R = 12 inches, the radius of the wheel.
The circumference of the wheel is
2πr = 2π(12) = 24π inches.
WHen the wheel makes 15 revolutions, the car will travel 15 times the circumference, which is
15*(24π) = 360π inches = 1131 inches.
Answer: 1131 inches
![\\ \text{Michael distrinutes 2345 packages of screws into 5 equal groups for shipping.}\\ \\ \text{we need to find the number of packages of screws in each group.}\\ \\ \text{so for this we use the unitary method.}\\ \\ \text{The number of packages of screws in 5 groups}=2345\\ \\ \text{so the number of packages of screws in 1 group}=\frac{2345}{5}\\ \\ =469](https://tex.z-dn.net/?f=%5C%5C%0A%5Ctext%7BMichael%20distrinutes%20%202345%20packages%20of%20screws%20into%205%20equal%20groups%20for%20shipping.%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bwe%20need%20to%20find%20the%20number%20of%20packages%20of%20screws%20in%20each%20group.%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bso%20for%20this%20we%20use%20the%20unitary%20method.%7D%5C%5C%0A%5C%5C%0A%5Ctext%7BThe%20number%20of%20packages%20of%20screws%20in%205%20groups%7D%3D2345%5C%5C%0A%5C%5C%0A%5Ctext%7Bso%20the%20number%20of%20packages%20of%20screws%20in%201%20group%7D%3D%5Cfrac%7B2345%7D%7B5%7D%5C%5C%0A%5C%5C%0A%3D469)
Hence the number of packages of screws in each shipping box is: 469
Hence option (B) is correct.
Answer:
![\bar X= \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
The reason is because we assume that each week have the same weight and replacing we got:
![\bar X= \frac{190+163+327+205}{4}= 221.25](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B190%2B163%2B327%2B205%7D%7B4%7D%3D%20221.25)
And the best option would be:
D. 221.25
Step-by-step explanation:
For this case we have the following data given
Week 1 2 3 4
Minutes 190 163 327 205
For this case we can find the mean with the following formula:
![\bar X= \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
The reason is because we assume that each week have the same weight and replacing we got:
![\bar X= \frac{190+163+327+205}{4}= 221.25](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B190%2B163%2B327%2B205%7D%7B4%7D%3D%20221.25)
And the best option would be:
D. 221.25
Answer:
The answer to your question is 24x³ - 46x²- 28x + 63
Step-by-step explanation:
Data
(6x² - x - 9)(4x - 7)
Process
1.- Multiply 4x by the three terms of the first polynomial
(6x² - x - 9)(4x) = 24x³ - 4x² - 36x
2.- Multiply -7 by the three terms of the first polynomial
(6x² - x - 9)(-7) = -42x² + 7x + 63
3.- Add both result
24x³ - 4x² - 36x -42x² + 7x + 63 = 24x³ - 46x²- 28x + 63