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AlekseyPX
3 years ago
8

El entrenamiento que hace marco 10 dias antes de su carrera es constante. Diariamente corre 45 minutos a una velocidad de 30 km-

h ¿Cuántos kilómetros ha corrido Marcos durante esos 10 dias?
Mathematics
1 answer:
alexira [117]3 years ago
6 0

Responder:

225 kilometros

Explicación paso a paso:

Dado que :

Velocidad de funcionamiento = 30 km / h

Duración = 45 minutos

Número de días = 10 días

Distancia total recorrida por día:

Duración en horas:

45 minutos / 60 = 0,75 hora (s)

Así, la distancia recorrida por día:

Tiempo de velocidad

30 km / h * 0,75 h = 22,5 km

Por lo tanto, la distancia total recorrida durante 10 días:

Distancia recorrida diariamente * 10

22,5 kilometros * 10 = 225 kilometros

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3 years ago
The number of letters in Stephanie's full name is sixteen less than twice the number of letters in Amy's full name. If the numbe
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Answer:

19 letters

Step-by-step explanation:

Suppose Number of Amy's full name letters = X

               Number of Stephanie's full name letters = Y

Now, according to given condition that ''number of letters in Stephanie's full name is sixteen less than twice the number of letters in Amy's full name'',

Equation 1 becomes

Y = 2X-16

Now, according to second condition, product of their names' letter is 418. So,

Equation 2 becomes

XY = 418

Deducing value of X from equation 1,

X = (Y+16)/2

Putting this value in equation 2 we get,

{(Y+16)/2}*Y = 418,

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(Y^2)+16Y-836=0

By breaking middle term (You can use quadratic formula here as well)

(Y^2)+38Y-22Y-836=0

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At this stage we have two values for Y,

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Now considering only positive value since the number of letters in a name can not be in negative number,

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3 years ago
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3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



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4 years ago
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Answer: (-3,-2)

Step-by-step explanation:

3 0
4 years ago
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