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2 years ago

Condense the expression log 2x + log(x² - 9) - log x +3

Mathematics

1 answer:

Rudik [331]2 years ago

4 0

Here the my solutions,hope you understand

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63% of 300 is what number

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Answer:

189

Step-by-step explanation:

You take 63% make it a decimal by dividing it by 100 and then you multiply it with 300. 300*.63=189.

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3 years ago

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Aloiza [94]

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3 years ago

Find the vertex of the parabola.<br> x² + y - 6x + 10 = 0

Drupady [299]

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2 years ago

A two digit number is written at random what is the probability that the number will be odd

leonid [27]

Answer:

50%

Step-by-step explanation:

So there are 90 2 digit numbers. There are 45 2 digit odd numbers.

The probability should be 50%

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3 years ago

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

nordsb [41]

Answer:

a) X is binomial with n = 10 and p = 0.3

Y is binomial with n = 10 and p = 0.7

b) The mean number of errors caught is 7.

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.

Binomial probability distribution

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

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

10 word errors.

This means that $n = 10$

(a) If X is the number of word errors missed, what is the distribution of X ?

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.

What is the mean number of errors missed?

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

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

2 years ago