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Nata [24]
3 years ago
5

Factor the sum of the term as a product of the GCF and a sum. GCF greatest common factor. 28+35

Mathematics
1 answer:
TiliK225 [7]3 years ago
3 0
I think its just.1for 1,2,4,7,14,28 an 1,3,5,35
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Which fraction is larger 10/13 or 11/14
dem82 [27]
10/13 would be the larger number and here is why

10/13

10/13 x 3780/3780

37800/49140

so your answer is 10/13


5 0
3 years ago
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Can someone pls help solve this
Len [333]

Answer:

YOU JUST NEED T APPLY THE FORMULA OF CIRCLE AND IRREGULAR SQUARE AS TRAPEZOIID

Step-by-step explanation:

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3 years ago
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
Luda [366]

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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Vladimir79 [104]

Answer:

o/3 or 1/1

Step-by-step explanation:

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2 and a half hours estimate
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