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Vanyuwa [196]
3 years ago
7

Look at the photo attached! the surface area of the cylinder is....??

Mathematics
1 answer:
sattari [20]3 years ago
8 0
The image is not loading for me, did you upload something ?
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HELP!!!!
Olin [163]
So if she picked 32 apples on Tuesday, and she picked 15 more apples than on Monday, then she picked 17 apple on Monday, because 17+15=32. A=17
6 0
3 years ago
What form is the equation below written in?<br> y = (x - 2)2 - 9
fgiga [73]

Answer:

linear equation

Step-by-step explanation:

it is because x has degree 1.

degree 2:- quadratic

,,. 3:- cubic

,,. 4:- bi quadratic

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3 0
3 years ago
How I divided a hexagon into 3 equal parts ?
Furkat [3]
It splits into 6 easerly so just make each part double walls as the pic shows

3 0
3 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

<em>Let </em>\mu<em> = true average percentage of organic matter in such soil</em>

SO, <u>Null Hypothesis</u>, H_0 : \mu = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}

<u>Alternate Hypothesis</u>, H_A : \mu \neq 3%   {means that the true average percentage of organic matter in such soil is different than 3%}

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

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

where,  \bar X = sample mean amount of organic matter = 2.481%

              s = sample standard deviation = 1.616%

              n = sample of soil specimens = 30

So, <u><em>test statistics</em></u>  =  \frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }  ~ t_2_9

                               =  -1.759

<u></u>

<u>Now, P-value of the test statistics is given by;</u>

       P-value = P( t_2_9 > -1.759) = <u>0.046</u> or 4.6%

  • If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
  • If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0
3 years ago
Pasty makes lemonade every morning. She uses 2 1/4 cup of sugar and 4 3/5 cups of lemon juice. How may cups of sugar and lemon j
Artyom0805 [142]
6 17/20 cups? Or 137/20 cups i guess
7 0
3 years ago
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