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lesantik [10]
2 years ago
7

If an event is very unlikely to occur would it’s probability be zero

Mathematics
1 answer:
tigry1 [53]2 years ago
5 0

Answer: No

Step-by-step explanation:

An event that cannot possibly happen has a probability of zero. If there is a chance that an event will happen, then its probability is between zero and 1. Thus, an unlikley event would be a low probability, not 0

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6 0
2 years ago
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Should i give up with math it to hard and i try every time
bekas [8.4K]

Answer:

No as hard as math is it’s a great feeling when you get that “OHH I GET IT” moment. And plus if your in school your kinda. Stuck with it anyway

Step-by-step explanation:

6 0
2 years ago
Before going to school, Eudora ran from her home to a secret laboratory at an average speed of 12 km. Since she was running late
Brrunno [24]

Answer:

Eudora spent 0.5 hours in running and 1.5 hours flying using her jet-pack.

Step-by-step explanation:

Let x represent time spent in running.

We have been given that the entire trip took 2 hours, so time spent in flying the jet-packs would be 2-x.

We know that distance is equal to speed times time.

\text{Distance}=\text{Speed}\times\text{Time}

Eudora ran from her home to a secret laboratory at an average speed of 12 km, so distance traveled by running would be 12x.

Eudora took one of her jet-packs and flew to her school at an average speed of 76 km, so distance covered by jet-packs would be 76(2-x).

We are told that Eudora traveled a total distance of 120 km. We can represent this information in an equation as:

12x+76(2-x)=120

12x+152-76x=120

Combine like terms:

-64x+152=120

-64x+152-152=120-152

-64x=-32

Divide both sides by -64:

\frac{-64x}{-64}=\frac{-32}{-64}

x=0.5

Therefore, Eudora spent 0.5 hours in running.

Upon substituting x=0.5 in expression for time spent in flying jet-packs, we will get:

2-x\Rightarrow 2-0.5=1.5

Therefore, Eudora spent 1.5 hours flying using her jet-pack.

7 0
3 years ago
Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur
stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Three heads.
  • Event B: Fair coin.

The probability associated with 3 heads are:

  • 0.5^3 = 0.125 out of 0.5(fair coin).
  • 1 out of 0.5(biased).

Hence:

P(A) = 0.125 + 0.5 = 0.625

The probability of 3 heads and the fair coin is:

P(A \cap B) = 0.5(0.125) = 0.0625

Then, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0
3 years ago
Lee was 2 under par on the first hole, 3 under par on the second and third hole, 2 over par on the fourth, 5 over par on the fif
Anika [276]
1 under par overall
3 0
3 years ago
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