Answer:
15 in.
Use this website : http://graysonmath.com/wp-content/uploads/2018/10/HM2_6.4-Practice-Key.pdf
Step-by-step explanation:
Last time I checked that was about right, but it's been a few years since then.
Answer:
well bcs we cool so appreciated <3 periodTH lolz
Step-by-step explanation:
9514 1404 393
Answer:
A: 66/21
E: 22/7
Step-by-step explanation:
The offered choices reduce to 3 1/7 or to 3. The best approximations of those given are the ones that reduce to 3 1/7.
A: 66/21 = 22/7 = 3 1/7
B: 60/20 = 3
C: 21/7 = 3
D: 45/15 = 3
E: 22/7 = 3 1/7
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<em>Comment on pi approximations</em>
The ratio 22/7 differs from π by about 0.04%. A better approximation is 355/113, which differs from π by about 0.000008%. The approximation 3.14 differs from π by about 0.05%.
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.