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Nookie1986 [14]
2 years ago
7

10w + 4w = -84 solve by combining like terms show all steps leading to awnser plz

Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
3 0
10w + 4w = - 84
14w = - 84
w = - 84 : 14
w = - 6  (  answer )
-----------------------------------
10 * (- 6) + 4 * (- 6) = - 84
            - 60 + (- 24) = - 84
                         - 84 = - 84
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pishuonlain [190]

Answer:

Finding the missing number is fairly simple and can be done in a few short steps. I will show you using the 21 / __ fraction.

First, divide the included number by its corresponding number in the original fraction.

21 ÷ 3 = 7

To find what the denominator of this would be, multiply 4 by 7.

4 * 7 = 28

The fraction would be 21 / 28.

This works with both numerators and denominators. If you need more of the answers just ask in the comments.


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Step-by-step explanation:

Using exponent rules an exponent to the power of a number will multiply by each other.

So the -4 and -9 will multiply.

The 6 in the numerator will now have a positive exponent of 36.

\frac{6^{36}}{6^6}

Now we can divide and when we do so the exponents will subtract from each other.

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Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that p = 0.12.

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so n = 18

This probability is:

P = P(X = 0) + P(X = 1)

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So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

There is a 34.60% probability that 0 or 1 of them is nearsighted.

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