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LuckyWell [14K]
3 years ago
14

Alan repairs televisions. His revenue is modeled by the function R(h)=10+30h for every h hours he spends repairing televisions.

His overhead cost is modeled by the function C(h)=5h^2−25 dollars.
After how many hours does he break even?

Enter your answer in the box.


Help with steps
Mathematics
1 answer:
RUDIKE [14]3 years ago
3 0

ANSWER

Alan breaks even after 7 hours.

EXPLANATION

Alan's revenue is modeled by the function

R(h)=10+30h

His overhead cost function is modeled by the function

C(h)=5 {h}^{2}  - 25

Alan will break even when his cost is equal to his revenue.

5 {h}^{2}  - 25 = 30h + 10

5 {h}^{2} - 30h - 35 = 0

{h}^{2}  - 6h  -7 = 0

Factorize:

(h - 7)(h + 1) = 0

h =  - 1

or

h = 7

We discard the negative value.

He breaks even after 7 hours.

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Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent fr
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Step-by-step explanation:

Given : The probability of having clear sunny skies in Seattle in July : p= 0.40

The number of days spent in Seattle  in July: n= 18

a) Using, Binomial probability formula : P(x)=^nC_xp^x(1-p)^{n-x}

The probability of having clear sunny skies on at least 13 of those days:-

P(x\geq13)=P(13)+P(14)+P(15)+P(16)+P(17)+P(18)\\\\=^{18}C_{13}(0.4)^{13}(0.6)^5+^{18}C_{14}(0.4)^{14}(0.6)^4+^{18}C_{15}(0.4)^{15}(0.6)^3+^{18}C_{16}(0.4)^{16}(0.6)^2+^{18}C_{17}(0.4)^{17}(0.6)^1+^{18}C_{18}(0.4)^{18}(0.6)^0

=\dfrac{18!}{13!5!}(0.4)^{13}(0.6)^5+\dfrac{18!}{14!4!}(0.4)^{14}(0.6)^4+\dfrac{18!}{15!3!}(0.4)^{15}(0.6)^3+\dfrac{18!}{16!2!}(0.4)^{16}(0.6)^2+(18)(0.4)^{17}(0.6)^1+(1)(0.4)^{18}

=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736

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P-value = P(x\geq13)=P(z\geq2.79)=1-P(z

=1-0.9973645=0.0026355\approx0.0026

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