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neonofarm [45]
2 years ago
14

Juan purchased an antique that had a value of \$200$200dollar sign, 200 at the time of purchase. Each year, the value of the ant

ique is estimated to increase 10\%10%10, percent over its value the previous year. The estimated value of the antique, in dollars, 222 years after purchase can be represented by the expression 200a200a200, a, where aaa is a constant. What is the value of aaa ?
Mathematics
1 answer:
garri49 [273]2 years ago
5 0

Answer:

<h2>The value of a is (\frac{11}{10} )^{222}.</h2>

Step-by-step explanation:

At the time of purchase, the value of the antique is $200.

After one year the value will increase 10%.

Hence, after one year, the value of the antique will be \frac{110}{100} \times200 = 220.

Similarly, after two year, the value will be 200\times (\frac{110}{100} )^{2}.

Thus, 222 years after purchase, the value of the antique will be 200\times (\frac{110}{100} )^{222} = 200\times (\frac{11}{10} )^{222}.

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Can someone please help me with my Imagine Math? My teacher assigned are class to do this even though we haven't learned about t
Daniel [21]

Answer:

first graph is not a function, second graph is a function, 3rd is not enough information, 4th is a function, 5th is not a function, 6th is a function

Step-by-step explanation:

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2 years ago
Lakya is using the Distributive Property to divide 128 by 4. Which does not show a
Gre4nikov [31]

Answer:

Step-by-step explanation:

5 0
2 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
2 years ago
In a survey of 2,300 people who owned a certain type of car, 1,610 said they would buy that type of car again. What percent of t
trapecia [35]

Answer:

In a survey of 2,300 people who owned a certain type of car, 1,610 said they would buy that type of car again. What percent of the people surveyed were satisfied with the car?

1610/2300x100=

70%

Step-by-step explanation:

total number of people for the survey= 2300

those that wanted to buy the car= 1610

percentage= 1610/2300x 100=0.7x 100

percentage= 70%

3 0
2 years ago
Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the b
Aloiza [94]

Answer:

a.) 0.1028

b.) 0.6477

c.) 0.0388

d.) 3

e.) 2.55

Step-by-step explanation:

Forming a binomial Probability distribution

n = 20

Probability of success for Weld metal failure = 85%

Probability of success for base metal failure = 15%

We use the probabilit distribution formula of combination to solve the problem.

P(x=r) = nCr * p^r * q^n-r

a.) if exactly 5 are base metal failures, then p = 15 and our solution becomes:

P(x=5) = 20C5 * 0.15^5 * 0.85^15

P(x=5) = 0.1028

b.) probability that fewer than 4 are base metal failure= P(x=0) + P(x=1) + P(x=2) + P(x=3)

P(x=0) = 20C0 * 0.15^0 * 0.85^20 = 0.0388

P(x=1) = 20C1 * 0.15¹ * 0.85^19 = 0.1368

P(x=2) = 20C2 * 0.15² * 0.85^18 = 0.2293

P(x=3) = 20C3 * 0.15³ * 0.85^17 = 0.2428

Probability that fewer than 4 are base metal failures becomes: 0.038 + 0.1368 + 0.2293 + 0.2428 = 0.6477

c.) probability that none of them are results of base metal failure = P(x=0). As earlier calculated,

P(x=0) = 0.0388

d.) mean of base metal failures = np = 20*0.15 = 3

e.) standard deviation of base metal failures = √np(1-p)

=3 * (1 - 0.15) = 3 * 0.85

= 2.55

4 0
2 years ago
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