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4 years ago

Is 6.7 mi more precise than 6 mi?

Mathematics

2 answers:

Nataly [62]4 years ago

8 0

The answer is . . . <u><em>no</em></u> (and the reason is because you haven't provided enough information data for one to determine the precision of the distance).

"6.7 mi" you might want to argue is more precise simply because it measures this distance to the tenth of a mile

"6 mi" happens to equal "6.0 mi" which is also measured to the tenth of a mile.

Precision of a value is determined on the basis of repeatability and/or reproducibility. If the distance is measure 3 times with results that closely match each other - that's precision.

Unfortunately, you didn't provide us with enough information, so there's no way anyone could determine whether or not this measurement is precise.

Arisa [49]4 years ago

7 0

Yes because precise means specific.

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In triangle KLM, if K is congruent to L, KL = 9x - 40, LM = 7x - 37, & KM = 3x + 23, find x & the measure of each angle.

timurjin [86]

This is hard to solve when we don't know the visual graph this question is trying to refer from

6 0

3 years ago

James has been assigned to provide candy for a class party he needs 5.3 pounds if candy cost $4.25 per pound at Harolds candy st

RoseWind [281]

Answer:

5.3 * 4.25 = 22.525

22.525 is about 22.5

so the answer is $22.50

3 0

2 years ago

Hem dibuixat tres rectangles. En el primer, la llargada mesura 3 cm més que l'amplada. El segon i tercer rectangle tenen unes di

quester [9]

Resposta:

Primer rectangle:

Amplada = 11

Longitud = 14

Segon rectangle:

Amplada = 12

Longitud = 15

Tercer rectangle:

Amplada = 13

Longitud = 16

Explicació pas a pas:

Donat que:

Primer rectangle:

Amplada = x

Longitud = x + 3

2n rectangle:

Augment de la dimensió d'1 cm respecte al primer rectangle;

Amplada = x + 1

Longitud = x + 4

3r rectangle:

Augment de la dimensió de 2 cm respecte al primer rectangle;

Amplada = x + 2

Longitud = x + 5

Suma dels tres perímetres del rectangle:

Perímetre d'un rectangle: 2 (l + O)

Primer rectangle:

2 (x + x + 3) = 2 (2x + 3) = 4x + 6

2n:

2 (x + 1 + x + 4) = 2 (2x + 5) = 4x + 10

3r:

2 (x + 2 + x + 5) = 2 (2x + 7) = 4x + 14

Suma de perímetres = 162

(4x + 6 + 4x + 10 + 4x + 14) = 162

12x + 30 = 162

12x = 162 - 30

12x = 130

x = 11

Per tant,

Primer rectangle:

Amplada = 11

Longitud = 11 + 3 = 14

2n rectangle:

Amplada = 11 + 1 = 12

Longitud = 11 + 4 = 15

3r rectangle:

Amplada = 11 + 2 = 13

Longitud = 11 + 5 = 16

8 0

3 years ago

The area of a rectangle is 108 square meters, the width is 9 meters. what is the lenght

hjlf

Answer: 12 meters

Step-by-step explanation: 108 / 9 = 12

7 0

2 years ago

Read 2 more answers

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".

- 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.

- 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