Criminal investigators use biometric matching for fingerprint recognition, facial recognition, and iris recognition. when matching fingerprints, each subject is given a mean match score for how well their fingerprint matches the given fingerprint, based on different criteria. there are six criteria used to calculate the match score. the mean match score used by a particular police department is 80. if the police department finds a higher match score than this number, they consider the person a fingerprint match and a suspect in the crime. the null hypothesis is that the mean match score is 80. the alternative hypothesis is that the mean match score is greater than 80. is the following a type i error or a type ii error or neither? the test shows that the mean match score is more than 80 when the person does not actually have a fingerprint match. group of answer choices type i error type ii error

Type I error.

Step-by-step explanation:

Let's remember the definition of Type I error and Type II error:

A type I error is the rejection of a true null hypothesis, this means that we would get a "false positive" with this error.

A type II error is the non rejection of a not true null hypothesis, this error would give us a "false negative".

In this problem, we are told that the mean match score to identify a suspect is 80. However, the test shows that the mean match score is more than 80 when the person doesn't have a fingerprint match (and therefore the person would not be a suspect). Therefore, this person would appear as a suspect when he/she really isn't one. This means that the test is giving a "false positive". Thus, this is a type I error.

mark branliest

step-by-step explanation:

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the answer is actually d. i just took the test

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.