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Burka [1]
3 years ago
11

The value of a collector’s item is expected to increase exponentially each year. The item is purchased for $500 and its value in

creases at a rate of 5% per year. Find the value of the item after 4 years
Mathematics
2 answers:
Reika [66]3 years ago
9 0

Answer:

The value of the item after 4 years is $607.75

Step-by-step explanation:

* Lets revise the exponential function

- The original exponential formula was y = ab^x, where a is the initial

 amount and b is the growth factor

- The new growth and decay functions is y = a(1 ± r)^x. , the b value

 (growth factor) has been replaced either by (1 + r) or by (1 - r).

- The growth rate r is determined as b = 1 + r

* Lets solve the problem

- The value of a collector’s item is expected to increase exponentially

  each year, so we will yes the exponential equation y = a(1 + r)^x ,

  where y is the value of the item after x years

- The item is purchased for $500

∵ The initial amount is 500

∴ a = 500

- Its value increases at a rate of 5% per year

∵ The rate of increasing is 5% per year

∴ r = 5/100 = 0.05

- To find the value of the item after 4 years replace x by 4

∵ x = 4

∴ y = 500(1 + 0.05)^4

∴ y = 500(1.05)^4 = 607.75

∵ y is the value of the item after 4 years

∴ The value of the item after 4 years is $607.75

Burka [1]3 years ago
3 0

Answer:607.81

Step-by-step explanation:that’s what I got believe me on this one guys

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5 0
2 years ago
2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation
Vesna [10]

Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

z = \dfrac{\bar x - \mu}{\sigma}

if x > 10.2

z = \dfrac{10.2- 10}{0.1}

z = \dfrac{0.2}{0.1}

z = 2

If x < 9.8

z = \dfrac{9.8- 10}{0.1}

z = \dfrac{-0.2}{0.1}

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

6 0
3 years ago
What are 2 methods for finding the sale price of an item that is discounted 30%?
anyanavicka [17]
First you have to know the price of item.

EX: Say a candy car costs $1.50. 
The candy bar was marked down 30%.
$1.50 x 30/100 = $.45
The item will decrease $.45.
$1.50 - $.45 = $1.05 





6 0
3 years ago
K divided by 11 is 7
Pani-rosa [81]
K/11 = 7
multiply 11 to both sides
K/11 (11) = 7(11)
multiply 7 and eleven together
K = 7(11)
Answer
K = 77


77 is your answer

hope this helps
5 0
3 years ago
Read 2 more answers
Find the t-value such that the area in the right tail is 0.005 with 28 degrees of freedom.
klasskru [66]

Answer:

t = 3.673900

Step-by-step explanation:

Given

df = 28 -- degree of freedom

Area = 0.005

Required

Determine the t-value

The given parameters can be illustrated as follows:

P(T > t) = \alpha

Where

\alpha = 0.005

So, we have:

P(T>t) = 0.005

To solve further, we make use of the attached the student's t distribution table.

From the attached table,

The t-value is given at the row with df = 28 and \alpha = 0.005 is 3.673900

Hence, t = 3.673900

4 0
2 years ago
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