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igomit [66]
2 years ago
6

The mass of a basket containing 3 balls was 3.64 kilograms. The mass of the same basket and 5 balls is 5.26 kilograms. Find the

mass of the empty basket in kilograms.
a. the mass of a empty basket is ___
b. explain your answer (in short)

please help
Mathematics
2 answers:
aksik [14]2 years ago
7 0

Answer:

mass of empty basket: 1.21 kg

Explanation:

the mass of basket is constant. Let the mass of basket be "b" and mass of each balls be "m"

Make equations:

  • b + 3(m) = 3.64  ___ equation 1
  • b + 5(m) = 5.26  ___ equation 2

<u>solve them simultaneously,</u>

  • 3.64 - 3(m) = 5.26 - 5(m)
  • -3(m) + 5(m) = 5.26 - 3.64
  • 2(m) = 1.62
  • m = 0.81 kg

<u>mass of basket,</u>

  • b = 3.64 - 3(m)
  • 3.64 - 3(0.81 )
  • 1.21 kg
jenyasd209 [6]2 years ago
4 0
  1. Bucket be x
  2. balls be y

Now

  • x+3y=3.64=>x=3.64-3y--(1)
  • x+5y=5.26=>x=5.26-5y

So

  • 3.64-3y=5.26-5y
  • -3y+5y=5.26-3.64
  • 2y=1.62
  • y=0.81kg

So

  • x=3.64-3(0.81)=3.64-2.43=1.21kg
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Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size
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Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

For this case we can use as estimator for the proportion \hat p =0.5, since we don't have any other previous info. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

8 0
3 years ago
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