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oksano4ka [1.4K]
3 years ago
5

104 divided 4 with distributive property

Mathematics
2 answers:
ollegr [7]3 years ago
6 0
The distributive property makes this a lot easier to answer. Dividing by 4 is the same as multiplying by 1/4, so:
1/4 x 104
104 equals 100 plus 4, so:
1/4 x (100 + 4)
now distribute the 1/4:
1/4 * 100 + 1/4 * 4
multiplying by 1/4 is the same thing as dividing by 4, so:
100/4 + 4/4
100 divided by 4 is 25, and 4 divided by 4 is 1:
25 + 1
26
therefore 104 divided by 4 is 26
Vikki [24]3 years ago
4 0
You can do 104 divided by 2 and get 52 and then you can divide it for 2 again and get 26
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Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but inc
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a) X is binomial with n = 10 and p = 0.3

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The mean number of errors missed is 3.

c) The standard deviation of the number of errors caught is 1.4491.

The standard deviation of the number of errors missed is 1.4491.

Step-by-step explanation:

For each typing error, there are only two possible outcomes. Either it is caught, or it is not. The probability of a typing error being caught is independent of other errors. So we use the binomial probability distribution to solve this question.

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Probability of exactly x sucesses on n repeated trials, with p probability.

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10 word errors.

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Human proofreaders catch 70 % of word errors. This means that they miss 30% of errors.

So for X, p = 0.3.

The answer is:

X is binomial with n = 10 and p = 0.3.

If Y is the number of word errors caught, what is the distribution of Y ?

Human proofreaders catch 70 % of word errors.

So for Y, p = 0.7.

The answer is:

Y is binomial with n = 10 and p = 0.7

(b) What is the mean number of errors caught?

E(Y) = np = 10*0.7 = 7

The mean number of errors caught is 7.

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E(X) = np = 10*0.3 = 3

The mean number of errors missed is 3.

(c) What is the standard deviation of the number of errors caught?

\sqrt{V(Y)} = \sqrt{np(1-p)} = \sqrt{10*0.7*0.3} = 1.4491

The standard deviation of the number of errors caught is 1.4491.

What is the standard deviation of the number of errors missed?

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.3*0.7} = 1.4491

The standard deviation of the number of errors missed is 1.4491.

6 0
3 years ago
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