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wariber [46]
3 years ago
15

The expression 1.08 + 1.02b predicts the end- of the year value if a financial portfolio where S is the value of the stocks and

B is the value of bonds in the portfolio at the beginning of the year
What is the predicted end-of value of a portfolio that begins the year with $200 in stocks and $100 in bonds?
Mathematics
1 answer:
iragen [17]3 years ago
4 0

Answer:

We have

Stock = $200

Bonds = $100

Substitute these values into 1.08s+1.02b, where 's' is stock and 'b' is bond

.() + .() =

Predicted end of year value is $318

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Afina-wow [57]

Answer:

The answer is

Step-by-step explanation:

Given: f(x) = 3 + 2x - x²

To compute f(3 + h) substitute x = 3 + h.

Similarly, to compute f(h) simply substitute x = h.

We are asked to compute $ \frac{f(3 + h) - f(h)}{h} $

f(3 + h) = 3 + 2(3 + h) - {(3 + h)²}

           = 3 + 6 + 2h - {9 + 6h + h²}

           = 9 + 2h - 9 - 6h - h²

           = - 4h - h²

Now, f(h) = 3 + 2h - h²

f(3 + h) - f(h) = -4h - h² - 3 - 2h + h²

                    = -6h - 3

$ \therefore \frac{f(3 + h) - f(h)}{h} = \frac{-3(2h + 1)}{h} $ which is the required answer.

6 0
3 years ago
Henrik the meerkat, the friend of Gottfried and Bernhardt, says that if an angle in radians is larger than 2pi, then the sine of
Naddik [55]

When angles are restricted to real numbers, Bernhard is correct. The sine function has a range of -1 to 1 (inclusive).


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When angles are allowed to be complex numbers, the magnitude of the sine of an angle may exceed 1. In this realm, both Henrik and Bernhard are incorrect.

3 0
3 years ago
Read 2 more answers
Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of
lora16 [44]

Answer:

The 90% confidence interval for the average bonus that all employees working for financial companies in New York City received last year is between $43,819 and $50,181

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 62 - 1 = 61

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 61 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.67

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.67\frac{15000}{\sqrt{62}} = 3181

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 47000 - 3181 = $43,819

The upper end of the interval is the sample mean added to M. So it is 47000 + 3181 = $50,181

The 90% confidence interval for the average bonus that all employees working for financial companies in New York City received last year is between $43,819 and $50,181

6 0
3 years ago
Please help me it is a quiz
fredd [130]

Answer:

Step-by-step explanation: I CAN HELP YOU what is the question

4 0
2 years ago
What is the length (magnitude) of the vector (5 -2) <br> a. √29<br> b. 5√5<br> c. 5<br> d.√2
malfutka [58]

Answer:

a. √29

Step-by-step explanation:

The formula for the magnitude of a vector is magnitude = sqrt(x^2 + y^2).

For vector (5, -2):

magnitude = sqrt(5^2 + -2^2)

magnitude = sqrt(25 + 4)

magnitude = sqrt(29)

Therefore, the answer is √29.

Hope this helped :D

8 0
3 years ago
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