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Helen
2 months ago
13

You can upgrade lighting at your factory to LED bulbs that cost $6.95 each and last an average of 5 years. It costs $3 in labor

to change a bulb. Over a 10-year period, about how much will it cost per year to install LED bulbs in 100 lamps and change the bulbs when they burn out?
Mathematics
2 answers:
babunello [11.8K]2 months ago
7 0

Answer: $200.00

Step-by-step explanation:

$6.95 plus $3.00 equals $9.95

Multiplying $9.95 by 100 bulbs gives $1,990

Finally, dividing $1,990 by $9.95 results in $200.00

To summarize, you start with the original cost of $6.95 along with $3 for labor, totaling $9.95. Then, multiply by 100 bulbs, leading to $1990, and divide by $9.95, yielding $200!

Inessa [12.5K]2 months ago
3 0
Each LED bulb, along with installation labor, is priced at
.. $6.95 +$3 = $9.95

For 100 bulbs over a span of 10 years, that equals (100*10) = 1000 bulb·years. At $9.95 per bulb, 5 bulb·years are obtained, and thus the projected total cost for 1000 bulb·years is
.. (1000 b·y)*($9.95/(5 b·y)) = $1990

In summary, for a decade, the installation and changes of 200 bulbs in 100 lamps amount to $1990. Therefore, the yearly cost is...
.. $1990/(10 yr) = $199/yr
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The probability density function for X is shown here:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

The probability of the temperature increase being under 20°C can be calculated as follows:

P(X

Consequently, the chance that the temperature increase will be below 20°C is 0.667.

(a2)

The probability of the temperature rise being in the range from 20°C to 22°C is computed as follows:

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This leads to the probability of the temperature increase being between 20°C and 22°C being 0.133.

(b)

To find the probability that the increase in temperature could be dangerous, we calculate:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

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E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

The expected value for the temperature increase computes to 17.5°C.

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