Which job here requires the least amount of training and education

Mathematics

2 answers:

gizmo_the_mogwai [7]2 years ago

4 0

Answer:

Cashier

Step-by-step explanation:

An elementary school teacher has to go to college, a college professor has to have a college degree, and although accountants are always required to go to college most places hire accountants that have gone to school.

marin [14]2 years ago

4 0

Answer:

C. Cashier

Step-by-step explanation:

Retail cashiers require no formal training.

To be an elementary teacher, you must have at least a Bachelor's degree.

To be a college professor, you must have at least a Master's degree.

To be an accountant, you must have at least a Bachelor's degree.

You might be interested in

X = |y| a function?

Lana71 [14]

Answer:

yes

Step-by-step explanation:

it is a function. it i think is a linear function

6 0

2 years ago

Mia practiced violin for 4⅛ hours per day for 5 days. How many hours did she practice?

Xelga [282]

Answer:

20 5/8

Step-by-step explanation:

convert mixed number into decimal by changing mixed number into an improper fraction then dividing

multiple by 5

boom you got your answer:p

5 0

2 years ago

Divide line segments

ser-zykov [4K]

tis noteworthy that the segment contains endpoints of A and C and the point B is in between A and C cutting the segment in a 1:2 ratio,

$\bf \textit{internal division of a line segment using ratios} \\\\\\ A(-9,-7)\qquad C(x,y)\qquad \qquad \stackrel{\textit{ratio from A to C}}{1:2} \\\\\\ \cfrac{A\underline{B}}{\underline{B} C} = \cfrac{1}{2}\implies \cfrac{A}{C}=\cfrac{1}{2}\implies 2A=1C\implies 2(-9,-7)=1(x,y)\\\\[-0.35em] ~\dotfill\\\\ B=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill$

$\bf B=\left(\cfrac{(2\cdot -9)+(1\cdot x)}{1+2}\quad ,\quad \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}\right)~~=~~(-4,-6) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -9)+(1\cdot x)}{1+2}=-4\implies \cfrac{-18+x}{3}=-4 \\\\\\ -18+x=-12\implies \boxed{x=6} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}=-6\implies \cfrac{-14+y}{3}=-6 \\\\\\ -14+y=-18\implies \boxed{y=-4}$