A long-time authorized user of an account makes 0.2 seconds between keystrokes. One day, data are recorded as someone typed the correct username and password. The average time between keystrokes was 0.3 with a standard deviation equal to 0.1183. The sample size was 25. 1. At a 1% level of significance, is this an evidence of an unauthorized attempt? Hint: Use a statistical test.
2. Can you apply the z-test and the normal approximation of the Z test statistic to this set of data? Are the conditions on the data for the test satisfied?

i can answer this question sweetie!

step-by-step explanation:

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

$649.60

step-by-step explanation: