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emmasim [6.3K]
3 years ago
7

Which is not true about the random sample method?

Mathematics
1 answer:
USPshnik [31]3 years ago
6 0

B. The sample is likely to represent one portion of the population.

This is impossible to say because it's random.

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Find the odds for and the odds against the event rolling a fair die and getting a 4, a 2, or a 1.
HACTEHA [7]

Answer: Both in favor and against: 3/6 OR 1/2 OR 50%.

Step-by-step explanation: Both in favor and against: 3/6 OR 1/2 OR 50%.

The probability of rolling one die and getting any digit is \frac{1}{6}. That is because the desired outcome is only 1 digit out of the 6 total digits on each of the six sides.

That means the probability of rolling either a 1, a 2, or a 4 <em>if you are looking for any of those numbers specifically, </em>is \frac{1}{6}. But! You are looking to roll a 1, a 2, or a 4, without any preference among those three. Therefore, you are going to add the desired outcomes.

\frac{1}{6} + \frac{1}{6}  + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} = 0.5 = 50%

Above, are added the probability of rolling a 1 PLUS the probability of rolling a 2 PLUS the probability of rolling a 4, yielding a 50% probability.

Think about it. There are six total digits on a die. You are okay with rolling three of those. Three is half of six. Therefore, the probability of rolling the number you'd like is one-half, fifty-percent.

Likewise, since the probability of all complement events happening is 1 (or 100%), The odds against rolling either 1, 2, or 4 is also fifty-percent. 100% (which is rolling <em>any </em>number on a die) MINUS 50% (which is rolling a 1, a 2, or a 4 with no preference) EQUALS TO 50% (which is rolling a 3, a 5, or a 6).

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3 years ago
Simplify the expression<br> (5x^2)^3
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Would that he the answer or option A ?
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Your answer is correct
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