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Effectus [21]
3 years ago
10

What is the difference between adding and subtracting fractions and multiplying and dividing fractions?

Mathematics
1 answer:
77julia77 [94]3 years ago
4 0
With adding and subtracting fractions, you must find a common denominator. In multiplication, you just multiply numerator by numerator; denominator by denominator. In division, you multiply the fraction by the reciprocal of the 2nd. In multiplication and division, finding a common denominator is not needed.
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A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
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Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;

                             T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = 31.8 ounces

             s = sample standard deviation = 0.48 ounces

             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0
3 years ago
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