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lubasha [3.4K]
3 years ago
14

Well Basically, it’s kinda a struggle to understand this math your explaining:( Well, 3/6 = 8/?

Mathematics
2 answers:
dangina [55]3 years ago
8 0
Remember, you can do anything to an equation as long as you do it to both sides


3/6=8/?
multiply both sides by ?
3?/6=8
mulitly both sides by 6
3?=(6)(8)
3?=48
divide both sides by 3
?=16


16 goes in the quation mark place
lord [1]3 years ago
7 0
16 will go where the question mark is located

So 3/6 = 8/16

Think of it like this: if you had three slices of pizza out of 6 total, then you had half the pizza. The same goes for if you had 8 slices out of 16 total

You can use the cross multiplication rule to find x
3/6 = 8/x
3*x = 6*8
3x = 48
3x/3 = 48/3
x = 16
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The 45 in snake is the shorter one.

Step-by-step explanation:

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3 years ago
HELPPPP <br><br> When y=4x-6 find y if x=1/2
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A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
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Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

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

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The expected value of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

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\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

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Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

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