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p(t)=100((1)/(2))^((t)/( 2.0373))
Radioactive isotopes are unstable atoms that decay into other atoms. Oxygen-15 is a radioactive isotope. The function models 
p, the percentage of oxygen-15 atoms remaining in a sample after 
t minutes. Which of the following statements is the best interpretation of the ordered pair 
(4.0746,25) ?
Choose 1 answer:
(A) About 25 percent of oxygen-15 atoms remain in a sample after 4.0746 minutes.
(B) About 4.0746 percent of oxygen-15 atoms remain in a sample after 25 minutes.
(C) About 25 percent of oxygen-15 atoms in a sample decay into other atoms in 4.0746 minutes.
(D) About 4.0746 percent of oxygen-15 atoms in a sample decay into other atoms in 25 minutes.

p(t)=100(12)t2.0373 p(t)=100\left(\frac{1}{2}\right)^{\frac{t}{2.0373}} \newlineRadioactive isotopes are unstable atoms that decay into other atoms. Oxygen15-15 is a radioactive isotope. The function models p p , the percentage of oxygen15-15 atoms remaining in a sample after t t minutes. Which of the following statements is the best interpretation of the ordered pair (4.0746,25) (4.0746,25) ?\newlineChoose 11 answer:\newline(A) About 2525 percent of oxygen15-15 atoms remain in a sample after 44.07460746 minutes.\newline(B) About 44.07460746 percent of oxygen15-15 atoms remain in a sample after 2525 minutes.\newline(C) About 2525 percent of oxygen15-15 atoms in a sample decay into other atoms in 44.07460746 minutes.\newline(D) About 44.07460746 percent of oxygen15-15 atoms in a sample decay into other atoms in 2525 minutes.

Full solution

Q. p(t)=100(12)t2.0373 p(t)=100\left(\frac{1}{2}\right)^{\frac{t}{2.0373}} \newlineRadioactive isotopes are unstable atoms that decay into other atoms. Oxygen15-15 is a radioactive isotope. The function models p p , the percentage of oxygen15-15 atoms remaining in a sample after t t minutes. Which of the following statements is the best interpretation of the ordered pair (4.0746,25) (4.0746,25) ?\newlineChoose 11 answer:\newline(A) About 2525 percent of oxygen15-15 atoms remain in a sample after 44.07460746 minutes.\newline(B) About 44.07460746 percent of oxygen15-15 atoms remain in a sample after 2525 minutes.\newline(C) About 2525 percent of oxygen15-15 atoms in a sample decay into other atoms in 44.07460746 minutes.\newline(D) About 44.07460746 percent of oxygen15-15 atoms in a sample decay into other atoms in 2525 minutes.
  1. Plug in t=4.0746t=4.0746: Plug in t=4.0746t=4.0746 into the function p(t)p(t) to see if it equals approximately 2525.\newlinep(t)=100(12)(t2.0373)p(t)=100\left(\frac{1}{2}\right)^{\left(\frac{t}{2.0373}\right)}\newlinep(4.0746)=100(12)(4.07462.0373)p(4.0746)=100\left(\frac{1}{2}\right)^{\left(\frac{4.0746}{2.0373}\right)}
  2. Calculate the exponent: Calculate the exponent first: (4.0746)/(2.0373)(4.0746)/(2.0373).\newline4.0746/2.037324.0746/2.0373 \approx 2
  3. Calculate (12)2(\frac{1}{2})^2: Now calculate (12)2(\frac{1}{2})^2.(12)2=14(\frac{1}{2})^2 = \frac{1}{4}
  4. Multiply the result: Multiply the result by 100100 to find the percentage.100×14=25100 \times \frac{1}{4} = 25
  5. Interpret the result: Since p(4.0746)=25p(4.0746) = 25, the ordered pair (4.0746,25)(4.0746,25) means that about 2525 percent of oxygen-1515 atoms remain after 4.07464.0746 minutes.

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