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LUCKY_DIMON [66]
3 years ago
6

How do you write a function of t that represents the situation(sales $10,000 increase by 65% each year)?

Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

  s(t) = 10000·1.65^t

Step-by-step explanation:

The general form of an exponential model is ...

  f(t) = (initial value) × (1 +(fractional change))^t

where the fractional change can be positive or negative and t is the number of time units. Each time unit will correspond to the period required for the fractional change.*

__

For your problem, you have (initial value) = 10,000 and (fractional change) = 65% = 0.65 with time units of 1 year. Your function for sales (s) will then be ...

  s(t) = 10000·1.65^t

_____

* The business regarding time units has to do with problems where you get a description like "increased 30% in 4 hours." Then, if t is in hours, the multiplier of the initial value looks like 1.30^(t/4).

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Consider random samples of size 80 drawn from population A with proportion 0.48 and random samples of size 66 drawn from populat
gogolik [260]

Answer:

Standard error = 0.070

Step-by-step explanation:

Formula for the standard error of the distribution of differences in sample proportions is;

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We are given;

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σ_(A - B) = √((0.48(1 - 0.48)/80) + (0.14(1 - 0.13)/66))

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List all elements of each set. (Note that sometimes you will have to put three dots at the end, because a set is infinite.)
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The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
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