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KengaRu [80]
2 years ago
11

Write the phrase as a numerical expression.

Mathematics
1 answer:
mariarad [96]2 years ago
7 0

3. 8+5

4. 2x - 8

Step-by-step explanation:

3. for example : 5 more than the number means

x+5 so that's how we get the answer as 8+5

4. This can be explained in two folds firstly, for example: 2 less than the number means x-2.

secondly;7 times the number means 7x so therefore we get our final answer as 2x - 8

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A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other 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 percentage of organic matter = 2.481%

             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

                                     =  -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0
3 years ago
I giveee brainlilsttt
EastWind [94]

Answer:

-5,1

Step-by-step explanation:

hope that helps

:))))))

5 0
2 years ago
What’s the fraction of 65 percent
Shtirlitz [24]
65% as a fraction is 65/100 which can also be reduced to 13/20
4 0
3 years ago
⚠️⚠️ Help I’m in desperate need of help ⚠️⚠️ the questions are basically in the top picture
Tom [10]

Answer:

1. 3^{3} * 3^{2}

2.6^{-4}

3. 5^{0} < 5^{1}

7 0
2 years ago
PLEASE LOOK AT PHOTO! WHOEVER ANSWERS CORRECTLY I WILL MARK BRAINIEST!!
Deffense [45]

Answer:23

Step-by-step explanation:

8 0
2 years ago
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