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MissTica
3 years ago
8

What is the GCF of 12b and 40b^4

Mathematics
1 answer:
Scilla [17]3 years ago
7 0
Finding the prime factors of each term to get the GCF as 4b 
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Solve by elimination <br> 5x-y=12 <br> 3x+2y=2
Ber [7]
(2)5x-y=12
   3x+2y=2
10x-2y=12
 3x+2y=2
13x=2
x=0.15
7 0
3 years ago
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What is the following quotient 9+sqrt 2/4-sqrt 7
babunello [35]

9 + √2 / 4 - √7

Multiply the numerator and denominator by the conjugate.

Exact form:

(36 + 9√ 7 + 4√2 + √14) / 9

or

about 7.69

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4 years ago
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How can i solved <br>?????????????????????????????
alexgriva [62]
1. figure c
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4 0
3 years ago
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A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

3 0
3 years ago
10 times as many as 1ten is
siniylev [52]

Answer:

1

Step-by-step explanation:

10 * 1/10 is 1

4 0
3 years ago
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