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lys-0071 [83]
3 years ago
13

A rectangular paperboard measuring 29in long and 20in wide has a semicircle cut out of it, as shown below.

Mathematics
1 answer:
katen-ka-za [31]3 years ago
6 0

Here, we are required to find the area of the paper board given after the semicircle is cut out of it

Area of the paper board thatremains is 423 in²

Length = 29 in

Width = 20 in

Area of a rectangle = length × width

= 29 in × 20 in

= 580 in²

Area of a semi circle = πr²/2

π = 3.14

r = diameter / 2 = 20 in / 2 = 10 in

Area of a semi circle = πr²/2

= 3.14 × (10 in)² / 2

= 3.14 × 100 in² / 2

= 314 in²/2

= 157 in²

The semicircle is cut out of the rectangle

Find the area of the paper board that remains after the semicircle is cut out of it by subtracting the area of a semi circle from the area of a rectangle

Area of the paper board that remains = Area of a rectangle - Area of a semi circle

= 580 in² - 157 in²

= 423 in²

brainly.com/question/16994941

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
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Answer:

Obese people

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

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Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

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If the two quadrilateral are congruent, then ABCD = PQRS is the correct congruence statement.
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