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4 years ago

Given the line of the best fit for a set of data points with equation y=-2.5+5x, what is the residual for the point (3,6)

1 answer:

6 0

The residual is the difference between the predicted value and the real value. Using the equation, we find for the value of y using the given value of x.

y = -2.5 + 5x
y = -2.5 + 5(3)
y = 12.5
The difference between this value and the given value of y is 12.5 - 6 = 6.5
Thus, the residual is equal to 6.5.

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I don't know how to do this question.

sergij07 [2.7K]

The perimeter is the measure around the shape, so just add all the sides together-
6+5x+x+2
6x+8
Since this is a right triangle, you multiply the legs (6 and x+2) and divide by 2
6(x+2)
6x+12
/2
3x+6

8 0

3 years ago

84,267 nearest thousands

AleksAgata [21]

So the answer would be 84,000. hope that helped

4 0

3 years ago

Read 2 more answers

A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati

I am Lyosha [343]

Answer:

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

7 0

4 years ago

PLEASE HELP ME!!!!!

suter [353]

The answer is c eight units

6 0

3 years ago

Which of the following can be used to calculate the volume of water inside the fish bowl?

Lunna [17]

The volume of a sphere can be calculated as

$V=\frac{4}{3}\pi r^3$

Where r is the radius of the sphere

We want to calculate half of the volume, then we must divide that volume by 2

$V^{\prime}=\frac{1}{2}\frac{4}{3}\pi r^3$

Now we must find the radius of our sphere, the segment AB is the diameter of the sphere, and the radius is half od the diameter, then

$r=\frac{AB}{2}=\frac{12}{2}=6$

Let's put it into our equation

$\begin{gathered} V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(r)^3 \\ \\ V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(6)^3 \end{gathered}$

The problem says to use

$\pi\approx\frac{22}{7}$

Then

$V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

Final answer:

The formula that can be used to calculate the volume of water inside the fish bowl is

$V=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

8 0

1 year ago