Answer:
B
Step-by-step explanation:
let f(x) = y and rearrange making x the subject
y =
( multiply both sides by 7 to clear the fraction )
7y = x + 2 ( subtract 2 from both sides )
7y - 2 = x
Change y back into terms of x with x the inverse function, that is
p(x) = 7x - 2
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>
The variable in this equation is b, therefore we have to calculate the value of b.
-38=2b+25-(-7b)
-38=2b+25+7b
2b+7b=-38-25
9b=-63
b=-63/9
b=-7
Answer: the value of the variable is b=-7
We can check it out this answer:
-38=2b+25-(-7b)
-38=2(-7)+25-(-7(-7))
-38=-14+25-49
-38=-38
It is easier to understand the problem if you create a number based on the criteria and then perform the computations. I am going to choose: 111 22 33 4
There are 10 options for the first "1" and only 1 option for the other two 1's
There are 9 remaining options for the first "2" and only 1 option for the other 2
There are 8 remaining options for the first "3" and only 1 option for the other 3
There are 7 remaining options for the "4"
10 x 1 x 1 x 9 x 1 x 8 x 1 x 7
10 x 9 x 8 x 7 = 5,040
Answer: 5,040
By inspection, it's clear that the sequence must converge to

because

when

is arbitrarily large.
Now, for the limit as

to be equal to

is to say that for any

, there exists some

such that whenever

, it follows that

From this inequality, we get




As we're considering

, we can omit the first inequality.
We can then see that choosing

will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that

.