2 is the answer
That’s the answer.
Answer:
Step-by-step explanation:
Given is a table showing the weights, in hundreds of pounds, for six selected cars. Also shown is the corresponding fuel efficiency, in miles per gallon (mpg), for the car in city driving.
Weight Fuel eff. x^2 xy y^2
X Y
28 20 784 560 400
3 22 9 66 484
35 19 1225 665 361
32 22 1024 704 484
30 23 900 690 529
29 21 841 609 441
Mean 26.16666667 21.16666667 797.1666667 549 449.8333333
Variance 112.4722222 1.805555556
Covariance -553.8611111
r -0.341120235
Correlaton coefficient =cov (xy)/S_x S_y
Covariance (x,y) = E(xy)-E(x)E(y)
The correlation coefficient between the weight of a car and the fuel efficiency is -0.341
Sorry dont know, any thing else you need help with?
Answer:
![(x^\frac{3}{8})^\frac{3}{4} = \sqrt[32]{x^9}](https://tex.z-dn.net/?f=%28x%5E%5Cfrac%7B3%7D%7B8%7D%29%5E%5Cfrac%7B3%7D%7B4%7D%20%20%3D%20%5Csqrt%5B32%5D%7Bx%5E9%7D)
Step-by-step explanation:
Given
Required
Convert to radical form
Evaluate the exponents
Split the exponent
Apply the following law of indices
![(x^a)^\frac{1}{b} = \sqrt[b]{x^a}](https://tex.z-dn.net/?f=%28x%5Ea%29%5E%5Cfrac%7B1%7D%7Bb%7D%20%3D%20%5Csqrt%5Bb%5D%7Bx%5Ea%7D)
So, we have:
D 28/4=7 that is the answer for the question
