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Tpy6a [65]
3 years ago
7

What is the least common multiple of 3,9, 15?

Mathematics
2 answers:
olya-2409 [2.1K]3 years ago
3 0
LCM\ is\ the\ least\ common\ number\ that\ is\ multiply\ of\ 3,9,15:\\\\ Factors\ of\ 3\\ 3:3\\
1:1\\\\ Factors\ of\ 9:\\ 9:3\\
3:3\\1:1\\\\ Factors\ of\ 15\\
15:3\\
5:5\\
1:1\\\\ LCM(3,9,15)=3^2*5=9*5=45\\\\ \textbf{LCM (3,9,15)=45}
solniwko [45]3 years ago
3 0
15: 15,30,45
9:9,18,27,45
3: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45
common multiples: 45
<span>LCM: 45</span>
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Answer:

I think that the answer is A) $0.50

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So, if you want to figure out how much the strawberries cost for year 0, all you have to do is go to the point (0,4). For the first year the price of the strawberries have increased to 4.50. I know this because when you look at the graph, it looks like the point is in the middle of 4, which makes 4.50. So the price of the strawberries in year 1 is (1, 4.50). The price for year two is (2, 5) and so forth. This is what the pattern would be (0,4) (1, 4.50) (2, 5) (3, 5.50) (4,6) (5, 6.50) (6, 7) (7, 7.50) (8,8) (9, 8.50) (10, 9).......etc.

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Construct a confidence interval that will capture the true population mean with 95% confidence. Population standard deviation is
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A confidence interval that will capture the true population mean with 95% confidence. The value is ( 23.6459    , 24.8293 ) .

What is population mean?

A population in statistics is a group of comparable objects or occurrences that are relevant to a particular topic or experiment. A statistical population can be a collection of real things or a hypothetical, possibly limitless collection of objects derived from experience.

Data: 18

for given data

$$\begin{aligned}\bar{X} & =\frac{\sum X i}{n} \\& =2448 / 101 \\& =24.2376\end{aligned}$$

as population variance is unknown so we used sample variance

$$\begin{aligned}S^2 & =\frac{1}{n-1}\left(\sum X i^2-n \times \bar{X}^2\right) \\S^2 & =\frac{1}{100}\left(60232-101 \times 24.2376^2\right) \\& =8.9830\end{aligned}$$

Now

\frac{\bar{X}-\mu}{S / \sqrt{n}} \sim t_{n-1}$$

hence 95 % of population mean is

As n=101 which is large so critical value is 1.98397

$$\begin{aligned}& =\left(\bar{X}-\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}, \bar{X}+\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}\right) \\& =\left(24.2376-\sqrt{\frac{8.9830}{101}} \times 1.98397,24.2376+\sqrt{\frac{8.9830}{101}} \times 1.98397\right) \\& =(23.6459,24.8293)\end{aligned}$$

To learn more about statistics visit:brainly.com/question/29093686

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