Find the cross product a ⨯ b. a = 2, 5, 0 , b = 1, 0, 9

Three years ago a school spent four hundred thirty-eight thousand, seven hundred sixty-nine dollars and seventy-nine cents build

alukav5142 [94]

Answer: $438,769.79
Please mark me Brainliest

4 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

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

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

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

3 0

3 years ago

Solve for x on each of these questions please!

IgorLugansk [536]

Answer:

1. x = 10

2. x = 4

Step-by-step explanation:

I use the angle ABC method:

AB² + AC² = BC²

6² + 8² = x²

x = 10

AB² + AC² = BC²

3² + x² = 5²

x = 4

HOPE THIS HELPS AND HAVE A NICE DAY <3

5 0

2 years ago

Help me with this problem

ryzh [129]

The answer is 60 since the other shape has same answer for that side

8 0

2 years ago

Read 2 more answers

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. Thi

Dovator [93]

Complete question:

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.

A) Find the mean and standard deviation for the number of girls in groups of 36 births.

B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.

C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?

Answer:

a) mean = 18

Standard deviation =3

b) low range = 12

High range = 24

c) The result of 33 girls is significantly high. Yes, the method is effective.

Step-by-step explanation:

Given:

p = 0.5

n = 36

a) The mean is the product of n and p

Mean u = np

u = 36 * 0.5 = 18

The standard deviation is the square root of the product of n and p&q.

S.d ó = $\sqrt{npq}$

$= \sqrt{np(1-p)}$

$= \sqrt{36(0.5)(1-0.5)} = \sqrt{9} = 3$

b) To find the range rule of thumb:

• For low range

Low range = u - 2ó

= 18 - (2 * 3)

= 12

• High range

= u + 2ó

= 18 + (2*3)

= 24

c) The result is significantly high, because 33 is greater than 24 girls.

A result of 33 girls would prove the method as effective.

8 0

3 years ago