The area of square is 36 square meters. what is the length (in meters) of one side in the square
1 answer:
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Answer:
(a) P = 2x + 2r + 2πr
(b) x = ½(P - 2r - 2πr)
(c) x = 123 feet
Step-by-step explanation:
See Attachment for sketch of the track
Solving for (a): Perimeter of the track
This is calculated by adding the perimeter of the rectangle to the 2 semicircles as follows.
P = Perimeter of Rectangle + Perimeter of 2 Semicircles
P = 2(x + r) + 2 * πr
[Where r is the radius of both semicircles]
Open bracket
P = 2x + 2r + 2πr
Solving for (b): Formula for x
P = 2x + 2r + 2πr
Make x the subject of formula
2x = P - 2r - 2πr
Divide both sides by 2
x = ½(P - 2r - 2πr)
Solving for (c): the value of x when r = 50 and P = 660
Substitute the given values in the formula in (b) above
x = ½(P - 2r - 2πr)
x = ½(660 - 2 * 50 - 2π * 50)
x = ½(660 - 100 - 100π)
x = ½(560 - 100π)
Take π as 3.14
x = ½(560 - 100 * 3.14)
x = ½(560 - 314)
x = ½ * 246
x = 123 feet
Answer:
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
In this problem, we have that:
What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile
This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.