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3 years ago

Rationalize

Mathematics

1 answer:

guapka [62]3 years ago

7 0

(5th root of 3 Times Square root of five) / 5

You do this by multiplying by sqrt5/sqrt5 which is 1

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The verticles of triangle A are (-1,3) B (7,9) and C (15,-3) Find the coordinates of the intersection of the medians.

Annette [7]

Step-by-step explanation:

Coordinates of A≡(−1,3)

Coordinates of B≡(7,9)

Coordinates of C≡(15,-3)

The median through the vertex C will meet at the mid point of AB.

Coordinate of mid-point on AB=[(-1+7)/2 , (3+9)/2]= (3,6)

Now,

Length of median through vertex C= Distance between (15,-3) and (3,6)

Therefore,

Length of median = √(3-15)²+(6-(-3))²=√144+81=15unit

Hence the length of median through the vertx C is 15 units.

6 0

3 years ago

A rational number between 1.7 and 1.8

vitfil [10]

1.75 would be the direct center between those two irrational numbers.

6 0

3 years ago

Okay can someone help me with 19 the answer is height = 5 but someone please explain the process to answer

sertanlavr [38]

Answer:

hope it helps you see the attachment for further information

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3 years ago

The population of a city is currently 2,395,326.

mylen [45]

I’m pretty sure the answer to this is 1,932,934.

3 0

3 years ago

A new, simple test has been developed to detect a particular type of cancer. The test must be evaluated before it is used. A med

g100num [7]

Answer:

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Test indicates cancer.

Event B: Person does not have cancer.

Probability of a test indicating cancer.

98% of 2%(those who have).

1% of 100 - 2 = 98%(those who do not have). So

$P(A) = 0.98*0.02 + 0.01*0.98 = 0.0294$

Probability of a test indicating cancer and person not having.

1% of 98%. So

$P(A \cap B) = 0.01*0.98 = 0.0098$

What is the probability of a randomly chosen person not having cancer given that the test indicates cancer?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0098}{0.0294} = 0.3333$

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

3 0

3 years ago