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Gennadij [26K]
3 years ago
13

In the measurement 0.503 l, which digit is the estimated digit? the 0 immediately to the left of the 3 5 the 0 to the left of th

e decimal point 3
Mathematics
2 answers:
BARSIC [14]3 years ago
5 0
I believe the answer is 3! Hope this helps!
babymother [125]3 years ago
5 0

Answer:

3.

Step-by-step explanation:

We are asked to find the estimated digit in the measurement 0.503 L.

We know that in each measurement the estimated digit is always the last digit.

We can see that the first digit in the given measurement is 5, the second digit is 0 and the last digit is 3.

Since the last digit is in our given measurement 3, therefore, the estimated digit is 3.

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Answer: x < 3

Step-by-step explanation:

Solve the rational inequality by combining expressions and isolating the variable x.

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3 years ago
Which is NOT a true statement?
Butoxors [25]
A

This is because 4x18 = 72
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6 0
2 years ago
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B

Step-by-step explanation:

6 0
2 years ago
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What is the LCM of 3,5,7
Lubov Fominskaja [6]

Answer:

The least common multiple would be 105.

Step-by-step explanation:

You would do list out the multiples for each. Or multiply

Since 3 x 30 is 90, and 3 x 5  is 15, This is 105

And since 5 x 20 is 100, and 5 x 1 is 5, This is 105 as well

Also, since 7 x 10 is 70, and 7 x 5 is 35, This is 105.

The LCM is 105

Have a great Day!

4 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
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