Answer:
wdym???
Step-by-step explanation:
You have an email with at white buffalos.net??? I thought that was only my school.
Answer:
13.89% of students are willing to report cheating by other students.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 180
Number of students who reported cheating, x = 25
We have to find the proportion of the students are willing to report cheating by other students.
Proportion of students can be calculate as
Thus, 13.89% of students are willing to report cheating by other students.
T is greater than -15 because numbers after -15, like -16... are less than -15, so when its decreasing it gets lower. Since the questions says the temp stayed above -15, that means the temp could be -14...-13 and so on.
Answer:
P(x) = 45/100 = 0.45
Mean sample distribution = probability x number sampled by the survey.
Mean sample distribution = 0.45 x 800 = 360.00 to two decimal places.
Step-by-step explanation:
Convert the percentage to decimal probability.
45%. P(x) = 45/100 = 0.45
If there are ranges of probability values, we construct a probability distribution table. This is not necessary in the case of one probability value(45%)
Multiply the probability by the number adults to be surveyed on whether they have received phishing emails.
0.45 x 800 = 360.
Here, we assume that the 45% recorded by 2005 data, is still valid for the recent trends.
Answer:
Step-by-step explanation:
Any number that times 17 is bigger than -17
