Answer:
c
Step-by-step explanation:
1/6. theres only one 2, and theres 6 numbers total.
The answer to this is question is C
Based on the amount that Steve Weatherspoon wants to withdraw every year beginning in June 30, 2024, and the interest rate, the balance on June 30th 2023 should be $45,203.
<h3>What should the balance be in 2023?</h3>
The fact that Steve Weatherspoon wants to be able to withdraw a particular amount every year, this makes this amount an annuity.
The value in 2023 would therefore be the present value of the annuity that will then accrue to the required amounts as the years go by.
The present value of an annuity is:
= Annuity amount per year x Present value interest factor of an annuity, 11%, 3 years between 2024 and 2027
Solving gives:
= 13,126.25 x 3.44371
= $45,203
In conclusion, the balance on the fund in 2023 should be $45,203 in order for Steve Weatherspoon to achieve his objectives.
Find out more on the present value of an annuity at brainly.com/question/25792915
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Based on the way these stations are connected, we can infer that there are <u>56 tubes. </u>
<h3>Number of tubes in Spaceland </h3>
Each pair of stations have 2 tubes. This means that every station is connected to each of the others.
We can therefore infer that each station has 7 tubes to the other stations.
The total number of tubes would therefore be:
= Number of stations x Number of tubes per station
= 8 x 7
= 56 tubes
In conclusion, there are 56 tubes.
Find out more on proportionality problems at brainly.com/question/24822176.
(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
