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Semmy [17]
3 years ago
12

Kevin works for a company that manufactures solar panels. In a large batch of solar panels, about 1 in 200 is defective. Suppose

that Kevin selects a random sample of six solar panels from this batch. What is the probability that none of the solar panels are defective?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

probability that none of the solar panels are defective = 0.97

Step-by-step explanation:

Proportion of solar panels that are defective, p = 1/200 = 0.005

Proportion of solar panels that are not defective, q = 1 - p = 1 - 0.005

q = 0.995

Sample size, n = 6

probability that none of the solar panels are defective, P(X=0)

Note: P(X=r) = nCr p^{r} q^{n-r}

P(X=0) = 6C0 * 0.005^{0} * 0.995^{6-0}\\P(X=0) = 6C0 * 0.005^{0} * 0.995^{6}\\P(X=0) = 1 * 1 * 0.9704\\P(X=0) = 0.97

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If KN =29, what is CN?
Nostrana [21]
29kn are equal to cn 2900000
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The length of a picture frame is 6 inches more than the width. For what values of x is the perimeter of the picture frame greate
zhannawk [14.2K]

Answer:

35

Step-by-step explanation:

35 is exactly 152 but since its asking for greater then you're right it is 36.

But since it says '>' it means not equal to so it would be:

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3 0
3 years ago
Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size
julsineya [31]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

For this case we can use as estimator for the proportion \hat p =0.5, since we don't have any other previous info. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

8 0
3 years ago
A triangle has angles measuring 23 and 48 degrees. What is the measure of the third angle?
icang [17]

Answer:

109

Step-by-step explanation:

Hope this helped :)

6 0
2 years ago
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ivann1987 [24]
-65, -13, -7, 7, 34, 55, 62.
Here’s your answer^
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Always negative will be least and positive will be the greatest always.
7 0
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