Answer:
57.62% of players weigh between 180 and 220 pounds
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:

What percent of players weigh between 180 and 220 pounds
We have to find the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 180.
X = 220



has a pvalue of 0.7881
X = 180



has a pvalue of 0.2119
0.7881 - 0.2119 = 0.5762
57.62% of players weigh between 180 and 220 pounds
Answer:
0.3
Step-by-step explanation:
To find this probability, we need to know the distance from point H to P and the distance from point L to K, and then the probability of a random point in HP being inside LK will be the division of the distance from L to K by the distance from H to P:
H to P: distance = 15 - 5 = 10
L to K: distance = 10 - 7 = 3
Probability: LK/HP = 3/10 = 0.3
That would be
4 feet - 28 feet = -24
The integer is -24
Answer:
6/8 of a mile,
Step-by-step explanation:
if one furlong is 1/8 of a mile, the 6 shall be 1/8 added to itself 6 times, therefore it shall be 6/8.