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Mrac [35]
3 years ago
12

A group of 500 transistors is known to contain one defective unit. What is the probability that a transistor selected at random

from the lot is the bad one? 1/500 499/500 1/499
Mathematics
2 answers:
lesya692 [45]3 years ago
5 0

Answer:

Option A is correct.

Step-by-step explanation:

Given: Number of transistors in a lot = 500

            Number of defective transistors in a lot = 1

To find: Probability of selecting a defective transistor from the lot.

Probability=\frac{Number\:of\:Favorable\:Outcome}{Total\:Number\:of\:Outcome}

Probability\:of\:getting\:a\:bad\:transistor=\frac{1}{500}

Therefore, Option A is correct.

sergejj [24]3 years ago
3 0
The answer is 1/500 because it will always be the probability of the event occurring/total events
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Gala2k [10]

Answer:0

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Step-by-step explanation:

5 0
3 years ago
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Evaluate the following expression.<br> 1/5^-2
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This is how the whole equation goes:
1/x^-a = x^a
So, 1/5^-2 is the same as 5^2, which is 25. 
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Grading managers. Some companies "grade on a bell curve" to compare the performance of their managers and professional workers.
Monica [59]

Answer:

\mu = 250, \sigma = 175.781

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the grades of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,\sigma)  

For this case we have two conditions given:

P(X

P(X>475) = 0.1 or equivalently P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

So we can find a value from the normal standard distribution that accumulates 0.1 and 0.9 of the area in the left, for this case the two values are:

z= -1.28, z=-1.28

We can verify that P(Z<-1.28) =0.1[/tex] and P(Z<1.28) =0.9[/tex]

And then using the z score we have the following formulas:

-1.28 = \frac{25 -\mu}{\sigma}   (1)

1.28 = \frac{475 -\mu}{\sigma}   (2)

If we add equations (1) and (2) we got:

\frac{25 -\mu}{\sigma} + \frac{475 -\mu}{\sigma} =0

We can multiply both sides of the equation by \sigma and we got:

25+ 475 -2 \mu = 0

\mu = \frac{500}{2}= 250

And then we can find the standard deviation for example from equation (1) and we got:

\sigma = \frac{25-250}{-1.28}=175.781

So then the answer would be:

\mu = 250, \sigma = 175.781

 

3 0
3 years ago
A line that includes the points (s, -9) and (5, -6) has a slope of -3. What is the value of s?
OleMash [197]

Answer:

s=6

Step-by-step explanation:

slope(m) = (y2-y1)/(x2-x1)

or, -3 = (-6+9)/(5-s)

or, -3(5-s) = 3

or, -15+3s = 3

or, 3s = 18

so, s = 6

5 0
2 years ago
A college basketball player makes 80% of his freethrows. Over the course of the season he will attempt 100 freethrows. Assuming
BARSIC [14]

Answer:

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

Step-by-step explanation:

According to the given data we have the following:

P(Make a Throw) = 0.80%

n=100

Binomial distribution:

mean:   np = 0.80*100= 80

hence, standard deviation=√np(1-p)=√80*0.20=4

Therefore, to calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:

P(X>80)= 1- P(X<80)

You could calculate this value via a normal distributionapproximation:

P(Z<(80-80)/4)=1-P(Z<0)=1-50=0.50

The probability that the number of free throws he makes exceeds 80 is approximately 0.50

3 0
3 years ago
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