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3 years ago

A data set contains three points and two of the residuals are -12 and 6. What is the third residual?

2 answers:

7 0

There may be a case in which all these data points will be plotted above the x-axis. Therefore, it may be noted that for this item, the maximum number of residual points that can be plotted above the x-axis is also 7.

Veronika [31]3 years ago

3 0

Answer:

the answer is -6

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Answer:

you need at least a score of 60

Step-by-step explanation:

anything 59 and below is 1.95 and every percent higher it adds .5 more

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2 years ago

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Bob’s heart rate is 68 beats per minute. a. If this is his average heart rate, how many times will his heart beat in 15 years? b

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2 years ago

ABC is a tangent to the circle below. O is the centre of the circle. Work out the size of angle 0. Give reasoning to justify you

Delvig [45]

Answer:

Θ = 46°

Step-by-step explanation:

the angle between a tangent and a radius at the point of contact is 90° , so

∠ ABO = 90°

since OB = OD ( radii of circle ) then Δ BOD is isosceles and

∠ OBD = ∠ ODB = 22°

the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ AOB is an exterior angle of the triangle , then

∠ AOB = 22° + 22° = 44°

the sum of the 3 angles in Δ AOB = 180° , then

Θ + 44° + 90° = 180°

Θ + 134° = 180° ( subtract 134° from both sides )

Θ = 46°

6 0

2 years ago

Please help as soon as possible

Leto [7]

Answer:

Step-by-step explanation:

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4 years ago

Read 2 more answers

HELP ME!! im so stumped. will give brainliest!

Olegator [25]

Answer:

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

Step-by-step explanation:

We have been given the parabola with vertex (1, -9) and y intercept at (0, -6).

Now we need to find the x-intercepts of that parabola. So first we begin by finding the equation of parabola using vertex formula:

$y=a\left(x-h\right)^2+k$

Vertex for this formula is given by (h,k)

Compare that with given vertex (1,-9), we get: h=1, k=-9

So plug these into vertex formula:

$y=a\left(x-1\right)^2-9$ ...(i)

Plug given point (0, -6). into (i)

$-6=a\left(0-1\right)^2-9$

$-6=a\left(-1\right)^2-9$

$-6=a\left(1\right)-9$

$-6+9=a$

$3=a$

Plug a=3 into (i)

$y=3\left(x-1\right)^2-9$

Now to find x-intercept, we just plug y=0 and solve for x

$y=3\left(x-1\right)^2-9$

$0=3\left(x-1\right)^2-9$

$9=3\left(x-1\right)^2$

$3=\left(x-1\right)^2$

$\pm \sqrt{3}=x-1$

$1+\pm \sqrt{3}=x$

Hence final answer are

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$

4 0

3 years ago