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2 years ago

Linear model function

Mathematics

1 answer:

sveticcg [70]2 years ago

5 0

Using equations of linear model function, the number of hours Jeremy wants to skate is calculated as 3.

<h3>How to Write the Equation of a Linear Model Function?</h3>

The equation that can represent a linear model function is, y = mx + b, where m is the unit rate and b is the initial value.

Equation for Rink A:

Unit rate (m) = (35 - 19)/(5 - 1) = 16/4 = 4

Substitute (x, y) = (1, 19) and m = 4 into y = mx + b to find b:

19 = 4(1) + b

19 - 4 = b

b = 15

Substitute m = 4 and b = 15 into y = mx + b:

y = 4x + 15 [equation for Rink A]

Equation for Rink B:

Unit rate (m) = (39 - 15)/(5 - 1) = 24/4 = 6

Substitute (x, y) = (1, 15) and m = 6 into y = mx + b to find b:

15 = 6(1) + b

15 - 6 = b

b = 9

Substitute m = 6 and b = 9 into y = mx + b:

y = 6x + 9 [equation for Rink B]

To find how many hours (x) both would cost the same (y), make both equation equal to each other

4x + 15 = 6x + 9

4x - 6x = -15 + 9

-2x = -6

x = 3

The hours Jeremy wants to skate is 3.

Learn more about linear model function on:

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Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

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$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

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(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

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Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

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