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TEA [102]
3 years ago
12

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a standard deviation of 1300 hours

and a mean life span of 15,000 hours. If a monitor is selected at random, find the probability that the life span of the monitor will be more than 17,340 hours.
Mathematics
1 answer:
matrenka [14]3 years ago
5 0

Answer:

0.03593

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 17,340

μ is the population mean = 15,000

σ is the population standard deviation = 1300

z score:

z = 17340 - 15000/1300

z =1.8

Probability value from Z-Table:

P(x<17340) = 0.96407

P(x >17340) = 1 - P(x<17340)

= 1 - 0.96407

= 0.03593

The probability that the life span of the monitor will be more than 17,340 hours is 0.03593

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a cold wave hit chicago when the temperature hit 62 f during the cold wave the temperature dropped 2 degrees every hour how many
Vladimir [108]
We start at 62 Fahrenheit. And every hour we drop two degrees. We want to know how long it took for the temperature to drop to 40 Fahrenheit.

If one hour passed, then the temperature dropped two degrees.
If two hours passed, then the temperature dropped 4 degrees.

See the pattern? We can define this as 2h. Where h represents time in hours.
We subtract 2h from 62. 

We can write this as a function. F(h) = 62 - 2h.
Where F is the temperature in Fahrenheit. And h is the hour(s).

Now that we have the formula, let's plug in the value 40 Fahrenheit to see how long it took for the temperature to drop to 40 degrees.

40 = 62 - 2h
Subtract 62 from each side

-22 = -2h
Divide both sides by 2

h = 11

So, it took 11 hours for the temperature to drop to 40 Fahrenheit.
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3 years ago
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Ms. Ache is paid $1250 per week but is fined $100 each day she is late to work. Ms. Ache wants to make at least $3,000 over the
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5 0
3 years ago
Appreciate answers and any help!
Savatey [412]

To find the expected value of the distribution, we multiply each outcome by it's probability. Doing this, we get that the expected value of defects on a skateboard is of \frac{4}{25}.

Outcomes and probabilities:

0 defects, 9/10 probability

1 defect, 1/20 probability

2 defects, 1/25 probability

3 defects, 1/100 probability.

Expected value:

E(X) = 0\frac{9}{10} + \frac{1}{20} + 2\frac{1}{25} + 3\frac{1}{100} = \frac{1}{20} + \frac{2}{25} + \frac{3}{100} = \frac{5 + 8 + 3}{100} = \frac{16}{100}

Dividing both numerator and denominator by 4:

\frac{4}{25}

Thus, the expected value of defects on a skateboard is of \frac{4}{25}.

A similar problem is given at: brainly.com/question/23156292.

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