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sergij07 [2.7K]
3 years ago
10

MODELING EXPONENTIAL GROWTH

Mathematics
1 answer:
romanna [79]3 years ago
8 0

Answer:

1. M = C*1.018^t

After 15 years: M = 4965.93

2. After 3 years: M = 2249.728

After 18 years: M = 4051.633

Time to achieve 3000: t = 10.338 years

Step-by-step explanation:

1. Since the money gets increased every year at a rate of 1.8% after one year it'll be the initial amount multiplied by 1.018, so:

After one year:

M = C*1.018

After two years:

M = C*1.018*1.018 = C*(1.018)²

After three years:

M = C*(1.018)²*1.018 = C*(1.018)³

And so on, therefore:

M = C*(1.018)^t

Where M is the final amount, C is the initial amount and t is the time elapsed in years. For this case we have:

M = 3800(1.018)^15 = 4965.92626

2. Applying the same line of thought as above, we have:

M = C*(1.04)^t

After 3 years:

M = 2000*(1.04)^3 = 2249.728

After 18 years:

M = 2000*(1.04)^18 = 4051.633

To obtain 3000:

3000 = 2000*(1.04)^t

2000*(1.04)^t = 3000

1.04^t = 3000/2000

1.04^t =1.5

log(1.04^t) = log(1.5)

t*log(1.04) = log(1.5)

t = log(1.5)/log(1.04) = 10.338 years

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mike and kim invest $14,000 in equipment to print yearbooks for schools. each yearbook costs $7 to print and sells for $35. how
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Hope this helps.
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4 years ago
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
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