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Vladimir [108]
3 years ago
5

Which characteristic best describes the data in the scatterplot?

Mathematics
2 answers:
denis23 [38]3 years ago
7 0

Answer:

A. Negative association

Step-by-step explanation:

The points are going down.

lbvjy [14]3 years ago
4 0
A. Negative Association
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Please Help extra points
Molodets [167]
<h2>A. $123.51</h2><h2 />

1^{st}: $24.48 per year + price * 0.85

2^{nd}: $36.83 per year + price * 0.75

Substitute each value into both equations as price.

A:

129.4635

129.4625

B:

98.158

101.84

C:

145.2055

143.3625

D:

155.703

152.615

A is the lowest value where the second value is lower than the first.

3 0
3 years ago
Some values of the linear function 'f' are shown in the table what is the value of 3?
Monica [59]
6 | 16 is the correct answer I believe
5 0
3 years ago
Solve the system if equations: 3x+y+3z=4 <br> 4x-y+2x=5 <br> X-2y-z=0
Ugo [173]

Answer:

The answer is in step-by-step explanation

Step-by-step explanation:

3 x + y + 3 z = 4

4 x - y + 2 z = 5

x - 2 y - z = 0

4 0
3 years ago
Which comparison symbol makes each inequality statement true? Enter &lt; , &gt; , or = to make each statement true. Enter your a
siniylev [52]

Answer:

−3.632 > −3.792

|−3.632| < |−3.792|

Step-by-step explanation:

Given :

−3.632 −3.792 |−3.632| |−3.792|

First comparison :

−3.632 −3.792

Using the knowledge of number line :

- 3.792 occurs farther to the left of a number line than - 3.632, and numbers which occurs farther to the left are lesser. Therefore,

−3.632 > −3.792

2.)

|−3.632| |−3.792|

The presence of the absolute value symbol annuls the negative sign in the digit given. Hence, we can view both numbers as being positive values as ; 3.632 3.792

Numbers further to the right of a number line are greater and 3.792 occurs farther to the right than 3.632.

Hence,

|−3.632| < |−3.792|

3 0
3 years ago
A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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