Who can help me and answer the 2 questions for me :(

The proof that point (1,root 3) lies on the circle that is centered at the origin and contains the point (0,2) is found in the t

AURORKA [14]

<h2>Answer:</h2>

D. Distance formula

<h2>Step-by-step explanation:</h2>

$The \ \mathbf{distance} \ d \ between \ the \ \mathbf{points} \ (x_{1},y_{1}) \ and \ (x_{2},y_{2}) \ in \ the \ plane \ is:\\ \\ d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

As in the statement:

<em>The distance form</em> $(0,0)$ to $(0,2)$ is $\sqrt{(0-0)^2+(2-0)^2}=\sqrt{2^2}=2$ <em>whose justification is the </em><em>Distance Formula. </em>

<em />

Here in this statement the justification is also the Distance Fromula, so we take the distance from $(0,0) \ to \ (1,\sqrt{2})$ whose result is also 2 and is the radius of the circle.

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Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will

ICE Princess25 [194]

Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Step-by-step explanation:

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

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

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .