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klemol [59]
3 years ago
7

What is the slope of the linear equation 4x - 5y = -30

Mathematics
1 answer:
MariettaO [177]3 years ago
6 0

Answer:

The slope of the line 4x – 5y = 12 is m = 4/5.

Step-by-step explanation:

To find the slope of this line we need to get the line into slope-intercept form (y = mx + b), which means we need to solve for y: The slope of the line 4x – 5y = 12 is m = 4/5.

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The manager of the motor pool wants to know if it costs more to maintain cars that are driven more often. Data are gathered on e
Liula [17]

Answer:

b. There's no statistically significant linear relationship between the number of miles driven and the maintenance cost

Step-by-step explanation:

The p-value for the slope estimate show us how strong is the certainty that there are a linear relationship between both variables. In this case, the p-value for the slopes shows if there is a significant relationship between the number of miles driven and the maintenance cost.

If we have a high p-value like 0.7 we can said that there is no certainty in the linear relationship. it means that there's no statistically significant linear relationship between the number of miles driven and the maintenance cost.

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2 years ago
Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
Bad White [126]

Answer:

The rate at which water is being pumped into the tank is 13,913.27 cubic centimetre per minute

Step-by-step explanation:

Given that, the height of the conical tank is 13 meters and diameter is 3 meters.

Radius = \frac32 meters

\therefore \frac{radius }{height}=\frac {\frac{3}{2}}{13}

\Rightarrow \frac{radius }{height}=\frac {3}{26}

\Rightarrow radius=\frac {3}{26}\times height   ......(1)

The volume of the cone is   V=\frac13 \pi r^2 h

Now putting r=\frac{3}{26} h

V=\frac13 \pi (\frac3{26}h)^2 h

\Rightarrow V=\frac{3}{676} \pi h^3

Differentiating with respect to t

\frac{dV}{dt}=\frac{3}{676}\pi .3h^2 \frac{dh}{dt}

\Rightarrow \frac{dV}{dt}=\frac{9}{676}\pi h^2 \frac{dh}{dt}

Given that, the water level is rising at a rate of 17 cm per minute when the height 1 meter= 100 cm .i.e \frac{dh}{dt}=17 cm/min

Putting \frac{dh}{dt}=17  

\frac{dV}{dt}=\frac{9}{676}\pi h^2 \times 17

\frac{dV}{dt}|_{h=100}=\frac{9}{676}\pi (100)^2 \times 17

             ≈7,113 cubic centimetre per minute

The volume of water increases 7,113 cubic centimetre per minute while 6,800 cubic centimetre per minute is leaking out.

It means the required rate at which water is being pumped into the tank is

=(7,113+6,800)cubic centimetre per minute

=13,913 cubic centimetre per minute

           

   

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