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Nookie1986 [14]

3 years ago

5

A trust fund worth $25,000 is invested in two different portfolios, this year, one portfolio is expected to earn 5.25% interest

and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1,150 in one year. How much money should be invested at each rate

2 answers:

Daniel [21]3 years ago

8 0

Answer:

Step-by-step explanation:

Let x be the investment at 5.25% and y at 4%

Then we have

$x+y =25000\\5.25x+4y = 115000$

We have to solve this system to get x and y

Multiply I equation by 4

$100000=4x+4y\\115000=5.25x+4y\\-----------------------------\\15000 = 1.25x\\x = 12000\\y =25000-12000 = 13000$

12000 should be invested in 5.25% and 13000 in 4% interest.

OLEGan [10]3 years ago

6 0

Answer:

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

Let

x ------> the amount of money that should be invested at the rate of 5.25%

25,000-x -----> the amount money that should be invested at the rate of 4%

in this problem we have

$t=1\ year\\ P_1=\$x\\P_2=\$(25,000-x)\\r_1=0.0525\\r_2=0.04\\I=\$1,150$

substitute in the formula above

$I=P_1(r_1t)+P_2(r_2t)$

$1,150=x(0.0525*1)+(25,000-x)(0.04*1)$

Solve for x

$1,150=0.0525x+1,000-0.04x$

$0.0525x-0.04x=1,150-1,000$

$0.0125x=150$

$x=\$12,000$

$|$25,000-x=\$13,000$

therefore

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

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A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)equals72 com

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Answer:

Part (A)

1. Maximum revenue: $450,000

Part (B)

2. Maximum protit: $192,500

3. Production level: 2,300 television sets

4. Price: $185 per television set

Part (C)

5. Number of sets: 2,260 television sets.

6. Maximum profit: $183,800

7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and price-demand equations correctly:</u>

Cost:

$C(x)=72,000+70x$

Price-demand:

$p(x)=300-\dfrac{x}{20}$

Domain:

$0\leq x\leq 6000$

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

$R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x$

Simplify

$R(x)=300x-\dfrac{x^2}{20}$

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

$R'(x)=300-\dfrac{x}{10}$

Solve for R'(x)=0

$300-\dfrac{x}{10}=0$

$3000-x=0\\\\x=3000$

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

<em>2. Part (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. </em>

i) Profit(x) = Revenue(x) - Cost(x)

Profit (x) = R(x) - C(x)

$Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)$

$Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000$

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

Profit' (x) = -x/10 + 230

-x/10 + 230 = 0

-x + 2,300 = 0

x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

p(x) = 300 - x/20

p(2,300) = 300 - 2,300 / 20

p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

<em>3. Part (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?</em>

i) Now you must subtract the $4 tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

Profit(x) = -x² / 20 + 230x - 4x - 72,000

Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

Profit'(x) = -x/10 + 226 = 0

-x/10 + 226 = 0

-x + 2,260 = 0

x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000

Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

p(2,260) = 300 - 2,260/20

p(2,260) = 187.

That is, the company should charge $187 per television set.

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3 years ago

When comparing answers on a group assignment, Jenny and Jeremy found that their answers did not match. Part 1: Who do you believ

vitfil [10]

PART 1:

Jeremy gives the correct answer.

The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....

Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value.

PART 2:

The long division method is shown in the picture below

PART 3:

As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct

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3 years ago

HELP PLEASE 20points!!!

labwork [276]

Answer:

61%

Step-by-step explanation:

We can see that out of all the people that were surveyed, 54% were 10th graders. Since 33% out of all the ones surveyed were 10th graders that chose robotics, the fraction would be 33/54 which is 0.611.

This is 61% approx.

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2 years ago

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LiRa [457]

Answer:

11.51% probability that it will take less than 450 days to train all the contestants.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of n variables, the mean is $\mu*n$ and the standard deviation is $s = \sigma\sqrt{n}$

In this question:

$n = 100, \mu = 100*5 = 500, s = 4\sqrt{100} = 40$

Approximate the probability that it will take less than 450 days to train all the contestants.

This is the pvalue of Z when X = 450.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{450 - 500}{40}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

11.51% probability that it will take less than 450 days to train all the contestants.

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Lelechka [254]

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