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zavuch27 [327]
2 years ago
9

Three identical circular coins are lined up in a row as shown (the 10s represent their value, not a length). The distance betwee

n the first and third coins is 3.2 cm. What is the diameter of one of these coins?​

Mathematics
1 answer:
son4ous [18]2 years ago
3 0

Answer:

Diameter of a circle = 1.6 Cm

Step-by-step explanation:

Given:

Distance between first and third coins = 3.2 cm

Find:

Diameter of the coins = ?​

Computation:

Number of circle = 3

In the given figure,distance between the first and third coins is 3.2 cm.

So,

Distance = 1st circle radius + 2nd circle diameter + 3rd circle radius

Distance = 1st circle radius + 2 (radius) + 3rd circle radius

Distance = R + 2R + R

3.2 Cm = 4 R

Radius = 0.8 Cm

Diameter of a circle = 2 × radius

Diameter of a circle = 2 × 0.8

Diameter of a circle = 1.6 Cm

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<span>-Both box plots show the same interquartile range.
 >Interquartile range (IQR) is computed by Q3-Q1. 
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</span><span>-Mr. Ishimoto had the class with the greatest number of students.
 >Mr. Ishimoto had 40 students, represented by the last data point of the whiskers.

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L+ (L-15)  + 2(L-15) = 83

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Laura 32

Kelly 17 (32-15)

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Brian buys a computer for £4300. It depreciates at a rate of 3% per year. How much will it be worth in 6 years? Give your answer
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The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

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