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S_A_V [24]
3 years ago
15

A victim of phishing would most likely be contacted in which of these ways. A. In person b. By computer c. By telephone. D by fa

x....
I think its a but not sure
Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0
Phishing is normally gettting your info
common things are like emails that say you have won
or emails from people claming to be a nigerian prince recently having come into posession of a large fortune...



phishing scams work best when the peson doing the scam is anonymus
internet is great
in person is terrible, you know who it is
internet is on computer
telephone, you can be tricked, but people are more likely to click on email than to be tricked by phone stuff
fax isn't as much


B is answer
kiruha [24]3 years ago
4 0
Ok so basically phishing is dome online so the only logical answer would be by computer option B.
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Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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Answer:

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I guess?

Step-by-step explanation:

Im really not sure

There's a pattern!!

You just need to count how many numbers passed by!

Clear?

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Hi plz help, if you can ill mark you 5 starz! :)
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Answer:

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