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AURORKA [14]
2 years ago
5

Please help! this ones pretty simple im just not sure how to do it

Mathematics
2 answers:
Masteriza [31]2 years ago
8 0

Step-by-step explanation: do what

Anika [276]2 years ago
4 0
Question pls . No pic here
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The exponential function f(x) has a horizontal asymptote at y=3 what is the end behaviour of f(x)?
tatuchka [14]

Answer:  C) as x → -∞, y → 3

                    as x→ ∞ , y → ∞

<u>Step-by-step explanation:</u>

see graph

Notice that as x approaches negative infinity (goes to the left), the y value approaches the asymptote of y = 3.

And as x approaches positive infinity (goes to the right), the y-value increases without bound so goes to infinity.

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The cost of making a cheese cake is RM45. In
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Miguel is making trail mix. He uses 5 1/3 ounces of nuts and 2 3/4 ounces of raisins. Which
Lady_Fox [76]
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4 0
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
2 years ago
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