Alligators captured in Florida are found to have a mean length of 2 meters and a standard deviation of 0.35 meters. The lengths of alligators are believed to be approximately normally distributed. What percent of alligators have lengths greater than 2.2 meters?

28.43% of alligators have lengths greater than 2.2 meters

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 question, we have that:

What percent of alligators have lengths greater than 2.2 meters?

This is 1 subtracted by the pvalue of Z when X = 2.2. So

has a pvalue of 0.7157

1 - 0.7157 = 0.2843

28.43% of alligators have lengths greater than 2.2 meters

the answer is a if my teacher taught me correct

60°

step-by-step explanation:

the measure angle of c in an isosceles triangle abc is 60° because in isosceles triangle,the measures of all the angles is 60°.