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3 years ago

HELP I NEED TO COMPLEYE THSISJI toDay

Mathematics

2 answers:

olchik [2.2K]3 years ago

8 0

The answer is A.

You have to take out 7 and x from each because they’re alike which simplifies it and put it on the outside. Sorry not super scientific answer but it’s right

Pachacha [2.7K]3 years ago

7 0

Answer: A. 7x(2x + 5)

Step-by-step explanation:

7x * 2x = 14x^2

7x * 5 = 35x

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1/X=5/25=<br> What’s the answer?

snow_lady [41]

Answer:

1/5

Step-by-step explanation:

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3 years ago

8 is 1/6 of what number

charle [14.2K]

8 is 1/6 of 8/6 = 1 1/3

125 is 5/10 of 625/10 = 62 1/2

32 is 4/10 of 128/10 = 12 4/5

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2 years ago

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How does 4 digit addends have a 5 digit sum

denpristay [2]

A 4 digit addends could have a 5 digit sum because of the rule of regrouping.
Regrouping is a situation where carrying and borrowing is involves. So let’s have an example

8999
<span>+5678
</span>14677

so in this example, we added 8999 to 5678.
9+8 = 17, so we bring down 7 and carry 1
9 + 7 = 16 + 1 = 17 , so again we bring down 7 carry 1
9 + 6 = 15 + 1 = 16, bring down 6 carry 1
8+5 = 13 + 1 =14, since this is our last number, bring down 14..
=> 14, 677

3 0

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

3 years ago

as the schools sign language interpreter kiran gets paid 35.50 for every parent-teacher conference that he attends. He also gets

yKpoI14uk [10]

<u>The interpreter attended </u><u>22 parent-teacher conference</u><u> and </u><u>5 school related assembly</u>

To solve this problem, we would write out two set of linear equations.

The data given on this problem are

parent-teacher conference = $35.50
school related assembly = $42
The total number of engagements = 27

Let x represent the number of parent-teacher conference

let y represent the number of school related conference

<h3>Equations</h3><h3> $x+y = 27...equation (i)\\35.50(x)+42y=991...equation(ii)$ </h3>

From equation (i)

$x+y = 27\\x = 27-y...equation(iii)\\$

Put equation(iii) into equation (ii)

$35.50x+42y=991\\35.50(27-y)+42y=991\\958.50-35.50y+42y=991\\6.50y=991-958.50\\y=5$

Put y = 5 into equation 1

$x+y=27\\y = 5\\x+5=27\\x=27-5\\x=22$

From the above calculations, he attended 22 parent-teacher conference and 5 school related assembly.

Learn more on linear equations here;

brainly.com/question/4074386

8 0

2 years ago

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