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Crank
3 years ago
14

Subtract one-half from 8 times w

Mathematics
1 answer:
NeTakaya3 years ago
3 0

The equation would be set up as 8w -1/2.

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The scatter plot below shows a linear association. Which is the best linear model for the data?
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ye its y=5x-10

Step-by-step explanation:

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PLZ I WILL GIVE BRAILIEST TO THE FIRST ANSWER
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Option B is the correct answer.

3 is greater than -5, -9, and -1.

Hint: Remember to distribute the negative when solving. A negative directly outside outside of and to the left the parenthesis times a negative number inside the parenthesis makes that number positive.

So for option B, you can simplify it to -3 +4 +2. And that equals 3.

A) -3+-4-(-2) = -5

B) -3-(-4)-(-2) = 3

C) -3+(-4)-2 = -9

D) -3-(-4)-2 = -1
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Which property can be used to show that if 3a = 4b, then 4b = 3a?
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12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
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)

So

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

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

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.

5 0
3 years ago
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