All categories

3 years ago

Its not C. SO WHAT IS IT???

Mathematics

1 answer:

Mice21 [21]3 years ago

8 0

The answer to your question is A

You might be interested in

Man Street Florist charges

Digiron [165]

Answer:

2 dozens, $41

Step-by-step explanation:

Given data

Man Street Florist charges

$18 for a dozen roses and a

$5 delivery fee

let the number of dozens be x

and the total charge for x dozens be y

hence

y= 18x+5--------------1

Blooms and Bouquets

charges $20 for a

dozen roses and a $1

delivery fee.

let the number of dozens be x

and the total charge for x dozens be y

hence

y= 20x+1 -------------2

Equate 1 and 2 we have

18x+5=20x+1

20x-18x= 5-1

2x= 4

x= 4/2

x= 2

Hence the number of dozens is 2

the cost is (put x= 2 in eqn 1 or 2)

y= 20(2)+1

y= 40+1

y= $41

6 0

2 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$