PLEASE HELP I’LL GIVE BRAINLIST!!!!!!!!

maw [93]

The answer would be 208 if you find the area of each quadrilateral and add them together.

6 0

3 years ago

Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two

mafiozo [28]

Answer:

The difference of the original two-digit number and the number with reversed digits is 18.

Step-by-step explanation:

Since it's a two digit number, let x represent the tens digit and let y represent the units digit.

Thus, the original two digit number is;

10x + y.

The reverse two digit number is;

10y + x.

We are told that five times the sum of the digits of the two-digit number is 13 less than the original number.

Thus;

5(x + y) = (10x + y) - 13

Multiplying out the bracket gives;

5x + 5y = 10x + y - 13

Rearranging gives;

10x - 5x + y - 5y = 13

5x - 4y = 13 - - - - (3)

Also,we are told that, four times the sum of its two digits is 21 less than the reversed two-digit number. Thus;

4(x + y) = (10y + x) - 21 - - - (4)

Simplifying gives;

4x + 4y = 10y + x - 21

>> 10y - 4y - 4x + x = 21

>> 6y - 3x = 21 - - - (4)

Solving eq(3) and (4) simultaneously gives;

x = 5 and y = 3

Thus,

Original number = 53

Reversed number = 35

Difference between original and reversed number = 53 - 35 = 18

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?

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

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