Tell us how the data is<br><br><br>

There was a 6 square foot piece of wrapping paper for a birthday present. It takes 3 3/8 feet of square feet of the paper to wra

Citrus2011 [14]

Answer: However, in real life it depends on the shape of the actual presents. If two sheets are used for the 2 presents. There should be enough left from those sheets to use for the third present. (They should be taped together.)

Step-by-step explanation: 6/3 then 3 over 8 ––> 3/8

6/ 27/8

= 6/1 / divide 27/8

= 6^2 /1 x 8/27 ^9 (^) –––> this mean square root of 2 or 9

6 cancels and 27 cancel

= 16/9

=1 7/9

However you will still need to buy 2 sheets

3 0

2 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?

So

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

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