Answer:
The owner should gather more.
Step-by-step explanation:
He should gather more so he can make sure he is correct.
" Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).1. = [3x + 2] + [4 – 5x] = 3x + 2 + 4 – 5x. = 3x – 5x + 2 + 4. = –2x + 6. (f – g)(x) 2. = f (x) – g(x)= [3x + 2] – [4 – 5x] = 3x + 2 – 4 + 5x. = 3x + 5x + 2 – 4. = 8x– 2. 3. ...= (3x + 2)(4 – 5x) = 12x + 8 – 15x2 – 10x. = –15x2 + 2x + 8. " - Google
Answer:
She worked 12 hours for the week
Step-by-step explanation:
Given that :
Hours worked :
Mondays = 40 hours
Tuesdays = 8 hours
If 1/4 of regular hours was worked in a certain week, the number of hours worked that week can be calculated thus ;
1/4 (Monday hours + Tuesday hours)
1/4(40 + 8)
(1/4 * 40) + (1/4 * 8)
10 + 2
= 12 hours
Option B is the correct answer because you start with 15 gallons (y-intercept), and the car consumes or decreases at 0.03 gal/per mile (your slope or m).
Answer: (751.05, 766.95)
Step-by-step explanation:
We know that the confidence interval for population mean is given by :-
,
where 
=population standard deviation.
= sample mean
n= sample size
z* = Two-tailed critical z-value.
Given : 
n= 42
We know that from z-table , the two-tailed critical value for 99% confidence interval : z* =2.576
Now, the 99% confidence interval around the true population mean viscosity :-
![759\pm (2.5760)\dfrac{20}{\sqrt{42}}\\\\=759\pm (2.5760)(3.086067)\\\\=759\pm7.9497=(759-7.9497,\ 759+7.9497)\]\\=(751.0503,\ 766.9497)\approx(751.05,\ 766.95)](https://tex.z-dn.net/?f=759%5Cpm%20%282.5760%29%5Cdfrac%7B20%7D%7B%5Csqrt%7B42%7D%7D%5C%5C%5C%5C%3D759%5Cpm%20%282.5760%29%283.086067%29%5C%5C%5C%5C%3D759%5Cpm7.9497%3D%28759-7.9497%2C%5C%20759%2B7.9497%29%5C%5D%5C%5C%3D%28751.0503%2C%5C%20766.9497%29%5Capprox%28751.05%2C%5C%20766.95%29)
∴ A 99% confidence interval around the true population mean viscosity : (751.05, 766.95)
