Answer:
A) The fraction of sum of money did each child receive is 
B) The sum of money did Jeff have $ 3200
Step-by-step explanation:
Given as :
Let The sum of money did Jeff have = $ x
The fraction of money did Jeff's wife get = 
of $ x
The remaining money Jeff will have = $ x - of $ x
I.e The remaining money Jeff will have = 
= 
A ) The remaining amount of money is divided equally among 4 children
So, The fraction of sum of money did each child receive = 
I.e The fraction of sum of money did each child receive =
B ) If each child will receive $ 600
∴, = $ 600
Or, 3 x = $ 600 × 16
Or, 3 x = $ 9600
∴ x = 
I.e x = $ 3200
So, The sum of money did Jeff have $ 3200
Hence ,
A) The fraction of sum of money did each child receive is
B) The sum of money did Jeff have $ 3200 Answer
Answer:
Color, favorite pet, genre of music
Step-by-step explanation:
Categorical variables are simply statistical variables which are non - numeric, usually employed in characterization and groupings based on a certain number of fixed attributes, categories. In the options atated above, the categorical variables fall under a heading with a limited and fixed attributes such as color, pet and genre of music. However, options such as number of siblings and profit are purely numeric variables which can take up any numeric digits and allow for direct numeric computation. They are called quantitative variables.
Answer:
<em>The correct option is: 82°</em>
Step-by-step explanation:
In the given diagram, two chords 
and 
are intersecting.
According to the <u>Angle of intersecting chord theorem</u>, "<em>If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle</em>."
That means here......
So, the measure of arc 
is 82°
(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
