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Phantasy [73]
3 years ago
13

Box 1 contains 1000 lightbulbs of which 10% are defective. Box 2 contains 2000 lightbulbs of which 5% are defective. (a) Suppose

a box is given to you at random and you randomly select a lightbulb from the box. If that lightbulb is defective, what is the probability you chose Box 1? (b) Suppose now that a box is given to you at random and you randomly select two light- bulbs from the box. If both lightbulbs are defective, what is the probability that you chose from Box 1? 4 Solve the Rainbow
Mathematics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

a) There is a 66.7% chance that you were given box 1

b) There is a 80% chance that you were given box 1

Step-by-step explanation:

To find this, we need to note that there is a 1/10 chance of getting a defective bulb with box 1 and a 1/20 chance in box 2.

a) To find the answer to this, find the probability of getting a defective bulb for each box. Since there is only one bulb pulled in this example, we just use the base numbers given.

Box 1 = 1/10

Box 2 = 1/2

From this we can see that Box 1 is twice as likely that you get a defective bulb. As a result, the percentage chance would be 2/3 or 66.7%

b) For this answer, we need to square each of the probabilities in order to get the probability of getting a defective one twice.

Box 1 = 1/10^2 = 1/100

Box 2 = 1/20^2 = 1/400

As a result, Box 1 is four times more likely. This means that it would be a 4/5 chance and have a probability of 80%

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The change in temperature for the day was 26 degrees F
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*place holder since picture has the work*
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A square.

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Evaluate A² for A = -3.<br> O-6<br> O 9<br> O 6<br> O-9
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Answer:

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3 0
1 year ago
A local high school has both male and female students.
Kitty [74]

Answer:

a. P(male) = 0.4

b. P(no sport and male) = 0.1

c. Unclear question (isn't it the same as b.?)

Step-by-step explanation:

The data below is what I've worked according to, which isn't very clear from the question so the answers are only correct if this is the correct table of data;

\left[\begin{array}{ccc}&No \ Sports&Sports\\Female&10&32\\Male&7&21\end {array}\right]

a.

P(male) = \frac{7 + 21}{70} \\\\ = \frac{28}{70} \\\\ = \frac{2}{5}

b.

Using the tree diagram in the picture;

P(no \ sport \ and \ male) = \frac{2}{5} * \frac{1}{4} \\\\ = \frac{1}{10}

8 0
2 years ago
Kevin and randy muise have a jar containing 87 ​coins, all of which are either quarters or nickels. the total value of the coins
boyakko [2]
Let x be the amount of nickels and y be the amount of quarters in a jar. Totally there are 87 coins, then x+y=87.

Since nickel is 5 cent coin and quarter is 25 cent coin, then 5x+25y=1255 ($12.55=1255 cents). Solve this system:x=87-y and 5(87-y)+25y=1255.
435-5y+25y=1255,
25y-5y=1255-435,

20y=820,
y=820÷20,
y=41,
x=87-41=46.
Answer: 46 nickels and 41 quarters.

4 0
2 years ago
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