 years ago
According to the last census (2010), the mean number of people per household in the United States is LaTeX: \mu = 2.58 Assume a
Veseljchak [2.6K]

Answer:

P(2.50 < Xbar < 2.66) = 0.046

Step-by-step explanation:

We are given that Population Mean, \mu = 2.58 and Standard deviation, \sigma = 0.75

Also, a random sample (n) of 110 households is taken.

Let Xbar = sample mean household size

The z score probability distribution for sample mean is give by;

             Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)

P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar \leq 2.50)

P(Xbar < 2.66) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{2.66-2.78}{\frac{0.75}{\sqrt{110} } } ) = P(Z < -1.68) = 1 - P(Z  1.68)

                                                              = 1 - 0.95352 = 0.04648

P(Xbar \leq 2.50) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{2.50-2.78}{\frac{0.75}{\sqrt{110} } }  ) = P(Z \leq  -3.92) = 1 - P(Z < 3.92)

                                                              = 1 - 0.99996 = 0.00004  

Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046

7 0
3 years ago
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