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Lady_Fox [76]
4 years ago
11

The scatter plot below can be used to find the approximate rate at which water flows through a garden hose. The line of best fit

for the scatter plot can be described by the equation y = 4/5x. If the rate, in gallons per minute, continues, approximately how many gallons of water will flow from the hose in 45 minutes?
36
39
56
24

Mathematics
1 answer:
creativ13 [48]4 years ago
6 0

Answer:

36\ \text{gallons per unit} will flow from the hose in 45 minutes.

Step-by-step explanation:

The line of best fit for the scatter plot can be described by the equation :

y=\dfrac{4}{5}x

y is amount of water in gallons

x is time

When x = 45 minutes, put x = 45 in above formula as :

y=\dfrac{4}{5}\times 45\\\\y=36\ \text{gallons per unit}

So, 36\ \text{gallons per unit} will flow from the hose in 45 minutes.

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Step-by-step explanation:

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Ron walks 0.5 mile on a track in 10 minutes. Stevie walks on the track in 6 minutes. Find the unit rate for each walker in miles
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7 0
3 years ago
3. Evaluate a + b for a = -46 and b= 34.<br> A:-12<br> B:80<br> C:-80<br> D:12
blsea [12.9K]

Answer:

<h2>A: -12</h2>

Step-by-step explanation:

Put the values of a = -46 and b = 34 to the expression a + b:

-46 + 34        <em>use the commutative property a + b = b + a</em>

= 34 + (-46) = 34 - 46 = -12

4 0
3 years ago
Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

7 0
3 years ago
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