The likelihood that at least one trip occurs before Isabella's birth is 0.7627.
Step-by-step explanation:
In this scenario, Isabella has invented a time machine, but she lacks control over where she travels. Each use of the device holds a 0.25 probability of leading her to a time preceding her birth. Over the initial year of trials, she operates her machine 5 times. If we assume every journey has an equal chance of going back in time, we can calculate the odds that at least one of these trips occurs before she was born. Here's the calculation:
The probability of traveling to a time prior to her birth is 0.25.
The chance of not traveling back in time, given that the machine is used 5 times:
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The probability that at least one trip goes before Isabella's birth is equal to 1 minus the probability of not traveling back to that period:
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Consequently, the chance that at least one trip travels before Isabella's birth is 0.7627.
H (t) = - 16t ^ 2 + 16t + 480
To address this, we first set the polynomial to zero to find its roots.
So we have:
0 = -16t ^ 2 + 16t + 480
This leads us to the roots of the polynomial:
t1 = -5
t2 = 6
We disregard the negative root since time cannot be less than zero.
Final answer:
Rose takes about
t = 6 seconds
To determine the number of bags, we simply compute the quotient, which yields:
(12 / 15 pounds) / (1 / 3 pounds) = 2.4 bags
Since bags can't be represented in decimals, only 2 bags can be filled. The remainder is 0.4 which can be expressed as:
0.4 * (1/3) = (2/5) * (1/3) = 2/15 pounds
Answer:
<span>2 bags are filled, with 2/15 pounds of granola remaining</span>