A fish story: The mean length of one-year-old spotted flounder, in millimeters, is 127 with standard deviation of 22, and the mean length of two-year-old spotted flounder is 158 with a standard deviation of 23. The distribution of flounder lengths is approximately bell-shaped. Part 1 of 4 (a) Anna caught a one-year-old flounder that was 155 millimeters in length. What is the z-score for this length

The z-score for this length is of 1.27.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

One-year-old flounder:

Mean of 127 with standard deviation of 22, which means that

Anna caught a one-year-old flounder that was 155 millimeters in length. What is the z-score for this length

This is Z when X = 155. So

The z-score for this length is of 1.27.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.