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MaRussiya [10]
3 years ago
13

Problema: Se guarda una moneda antigua en una caja cubica de tal forma que el contorno de la moneda toca las 4 paredes de la caj

a, si la base de la caja tiene un perímetro de 24 cm ¿Cual es el área de la moneda?
Mathematics
1 answer:
nikdorinn [45]3 years ago
3 0

Answer: 50.24 cm^2

Step-by-step explanation:

This can be translated to:

An old coin is kept in a cubic box in such a way that the outline of the coin touches the 4 walls of the box, if the base of the box has a perimeter of 24 cm. What is the area of ​​the coin?

The fact that the coin touches the interior of the box means that the diameter of the coin is equal to the side lenght of the box.

The perimeter of the box is 24 cm, and the perimeter of a square is equal to:

P = 4*L

where L is the side lenght of the square.

24 cm = 4*L

L = 24cm/4 = 8cm

Now we know that the diameter of the coin is 8cm

Now, the area of a circle (the coin) is equal to:

A = 3.14*(d/2)^2

where d is the diameter, so we have:

A = 3.14*(4cm)^2 = 50.24 cm^2

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Answer:

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

df=n_1 +n_2 -2=13+19-2=30  

Since is a right tailed test the p value would be:  

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

\bar X_{1}=810 represent the mean for sample 1  

\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

\alpha=0.05 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

P value  

Since is a right tailed test the p value would be:  

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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Answer:

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Step-by-step explanation:

Divide the circumference by π, or 3.14 for an estimation. The result is the circle's diameter.

<em>72.22 ÷ 3.24 = 23</em>

diameter = 23

Divide the diameter by 2.

<em>23 ÷ 2 = 11.5</em>

There you go, you found the circle's radius

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