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4 years ago

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The curve the order pairs (0,10) and (1,5) and (2,2.5) can be represented by the function f(x)=10(0.5)x What is the multiplicati

ve rate of change if the functions

1 answer:

max2010maxim [7]4 years ago

8 0

Answer:

The multiplicative rate of change is 0.5.

Step-by-step explanation:

The given function is

$f(x)=10(0.5)^x$ .... (1)

It is given that the curve is passing through (0,10), (1,5) and (2,2.5).

The exponential function is defined as

$f(x)=ab^x$ .... (2)

Where, a is initial value and b is growth factor of multiplicative rate of change.

On comparing (1) and (2), we get

$a=10$

$b=0.5$

Therefore the multiplicative rate of change is 0.5.

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-3x+4y=-10 in y=mx+b

belka [17]

Y = 3/4x -5/2

Hope this helps!

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3 years ago

The answer is not A can someone plz help me I’m not understanding it thank you it would be appreciated

jeka94

nswer:

209.8

Step-by-step explanation:

the bottom is = 11 * 5

two of the sides are 11 * 9.3 / 2 because they are triangles, and then times 2 because there are two of them

the other two sides are 10.5 * 5 / 2 , and the * 2 because there are two of them again.

(11 * 5) + (11 * 9.3 * 2 / 2) + (10.5 * 5 * 2 / 2) =209.8

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3 years ago

What is the volume of this cone?

Ksju [112]

Answer:

Its neither because the answer is 167.55

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3 years ago

Solve the following system using the substitution method.

Liono4ka [1.6K]

X = (15 - 8y)/9

-5[(15 - 8y)/9] + 12y = -107

(-75/9) + (40/9) + 12y = -107

y = -8.59

x = [15 - 8(-8.59)]/9

x = 9.3
(x,y) = (9.3, -8.59)

3 0

3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

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

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

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

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

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

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

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

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

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(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

c)

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

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

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

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

d)

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

= 0.4866

6 0

3 years ago