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Alborosie
2 years ago
8

What is the measure of the apothem, rounded to the nearest hundredth of a centimeter?

Mathematics
1 answer:
malfutka [58]2 years ago
5 0

The measure of apothem is 1.75 cm.

The complete question is

The area of the regular octagon is 10.15 cm2.

What is the measure of the apothem, rounded to the nearest hundredth of a centimeter?

<h3>What is an Octagon ?</h3>

A polygon with eight sides , eight angles and eight vertices is called an octagon.

If a is the Side of the octagon.

10.15 = 2(1+√2) a²

a² = 10.15 / [2×(1+1.414)]

a² = 2.10

a = 1.45 cm

Therefore the side = 1.45 cm

A = 8 × (a × h)/2

If we consider  8 equilateral triangles with

a = face of octagon

h = Apothem of octagon

10.15 = 8 × (1.45 × h)/2

h = 10.15 /4/ 1.45

h = 1.75  cm

Therefore the measure of apothem is 1.75 cm.

To know more about Octagon

brainly.com/question/16543440

#SPJ1

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The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
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Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

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