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love history [14]
3 years ago
10

a machine packs boxes at a constant rate of 2/3 of a box every 1/2 minute. what is the number of boxes per minute that the machi

ne packs
Mathematics
1 answer:
Kruka [31]3 years ago
8 0
I think it's 2/3 x 2 = <em>4/6</em>
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Help with math please ​
MA_775_DIABLO [31]

Answer:

B

Step-by-step explanation:

its kinda obvious

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3 years ago
What's the volume of this composite fuigure.​
Eddi Din [679]
3.14(radius)squared(height)(3)
3.14(9)(8)(3)
3.14(72)(3)
3.14(216)
678.24
4 0
3 years ago
Help please asp what does this equal:<br> 1/5.1/5.1/5
motikmotik

Answer:

0.00768935024

((1/5.1)/5.1)/5 = 0

(0.19607843/5.1)/5 = 0

0.03844675/5 = 0

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2 years ago
Determine the median of the data. 2, 3, 3, 4, 5, 7, 9
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3 0
3 years ago
Read 2 more answers
Depreciation
hammer [34]

Answer:

a) y=-3950\cdot{x}+21500

b) f(x) =21500\cdot{(0.7953)^x}

c) Year 1: Linear $17550, exponential $17099

Year 4: Linear $5700, exponential $8601

d) Exponential model

e) The linear model depreciates the value quicker than exponential model long term around 4 years

Step-by-step explanation:

a) At year 0 the price is 21500 and at year 2 the price is 13600

WE can use points (0,21500) and (2,13600)

We can determine the gradient

m=(13600-21500)/(2-0)=-3950

We can use the point slow formula:

y-y_1=m\cdot{(x-x_1)}

y-21500=-3950\cdot{(x-0)}

y=-3950\cdot{x}+21500

b) We can use the following equation:

f(x) =a\cdot{(1+r)^x}

f(x) is the depreciation value after amount of time t, a is the new value, r is the rate of depreciation and x is the time.

13600=21500\cdot{(1+r)^2}

r=-0.2047

The depreciation rate is 20.47% and is negative because it decreases the new value of the car

c) Year 1:

y=-3950\cdot{1}+21500=17550

f(x) =21500\cdot{(0.7953)^1}=17099

Year 4:

y=-3950\cdot{4}+21500=5700

f(x) =21500\cdot{(0.7953)^4}=8601

d) Year 2

y=-3950\cdot{2}+21500=13600

f(x) =21500\cdot{(0.7953)^2}=13598

3 0
3 years ago
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