True or False: In scientific notation a number is written m X 10^w where m is between 1.000 and 9.999...

Mathematics

1 answer:

Bogdan [553]3 years ago

8 0

Answer: True

Step-by-step explanation:

We try to keep the number greater than 1 and lower than 10, so between 1, and 9.

For example: 3545346 would be 3.5x10^6

you can see 3.5 is between 1 and 9

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The Powerball lottery is open to participants across several states. When entering the powerball lottery, a participant selects

mr Goodwill [35]

Answer and Step-by-step explanation:

Number of ways = 59C5 × 35= 59!/(54!*5!)* 35= 175223510

Odds are thus 1 in 175223510

b) 1 million - Number of ways = 59!/(54!*5!) =5006386*35/34 = 5153632

Odds are thus 1 in 5153632

6 0

3 years ago

Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$