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2 years ago

Hiiii can u please pls pls pls

Mathematics

2 answers:

Shkiper50 [21]2 years ago

4 0

Answer:

x | y

0 | 0

6 | 2

12 | 4

Step-by-step explanation:

Multiply each x value in the table by 1/3 to get 0, 2, and 4 for your y values.

kotegsom [21]2 years ago

3 0

Answer:

x y

0 0

6 2

12 4

Step-by-step explanation:

Is means equals

The equation is

y = 1/3 x

Let x = 0

y = 1/3 (0) = 0

Let x = 6

y = 1/3 (6) = 2

Let x = 12

y = 1/3 (12) = 4

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What is 57.2 times 56.4

Rainbow [258]

Answer:

3226.08

Step-by-step explanation:

57.2 x 56.4

6 0

3 years ago

Read 2 more answers

Solve the system of linear equations.

icang [17]

Answer: The set does not have a solution

Step-by-step explanation:

Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5

Putting this value of x in eq. 2 we get

-2y + 2z = -1-(7/5) or

2y - 2z = 12/5 or 5y - 5z = 6

Multiplying eq. 1 by 2 we get

4x + 2y - 2z = 6

adding this with eq. 2 we get 5x = 5 or x = 1

As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.

Same can be verified by using different sets of two equations also.

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?

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

If T(n)=6n-1what is the first term

OleMash [197]

So if you find the 1st term, you should write 1 instead of n
T1 = 6.1 -1
T1= 6-1
T1= 5
Hope it helps!
#MissionExam001

3 0

3 years ago

11. Give two examples of each type of number: real, natural, integers, rational and irrational.

zaharov [31]

Answer:

real:√2, 3,-1, 1/2 etc

natural: 1,2,3,4,5,6.....(0 is not included)

integers:. .............-4,-3,-2,-1,0,1,2,3,4........ etc

rational: nos. which are in p/q form

:1/2,3/4,4/9 etc

irrational: nos. which cannot be written in p/q form

: √2,√3... etc

irrational: √2, √3, √5, √11, √21, π(Pi)

7 0

2 years ago