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ikadub [295]
3 years ago
15

Round 981.992 to the nerest whole number

Mathematics
2 answers:
kogti [31]3 years ago
6 0
982 or 980 depending on where you are rounding to...
zepelin [54]3 years ago
5 0
999.68 would be 1000
999.32 would be 999
Therefore 981.992 would be 982 if you were rounding up
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Graph the line -3x + y = -6
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Step-by-step explanation:

Step 1: Rearrange -3x + y = -6 into y  = mx + b

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Step 2: Find the y-intercept

The y-intercept is "b" so in this case the y-intercept is -6

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We start at (0,3) the y-intercept and go up by 3 and left by 1 (\frac{rise}{run} =\frac{3}{1})

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What is the slope of a line perpendicular to the line whose equation is x +y= 2.
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Geologist has collected 10 specimens of basaltic rock and 10th specimens of granite. The geologist instructs a laboratory assist
olya-2409 [2.1K]

Answer:

p(5)=0.00136

(a) p(x)=\frac{(10Cx)*((10)C(15-x))}{20C10}

(b)p(10)+p(5)=0.00272

(c)p(7)+p(8)=0.0584

Step-by-step explanation:

If we have N elements with k elements that we consider success and N-k elements that we consider fail, and we take a sample of n elements, the probability that there are x successes in the sample follows a hypergeometric distribution, so it is calculated as:

p(x)=\frac{(kCx)*((N-k)C(n-x))}{NCn}

In this case, N is equal to 20, k is equal to 10 specimens of granite rock, N-k is equal to 10 specimens of basaltic rock and n is 15. So, the probability that there are x specimens of granite rock in the sample or the pmf of the number of granite specimens selected for analysis is:

p(x)=\frac{(10Cx)*((10)C(15-x))}{20C10}

Now, the probability of getting exactly 5 granite specimens selected for analysis is:

p(5)=\frac{(10C5)*((10)C(15-5))}{20C10}=0.00136

On the other hand, the probability that all specimens of one of the two types of rock are selected  for analysis is the probability to select 10 granite specimens or 10 basaltic specimens. Where the probability to take 10 basaltic specimens is equal to the probability to take 5 granite specimens.

Then, the probability that all specimens of one of the two types of rock are selected  for analysis is:

p(10)+p(5)=\frac{(10C10)*((10)C(15-10))}{20C10}+\frac{(10C5)*((10)C(15-5))}{20C10}\\p(10)+p(5)=0.00136+0.00136=0.00272

For the hypergeometric distribution, the mean E(x) and standard deviation S(x) are:

E(x)=\frac{nk}{N}=\frac{15*10}{20}=7.5\\S(x)=\sqrt{\frac{nk}{N}*(1-\frac{k}{N})*(\frac{N-n}{N-1})}\\S(x)=\sqrt{\frac{15*10}{20}*(1-\frac{10}{20})*(\frac{20-15}{20-1})}=0.9934

Then, the number of granite specimens within 1 standard deviation of its mean value is:

  E(x) - S(x) ≤ mean ≤ E(x) + S(x)

7.5-0.9934 ≤ mean ≤ 7.5+0.9934

     6.5066 ≤ mean ≤ 8.4934

               7 ≤ mean ≤ 8

Finally, the probability that the number of granite specimens selected for analysis is  within 1 standard deviation of its mean value is:

p(7)+p(8)=\frac{(10C7)*((10)C(15-7))}{20C10}+\frac{(10C8)*((10)C(15-8))}{20C10}\\p(10)+p(5)=0.0292+0.0292=0.0584

8 0
4 years ago
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