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MrMuchimi
3 years ago
9

mike is a parking meter attendant. whenever he approaches a parked car at a meter, there is a 29% chances that the meter has exp

ired. what is the probability that once michael starts his shift, the first car the he finds parked at an expired mether will be the 5th car he approaches?
Mathematics
2 answers:
nydimaria [60]3 years ago
8 0
Well. If you get 29% of 5 cars that is 145%. This was a little confusing but I am defiantly sure it is correct. I like to use simpler numbers to see if I am doing the work right. So I said if he has a 50 % likely hood to find a car that was expired and had 1 car. It would be 50 percent. Now if he had 2nt got it the first time it would be a 100 % chance to find the car expired . Hope I didn’t co fuse you more
Whitepunk [10]3 years ago
3 0

Answer:

Solids and liquid

Step-by-step explanation:

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Square root of 120 over the square root of 30. what is the following quotient.
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3 0
2 years ago
Find the intercepts and asymptotes, use limits to describe the behavior at the vertical asymptotes. (37 and 39)
n200080 [17]
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Its always best to factorise where you can, in most questions it tends to be a necessary step and even if its not, it usually makes things easier;
In this case, you can factorise the denominator:
h(x) = \frac{x \ - \ 1}{ x^{2} \ - \ x \ - \ 12} \\\\ = \frac{x \ - \ 1}{(x \ - \ 4)(x \ + \ 3)}

To get y-intercept/s, set x = 0 and solve:
h(0) = \frac{(0) \ - \ 1}{((0) \ - \ 4)((0) \ + \ 3)} \\\\ =  \frac{-1}{-12} \\\\ =  \frac{1}{12}
y-intercept: (0, 1/12)

To get x-intercept/s, set y = 0 and solve:
\frac{x \ - \ 1}{(x \ - \ 4)(x \ + \ 3)} = 0 \\\\ x \ - \ 1 = 0 \\\\ x = 1
x-intercept: (1, 0)

To find the asymptotes, it is necessary for us to have factorised the denominator so its a good thing that I did so at the beginning;

To get the vertical asymptote/s, set the denominator equal to 0, then just rearrange to get x:
(x \ - \ 4)(x \ + \ 3) = 0 \\ x \ - \ 4 = 0 \\ x = 4 \\ and \\ x \ + \ 3 = 0 \\ x = -3
vertical asymptotes: x = 4  and  x = -3

There are no horizontal asymptotes.
8 0
4 years ago
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