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Rashid [163]
3 years ago
11

Differentiate y=x/5-5/x

Mathematics
1 answer:
igomit [66]3 years ago
7 0
I’m trying to find this too
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2 years ago
A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch
ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

P = a person passes the security system

H = a person is a security hazard

Given:

P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01

Then,

P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The value of P (P) is:

P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

7 0
3 years ago
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