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2 years ago

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There are 9.5 ounces of juice in a container. An additional 1.75 ounces of juice are poured into the container each second. How

many ounces of juice are in the container after 6 seconds? Enter your answer in the box.

1 answer:

VLD [36.1K]2 years ago

7 0

There would be 20 ounces of juice in a container in 6 seconds.
Hope This Helps!

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Work out the difference between 11/12 and 3/12.Give your answer in simplest form.

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ΔEFG ~ ΔLMN. Find LM.

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EFG ~ LMN, if EF ~ LM, then whatever the number is for EF times whatever the others are multiplied by, you will get LM
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FG (10) ~ MN (20)

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Another example:

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lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

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Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

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Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$ </span>

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Step-by-step explanation:

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