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alex41 [277]
3 years ago
15

A sidewalk around a circular park covers a distance of 65 meters. A new sidewalk is being constructed across the park through it

s center. About how long will the new sidewalk be? Use 3.14 for π π . Round to the nearest meter.
21 m
204 m
102 m
10 m
Mathematics
2 answers:
Tpy6a [65]3 years ago
6 0

Answer:

102


Step-by-step explanation:


luda_lava [24]3 years ago
4 0

<u>Answer:</u>

21 m

<u>Step-by-step explanation:</u>

A sidewalk around a circular park covers a distance of 65 meters which means its the circumference of the circular park.

We know the formula for the circumference of a circle so we can write it as:

<em>Circumference of circular park = 2πr</em>

65 = 2πr

Solve this to find r:

65 = 2 x 3.14 x r

r = 65 / 6.28

r = 10.35

Now to find the find the distance of the new sidewalk, multiply r with 2 (to get the diameter):

2 x 10.35 = 20.7 ≈ 21 m

Therefore, the distance of new side walk, rounded to the nearest meter, is 21 m.

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a standard deck of 52 playing cards contains 13 cards in each of 4 suits. what is the appropriate probability that exactly three
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2%

Step-by-step explanation:

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12. The width of a rectangle is 16.55 inches. The length of the rectangle is half its width.
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Step-by-step explanation:

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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
Ksenya-84 [330]

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.

8 0
3 years ago
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