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3 years ago

Please help. I will mark BRAINLIEST! Find an equation for the graph sketched below.

Mathematics

1 answer:

djverab [1.8K]3 years ago

7 0

9514 1404 393

Answer:

f(x) = 2(3^x) -5

Step-by-step explanation:

The function is apparently exponential, with a vertical offset. The form of it will be ...

f(x) = a(b^x) +c

For large negative x, the value of f(x) approaches c, which appears to be -5 on the graph. For x=0, the value of f(x) is a+c, so we have ...

a -5 = -3

a = 2 . . . . . . add 5

For x=1, it appears that f(1) = 1. This tells us ...

f(x) = 2(b) -5 = 1

2b = 6 . . . . . . . . . . add 5

b = 3 . . . . . . . . divide by 2

Then the function equation is ...

f(x) = 2(3^x) -5

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1. Requires us to compute

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2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).