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deff fn [24]
3 years ago
9

0.8 is 10 times as great as wich decimal?

Mathematics
1 answer:
cestrela7 [59]3 years ago
6 0
I don't know what to say, I've been trying to explain this for like... 10 minutes but I can't think of how to explain it. All I can say is to just divide .8 by 10. The answer is .08.
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Mark is in training for a bicycle race. He needs to ride 50 km a week as part of his training. He rode 18.23 km on Monday and 13
konstantin123 [22]
The answer is B-about 18km


This is because you round 18.23 to 18km


then you round 13.94 to 14km


you add these rounded numbers together


you take it away from 50km


the answer would then me 18km


but it be ABOUT because that was only a rough estimate because it was rounded
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Step-by-step explanation:

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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_%        

                                                           

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daser333 [38]

The correct answer is B.

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Which is the graph of<br><img src="https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7Bx%20%2B%201%7D%20-%202" id="TexFormula1" t
Anna11 [10]

ANSWER

See attachment.

EXPLANATION

The given function is

y =  \sqrt[3]{x + 1}  - 2

The parent function is

y =  \sqrt[3]{x}

When we shift the parent graph to the left one unit, and down 2 units, we obtain the graph of the given function.

The graph of this function is shown in the attachment.

5 0
3 years ago
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