Which equation can be used to find the side lengths if the longest side measures 6.2cm?

cupoosta [38]

Answer:

(7, 5)

Step-by-step explanation:

AC is the resultant. Point C is 7 units right and 5 units up from point A. If those are what go in your boxes, the resultant is ...

$\displaystyle\left(\,7\,\above\,5\,\right)$

_____

<em>Comment on the question</em>

Vectors can be represented a number of different ways. Components can be given in rectangular or polar form, and the presentation can be made as a row vector, column vector, sum of orthogonal unit vectors, and other ways. We assume this is supposed to be a column vector of the form ...

$\displaystyle\left(x\above y\right)$

7 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?

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

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