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Strike441 [17]
1 year ago
5

Christmas Tree lights are used not only for Christmas trees, but sometimes to decorate college dorm rooms. Suppose that a string

has 20 light bulbs. These lights are all wired in series, so that if one light bulb fails, the whole string fails. The probability that any one light fails is 0.02. Each of the individual lights are independent of each other. What is the probability that the string will NOT fail?
a. 0.02
b. 0.40
c. 0.668
d. 0.333
Mathematics
1 answer:
puteri [66]1 year ago
6 0

Answer: Hello!

Ok, because the bulbs are wired in series, then if only one fails, all the string fails.

Then we need to see the probability for the 20 bulbs to not fail.

If the probability for each bulb to fail is 0.02, then the complement (or the probability of working fine) is 1 - 0.02 = 0.98

then we have 20 bulbs, and each one has a probability of 0.98 of working alright, then the probability for all them to work alright is the multiplication of this probabilities, this is

0.98^{20} = 0.6676

rounded up in the decimal, we have 0.668

then the correct answer is c.

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Answer:

The goat population reaches 1000 in 12.4 years

Step-by-step explanation:

After <em>t </em>years, the number of goats is given by

N = N_0e^{bt}

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From the question,

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1000 = 2e^{0.5t}

e^{0.5t} = 500

0.5t = \ln 500 = 6.214

t = \dfrac{6.2}{0.5} = 12.4

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Answer:

answer is

-0.245 \pm2.160(0.205)

Step-by-step explanation:

After working this way for 6 months he takes a simple random sample of 15 days. He records how long he walked that day (in hours) as recorded by his fitness watch as well as his billable hours for that day as recorded by a work app on his computer.

Slope is -0.245

Sample size  n = 15

Standard error is 0.205

Confidence level 95

Sognificance level is (100 - 95)% = 0.05

Degree of freedom is n -2 = 15 -2 = 13

Critical Value =2.16 = [using excel = TINV (0.05, 13)]

Marginal Error = Critical Value * standard error

= 2.16 * 0.205

= 0.4428

-0.245 \pm2.160(0.205)

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Two armadillos and three aardvarks sat in a five-seat row at the Animal Auditorium. What is the probability that the two animals
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Answer:

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The number of different ways in which the<em> two armadillos</em> would be at the ends of the row is 2:

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Now calculate the total number of different ways in which the animals can sit. It is P(5,5):

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