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arsen [322]
2 years ago
6

on halloween kaleb scored 34 pieces of gummi candy. he got 8 times as much sugar candy and 4 times as much chocolate candy. how

many pieces of candy did he get
Mathematics
1 answer:
Lady bird [3.3K]2 years ago
7 0
We can first find the number of different types of candy he got:

The number of sugar candy he got:

34×8
=272

The number of chocolate candy he got:

34×4
=136

We can then add them all up to find the total pieces of candy he got:

272+136+34
=442

Therefore, the answer is 442.

Hope it helps!
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a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou
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Answer:

\displaystyle P(A)=\frac{6}{11}

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

\Omega=\{WW,WN,NW,NN\}

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\displaystyle \frac{5}{11}

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\displaystyle \frac{6}{10}

Thus the option WN has the probability

\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}

Now for the second option NW. The initial probability to pick a non-white counter is

\displaystyle \frac{6}{11}

The probability to pick a white counter is

\displaystyle \frac{5}{10}

Thus the option NW has the probability

\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}

The total probability of event A is the sum of both

\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}

\boxed{\displaystyle P(A)=\frac{6}{11}}

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2 years ago
3. Scout is conducting a research project on traffic patterns
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Answer:

7:45

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