Upon asking I got the rest of the information needed to answer this question.
-> Which one becomes a repeating decimal?
Your answer is D. 1/9.
1/9 = 0.111111111111111111111111111111111 ... or 0.1 repeating. It can be written with a line over the 1 asyou can see in the attached.
This means that D is the answer to your question.
Given:
w be the weight in pounds of a baby tigers.
To find:
The inequality if baby tigers can be no larger than 4 pounds.
Solution:
Baby tigers can be no larger than 4 pounds. It mean weight of baby tigers cannot be greater than 4. In other words, the weight of the baby tigers must be less than or equal to 4.
Let w represents weight in pounds and the weight of the baby tigers must be less than or equal to 4. So,
Therefore, the required inequality is .
Answer:
a) The percentage of snails that take more than 60 hours to finish is 4.75%.
b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.
c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.
d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish
e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.
f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.
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:
a. The percentage of snails that take more than 60 hours to finish is
This is 1 subtracted by the pvalue of Z when X = 60.
has a pvalue 0.9525
1 - 0.9525 = 0.0475
The percentage of snails that take more than 60 hours to finish is 4.75%.
b. The relative frequency of snails that take less than 60 hours to finish is
This is the pvalue of Z when X = 60.
has a pvalue 0.9525
The relative frequency of snails that take less than 60 hours to finish is 95.25%.
c. The proportion of snails that take between 60 and 67 hours to finish is
This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.
X = 67
has a pvalue 0.9977
X = 60
has a pvalue 0.9525
0.9977 - 0.9525 = 0.0452
The proportion of snails that take between 60 and 67 hours to finish is 4.52%.
d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)
This is 1 subtracted by the pvalue of Z when X = 76.
has a pvalue of 1
1 - 1 = 0
0% probability that a randomly-chosen snail will take more than 76 hours to finish
e. To be among the 10% fastest snails, a snail must finish in at most hours.
At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.
To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.
f. The most typical 80% of snails take between and hours to finish.
From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.
10th percentile
value of X when Z has a pvalue of 0.1. So X when Z = -1.28.
90th percentile.
value of X when Z has a pvalue of 0.9. So X when Z = 1.28
The most typical 80% of snails take between 42.32 and 57.68 hours to finish.