Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
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Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
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Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
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Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
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Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
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Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
Answer:
c) d is greater than or equal to 530
Step-by-step explanation:
Hope this helps!
<u>Answer:</u>
Amount collected in 2012 = 3.49
and in 2013 = 3.35
<u>Explanation:</u>
Let x be the baggage fees collected in 2013 and
y be the baggage fees collected by airlines in 2012
From the given question we can obtain two equations as follows:
x + y = 6.84 $ - eq 1
and
y = 0.14 + x
as fees in 2012 exceeds that in 2013 by 0.14$
Now, solving these two equations by substitution,
we substitute y = 0.14 + x in eq 1
and solve for x,
which gives us x = 3.35 and
substituting this value in eq 1
we get y = 3.49
Answer:
How many of these passwords contain at least one occurrence of at least one of the five special characters?
If they are just looking for the “inverse” which means flip it it’ll be 5radical (which the the v shape looking bar) x-4 = h(x)
