![3\sqrt[]{2y^3}.7\sqrt[]{18y}](https://tex.z-dn.net/?f=3%5Csqrt%5B%5D%7B2y%5E3%7D.7%5Csqrt%5B%5D%7B18y%7D)
We will start with 3 x 7 = 21
![\sqrt[]{2y^3}\times\sqrt[]{18y}=\sqrt[]{2y^3\times18y}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B2y%5E3%7D%5Ctimes%5Csqrt%5B%5D%7B18y%7D%3D%5Csqrt%5B%5D%7B2y%5E3%5Ctimes18y%7D)
Multiply the values under the root
![\sqrt[]{(2\times18)(y^3\times y)}=\sqrt[]{36y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%282%5Ctimes18%29%28y%5E3%5Ctimes%20y%29%7D%3D%5Csqrt%5B%5D%7B36y%5E4%7D)
Square root 36 = 6 and square root y^4 = y^2
![\sqrt[]{36y^4}=6y^2](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B36y%5E4%7D%3D6y%5E2)
Then multiply 21 by 6y^2
![21\times6y^2=126y^2](https://tex.z-dn.net/?f=21%5Ctimes6y%5E2%3D126y%5E2)
The answer is D
Well, he reads 48 pages in those 12 days, that is 12*4 pages.
He has already read 48 pages and he still has 82, so he has a 130=48+82 page book.
Answer:
y = -(1/2)x + 16
Step-by-step explanation:
See attached image.
The probability that Nins goes 0 days is 0.2, the probability that she goes 1 day is 0.50, and the probability that she goes 2 days is 0.30. You can write these data as a table:
![\begin{array}{cccc} x_i & 0 & 1 & 2 \\ p_i & 0.2 & 0.5 & 0.3 \end{array}](https://tex.z-dn.net/?f=%20%5Cbegin%7Barray%7D%7Bcccc%7D%20%20%20%20x_i%20%26%200%20%26%201%20%26%202%20%5C%5C%20%20%20%20p_i%20%26%200.2%20%26%200.5%20%26%200.3%20%20%20%5Cend%7Barray%7D%20)
where
are numbers of days and
are probabilities.
The expected value of the number of days per week on which Nina goes to the supermarket is
.
Answer: the expected value of the number of days per week on which Nina goes to the supermarket is 1.1.
Answer:
67
Step-by-step explanation: