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2 years ago

WILL GIVE BRAINLESTthis table represents a quadratic function.....

Mathematics

1 answer:

KatRina [158]2 years ago

5 0

Everything on the y-axis side is being added by odd numbers. Ex:(1,3,5,7,9, etc).

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PLZ HELP ASAP The numbers of motor vehicle fatalities in the United States for the years 2010 and 2011 are listed in the table b

Nonamiya [84]

To find which measure of variability is greater and which average number of monthly fatalities is higher, you will need to calculate the mean and the mean absolute deviation for both years

The mean will tell us which is generally higher, and the mean absolute deviation will tell us which has a greater variability.

The correct answer is D.

Please see the attached picture for the work.

5 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

The first and third quartiles

aalyn [17]

1. You'll need to download this data, or copy it down by hand.
2. Rearrange the data from lowest to highest values.
3. You have 24 data points (an even number).
In this case, to find the 1st quadrant, take the left half (that is, the left 12) data points. Since 12 is an even number, you must find the average of the middle two of these 12 data points. Your result is the 1st quadrant.
To find the 3rd quadrant, find the middle two data points of the right-hand 12 data points. Average these two points. The result is the 3rd quadrant.

6 0

2 years ago

-x+2y=11 for 3 points

SOVA2 [1]

2y-x=11
-x=11-2y
x=-11+2y
x=2y-11

6 0

2 years ago

It is 4 kilometers from Ravi's house to the nearest mallbox. How far is it in meters

DIA [1.3K]

Answer:

4,000 meters

Step-by-step explanation:

The conversion from kilometers to meters is 1 kilometer equals 1,000 meters. So times the conversion by 4 to get 4,000 meters.

8 0

2 years ago