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blagie [28]
2 years ago
6

1/3p multiplied by 9q is what?

Mathematics
1 answer:
Makovka662 [10]2 years ago
4 0

Answer:

3pq

Step-by-step explanation:

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6 0
3 years ago
Let X be the temperature in at which a certain chemical reaction takes place, and let Y be the temperature in (so Y = 1.8X + 32)
Black_prince [1.1K]

Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The  temperature in at which a certain chemical reaction takes place.

                             Y = 1.8*X + 32

- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

                             P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

                             P ( Y ≤ 1.8*η + 32 )

- The left hand side of the inequality can be replaced by the linear relation:

                             P ( 1.8*X + 32 ≤ 1.8*η + 32 )

                             P ( 1.8*X ≤ 1.8*η )   ..... Cancel "1.8" on both sides.

                            P ( X ≤ η ) = 0.5 ...... Proven

Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

b)

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

                             P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

          P ( Y ≤ 1.8*p_0.9 + 32 ) = P ( 1.8*X + 32 ≤ 1.8*p_0.9 + 32 )

                                                 = P ( 1.8*X ≤ 1.8*p_0.9 )

                                                 = P ( X  ≤ p_0.9 )

                                                 = 0.9

- So from conjecture we saw that the 90th percentile of "X" distribution is also the 90th percentile of "Y" distribution.

c)

- The more general relation between two random variate "Y" and "X" is given:

                            Y = aX + b

Where, a : is either a positive or negative constant.

- Denote, (np) as the 100th percentile of the X distribution, so the corresponding 100th percentile of the Y distribution would be : (a*np + b).

- When a is positive,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≤ p_% )

                                                 = np_%        

- When a is negative,

                   P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

                                                 = P ( a*X ≤ a*p_% )

                                                 = P ( X  ≥ p_% )

                                                 = 1 - np_%        

                                                           

4 0
3 years ago
O,my god if you do this 2 pages I will be done with my homework
katovenus [111]

Answer:

1. 8% of 50 is 4

2. 95% of 40 is 38

3. 42% of 263 is 110.46

4. 110% of 70 is 77

5. 115% of 20 is 23

6. 130% of 78 is 101.4

7. 6.5% of 50 is 3.25

37. 15% of 50 is 7.5

38. 5% of 19 is 0.95

39. 8% of 275 is 22

I hope this helps! Can I plz have a Brainliest??

Step-by-step explanation:

6 0
2 years ago
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