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Dmitrij [34]
3 years ago
9

Help if yk plz and thanks

Mathematics
2 answers:
dalvyx [7]3 years ago
8 0

Answer:

3f

Step-by-step explanation:

<u>The expression given is f + f + f.</u>

f + f + f = f × 3

<u>The correct way to write an expression that is being multiplied is to put the number before it with no symbols or spaces in between them:</u>

f + f + f =

f × 3 =

3f

den301095 [7]3 years ago
7 0

Answer:

I am pretty sur it is 3

Step-by-step explanation:

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Rounding down from 100 to the nearest thousand is 0.

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2. 9 x 2 + 6 - 4 - 1
3. 18 + 6 - 4 - 1
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rosijanka [135]

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The slope is 1.

Step-by-step explanation:

Remember:

m = \frac{y_2 - y_1 }{x_2 - x_1}

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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
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