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alexandr1967 [171]
3 years ago
7

Sammy had 30 minutes to do a three problem quiz. She spent 10 1/3 minutes on problem 1 and 5 4/5 minutes on problem 2. How much

time did she have time for problem 3? Write your answer as a mixed number.
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0

The time spent on problem 3 is 13\frac{13}{15} \text{ minutes }

<em><u>Solution:</u></em>

Given that Sammy had 30 minutes to do a three problem quiz

To find: Remaining time left for problem 3

From given question,

Total time = 30 minutes

Time spent on problem 1 = 10\frac{1}{3} \text{ minutes }

\rightarrow 10\frac{1}{3} = \frac{3 \times 10 + 1}{3} = \frac{31}{3} \text{ minutes }

Time spent on problem 2 = 5\frac{4}{5} \text{ minutes }

\rightarrow 5\frac{4}{5} = \frac{5 \times 5 + 4}{5} = \frac{29}{5} \text{ minutes }

Therefore,

Time spent on problem 3 = total time - (time on problem 1 + time on problem 2)

\text{ Time spent on problem 3 } = 30 - (\frac{31}{3} + \frac{29}{5})\\\\\rightarrow 30 - \frac{31 \times 5 + 29 \times 3}{3 \times 5}\\\\\rightarrow 30 - \frac{242}{15}\\\\\rightarrow \frac{30 \times 15 - 242}{15}\\\\\rightarrow \frac{208}{15} = 13\frac{13}{15}

Thus time spent on problem 3 is 13\frac{13}{15} \text{ minutes }

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Step-by-step explanation:

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f(x)=e^{-x}

For a)

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P(T>2)=1-P(T\leq 2)

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P(T>2)=1-(-e^{-2}+1)

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For b)

P(T\geq 10/T>9)

The event (T ≥ 10 / T > 9) is equivalent to the event T ≥ 1 so they have the same probability of occur

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3 years ago
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This means that the point (3, 0) is a solution for the line equation, so when x = 3, we also have y = 0.

Replacing these values in our equation we get:

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Dima020 [189]
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Romashka-Z-Leto [24]
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3 years ago
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