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dlinn [17]
3 years ago
14

Eight cards are marked 3, 4, 5, 6, 7, 8, 9, and 10 such that each card has exactly one of these numbers. A card is picked withou

t looking. Find each probability.
Prompt 1P(9)
Answer for prompt 1 P(9)
Prompt 2P(3 or 4)
Answer for prompt 2 P(3 or 4)
Prompt 3P(greater than 5)
Answer for prompt 3 P(greater than 5)
Prompt 4P(less than 3)
Answer for prompt 4 P(less than 3)
Prompt 5P(odd)
Answer for prompt 5 P(odd)
Prompt 6P(4, 7, or 8)
Answer for prompt 6 P(4, 7, or 8)
Prompt 7P(not 6)
Answer for prompt 7 P(not 6)
Prompt 8P(not 5 and not 10)
Answer for prompt 8 P(not 5 and not 10)
Mathematics
1 answer:
katrin2010 [14]3 years ago
3 0

Answer:

I could only find 2 for my brain.

Prompt 8P: 0.75, 75% or 3/4

Prompt 6P: 0.375, 37.5% or 3/8

Step-by-step explanation:

meh

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Answer:

5 years and 5 months

Step-by-step explanation:

<u />

<u>Compound Interest Formula</u>

\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • A = $17,474.00
  • P = $7,790.00
  • r = 15% = 0.15
  • n = 12
  • t = number of years

Substitute the given values into the formula and solve for t:

\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}

\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)

\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}

\implies \sf t=5.419413037...\:years

Therefore, the money was in the account for 5 years and 5 months (to the nearest month).

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Step-by-step explanation:

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