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raketka [301]
3 years ago
14

This net can be folded to make a square pyramid. What is the surface area of the pyramid?

Mathematics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer:  

85in^2

Step-by-step explanation:

to find the surface area we need to find the followng areas:  

  • area of the square  
  • area of a triangle and multiply it by 4 (because there are 4 triangles)

And once we have those areas, we add them to find the surface area.

Area of the square:

the formula to find the area of a square is:

a_{square}=l^2

where l is the length of the side: l=5in

thus the area of the square is:

a_{square}=(5in)^2\\a_{square}=25in^2

Area of the triangles:

the are of 1 triangle is given by

a_{triangle}=\frac{b*h}{2}

where b is the base of the triangle: b=5in (the base of the triangle is the side of the square)

and h is the height of the triangle: h=6in

thus, the area of 1 triangle is:

a_{triangle}=\frac{(5in)*(6in)}{2}\\ a_{triangle}=\frac{30in^2}{2}\\ a_{triangle}=15in^2

the area of the 4 triangles is (we multiply by 4):

a_{4-triangles}=4(15in^2)\\a_{4-triangles}=60in^2

finally we add the area of the square and the area of the 4 triangles to find the total surface area:

Surface=25in^2+60in^2\\Surface=85in^2

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Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

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Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

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