According to Internet security experts, approximately 90% of all e-mail messages are spam (unsolicited commercial e-mail), while the remaining 10% are legitimate. A system administrator wishes to see if the same percentages hold true for the e-mail traffic on her servers. She randomly selects e-mail messages and checks to see whether or not each one is legitimate. (Unless otherwise specified, round all probabilities below to four decimal places; i. e. your answer should look like 0.1234, not 0.1234444 or 12.34%.) Assuming that 90% of the messages on these servers are also spam, compute the probability that the first legitimate e-mail she finds is the fourth message she checks: Compute the probability that the first legitimate e-mail she finds is the fourth or fifth message she checks: Compute the probability that the first legitimate e-mail she finds is among the first four messages she checks: On average, how many messages should she expect to check before she finds a legitimate e-mail?

3

step-by-step explanation:

in standard form:

a= dictionaries

b= maps

you have to buy 3 dictionaries, so put a 3 next to the a (dictionaries) so that you can multiply them by each other.

a3+b= 60 (the amount of money you have)

we already know the cost of a=15 so replace a with $15

($15 x 3)+b=60

multiply 15 and 3 to get $45 and replace it with $45

45+b=60

and then replace b with 15 to get the total:

45+15=60

that means you have $15 left after buying 3 dictionaries.

15 divided by the cost of a map ($5) equals 3.

so you can buy 3 maps.

sorry if it was confusing!

c) yes, because of asa or aas

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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