So the waiting time for a bus has density f(t)=λe−λtf(t)=λe−λt, where λλ is the rate. To understand the rate, you know that f(t)dtf(t)dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.

So define X=min(B1,B2)X=min(B1,B2), where B1B1 is exponential with rate 33 and B2B2 has rate 44. It's easy to show the minimum of two independent exponentials is another exponential with rate λ1+λ2λ1+λ2. So you want:

P(X>20 minutes)=P(X>1/3)=1−F(1/3),P(X>20 minutes)=P(X>1/3)=1−F(1/3),

where F(t)=1−e−t(λ1+λ2)F(t)=1−e−t(λ1+λ2).

8 0

2 years ago

Which expressions are equivalent to the expression below?

Dmitriy789 [7]

Given :-

-1/3 - ( -4 + 1/6 )

To Find :-

The expression that is equivalent to one of the choices given .

Solution :-

As we know that ,

→ (-) × (-) = (+)

→ (-) × (+) = (-)

→ (+) × (-) = (-)

→ (+) × (+) = (+)

On using these open the brackets ,

→ -1/3 - 1( -4 + 1/6 )

→ -1/3 - 1(-4) + -1(+1/6)

→ -1/3 (-)(-)(1* 4) (+)(-) (1*1/6)

On using now above stated rules ,

→ -1/3 +4 -1/6

Somewhat rearrange ,

→ 4 -1/3 -1/6

Take (-) as common,

→ 4 - (1/3 +1/6)

Hence Option (d) & (f) are correct .

I hope this helps.

8 0

1 year ago

Quesuon 10 of 10

Bad White [126]

Answer:

(p, q) is (2, 5)

Step-by-step explanation:

Given the expression

x^2 + 3x - 10

Factorize;

x^2 + 5x - 2x - 10

x(x+5) - 2(x+5)

(x-2)(x+5)

Comparing with (x + p)(x + 9).

x+2 = x+ p

x-x+2 = p

2 = p

p = 2

Similarly;

x+5 = x+q

x-x+5 = q

5 = q

q= 5

Hence (p, q) is (2, 5)

7 0

3 years ago