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3 years ago

Write a real world scenario that matches the following equation. 15+3h=30 Then solve for the equation

1 answer:

7 0

Bowling costs 15 dollars per person and 3 is amount of money per game it cost to play. If you play 5 games then you pay a total of 30 dollars.

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Help quickly plsss Hsjsjdjjd

I am Lyosha [343]

Answer:

answer is 75

Step-by-step explanation:

6÷8x100=75%

from 58 to 66 its have 6 colums..thats why 6/8

5 0

3 years ago

Find the product of largest 4 digit number and smallest 3 digit number.

Afina-wow [57]

Largest 4 digit number =9999
Smallest 3 digit number= 100
So the product of 9999 x 100 = 999,900

3 0

3 years ago

Read 2 more answers

In everyday English, interpret the financial meaning of the

Gemiola [76]

Answer:

The y-intercept refers to the y-coordinate of a point where a curve or a line, intersects the y-axis.

In the given equation, the y-intercept is ( 0, 9.5).

Step-by-step explanation:

We have been given the equation;

y=0.10x+9.50

The y-intercept is simply the y-coordinate of a point where the line intersects the y-axis. At this point, the value of x is usually zero.

y = 0.10 (0) + 9.50

y = 0 + 9.50

y = 9.50

is our y-intercept

8 0

3 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

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