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andrey2020 [161]
2 years ago
11

Can someone pls help, I know this is kind of a lot but I’ve been stuck on this for a while now.

Mathematics
1 answer:
Greeley [361]2 years ago
3 0

9514 1404 393

Answer:

  (a, b, c) = (-0.425595, 11.7321, 2.16667)

  f(x) = -0.425595x² +11.7321x +2.16667

  f(1) ≈ 13.5

Step-by-step explanation:

A suitable tool makes short work of this. Most spreadsheets and graphing calculators will do quadratic regression. All you have to do is enter the data and make use of the appropriate built-in functions.

Desmos will do least-squares fitting of almost any function you want to use as a model. It tells you ...

  a = -0.425595

  b = 11.7321

  c = 2.16667

so

  f(x) = -0.425595x² +11.7321x +2.16667

and f(1) ≈ 13.5

_____

<em>Additional comment</em>

Note that a quadratic function doesn't model the data very well if you're trying to extrapolate to times outside the original domain.

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Answer:

None of the above.........

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Simplify 6(x + y) + (x - y).<br><br> 6x<br> 7x<br> 7x + 6y<br> 7x + 5y
Law Incorporation [45]
7x+5y is the answer
yepyep
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3 years ago
point I is on line segment HJ. GIVEN IJ = 3× + 3, HI = 3× - 1, and HJ = 3× + 8, determine the numerical length of HJ.
Vedmedyk [2.9K]

Given, IJ = 3x + 3, HI = 3x - 1, and HJ = 3x + 8.

Since I is a point on line segment HJ, we can write

HJ=HI+IJ\begin{gathered} 3x+8=(3x-1)+(3x+3) \\ 3x+8=6x+2 \\ 8-2=6x-3x \\ 6=3x \\ 2=x \end{gathered}

Put x=2 in HJ=3x+8.

\begin{gathered} HJ=3\times2+8 \\ HJ=6+8 \\ =14 \end{gathered}

Therefore, the numerical length of HJ is 14.

4 0
1 year ago
Your dog is 8 years younger than your friend. In 2 years, your friend will be three times as old as your dog. How old is your do
ra1l [238]

Answer:

Dog is 2 years old

Step-by-step explanation:

Let dog's age = d

Let friend's age = f

Your dog is 8 years younger than your friend:

d +8 = f

In 2 years, your friend will be three times as old as your dog:

3(d+2) = (f+2)

3d+6 = f+2

3d = f - 4

(sub in f=d+8 from above)

3d = f - 4

3d = d+8 - 4

2d = 4

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5 0
2 years ago
A soft-drink machine can be regulated so that it discharges an average of μ oz. per cup. If the ounces of fill are Normally dist
sattari [20]

Answer:

μ = 5.068 oz

Step-by-step explanation:

Normal distribution formula to use the table attached

Z = (x - μ)/σ

where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.

98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].

From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives

Z = (x - μ)/σ

2.33 = (6 - μ)/0.4

μ = 6 - 2.33*0.4

μ = 5.068

This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.

4 0
3 years ago
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