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ivann1987 [24]
2 years ago
15

One large jar and five small jars can hold 26 ounces of jam. One large jar minus one small jar can hold 2 ounces of jam. Use the

given matrix equation to solve for the number of ounces of jam that each jar can hold. Explain the steps that you took to solve this problem.
A matrix with 2 rows and 2 columns, where row 1 is 1 and 5 and row 2 is 1 and negative 1, is multiplied by matrix with 2 rows and 1 column, where row 1 is l and row 2 is s, equals a matrix with 2 rows and 1 column, where row 1 is 26 and row 2 is 2.
Mathematics
1 answer:
Lesechka [4]2 years ago
4 0

Using a system of equations, it is found that one large jar holds 6 ounces and one small jar holds 4 ounces.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

  • Variable l: Weight that a large jar holds.
  • Variable s: Weight that a small jar holds.

One large jar and five small jars can hold 26 ounces of jam, hence:

l + 5s = 26, which is the first equation in matrix form.

Then:

l = 26 - 5s.

One large jar minus one small jar can hold 2 ounces of jam, hence:

l - s = 2, which is the second equation in matrix form:

Then:

l = 2 + s = 26 - 5s

2 + s = 26 - 5s

6s = 24

s = 4.

l = 26 - 5s = 6.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

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Answer:

n=206

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

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:

t_{\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 0.09, we can use as prior estimate of p 0.5, since we don't have any other info provided, 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)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.09}{2.58})^2}=205.44  

And rounded up we have that n=206

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