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3 years ago

I need help. !!! Please explain this to me

Mathematics

1 answer:

Airida [17]3 years ago

5 0

It seems appropriate to use the linear regression function of your calculator for this. Looking at the data, you know the slope is generally positive, so only the 1st and 3rd choices are potentially viable.

A calculator shows the correlation coefficient (r) rounds to 0.816, so the appropriate selection is ...
A. 0.816

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45 tens is the same as

MakcuM [25]

Tens refers to the unit 10,
This means that 45 10s is equal to:
45*10 = 450

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

A number line is shown. Which expression represents the distance between two points, X and Y, on the number line?

IrinaK [193]

Answer: C. |-4 - 3|

<u>Step-by-step explanation:</u>

x = -4

y = 3

The distance between them is the difference. The mathematical operation for difference is subtraction.

Distance is a unit of measurement. Measurements cannot be negative so the absolute value sign is used to ensure the difference is positive.

4 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: