Now, if the customers left 8.23 extra, based on a percentage of the cost of the meal, namely, the 8.23 is a percent of 49.28, then

if we take 49.28 to be the 100%, how much of a percentage is 8.23?

$\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
49.28&100\\
8.23&t
\end{array}\implies \cfrac{49.28}{8.23}=\cfrac{100}{t}\implies t=\cfrac{8.23\cdot 100}{49.28}$

8 0

2 years ago

Discussion Prompt: A fair coin is tossed twice and the number of heads is recorded. Jose says the P(1 head) = 1/2, while Jeff st

NemiM [27]

Let H represent heads and T represent tails.

Writing HH means we get two heads and HT means we get heads first, then tails second, and so on.

We have these four possible outcomes when flipping two coins

Of those four outcomes, two of them have exactly one head show up (HT and TH). The probability of getting exactly one head is 2/4 = 1/2, so this is why Jose is correct.

8 0

2 years ago