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2 years ago

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Mrs. Powell has $800 invested in stocks and $500 invested in a Certificate of Deposit. The CD has an interest rate twice as grea

t as the stocks. If her annual interest on these two investments is $45, at what rate of interest is the $500 invested?

2 answers:

Mashutka [201]2 years ago

7 0

Let's use the variable "r" to represent the investment rate of the stocks:
2r(500) + r(800) = 45
1000r + 800r = 45
1800r = 45
r = 0.025
2r = 0.05
The $500 is invested at 5%

vodomira [7]2 years ago

6 0

Answer:

$500 at 5%

Step-by-step explanation:

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A standard six-sided die is rolled $6$ times. You are told that among the rolls, there was one $1,$ two $2$'s, and three $3$'s.

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Answer:

$Sequence = 120$

Step-by-step explanation:

Given

6 rolls of a die;

Required

Determine the possible sequence of rolls

From the question, we understand that there were three possible outcomes when the die was rolled;

The outcomes are either of the following faces: 1, 2 and 3

Total Number of rolls = 6

Possible number of outcomes = 3

The possible sequence of rolls is then calculated by dividing the factorial of the above parameters as follows;

$Sequence = \frac{6!}{3!}$

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Find solution to the system of linear equations. 5x1 + x2 = 0 , 25x1 + 5x2 = 0

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Answer:

the system of equation has infinite solution

Step-by-step explanation:

$5x_1 + x_2 = 0$ , $25x_1 + 5x_2 = 0$

Solve the first equation for x_2

Subtract 5x1 on both sides

$5x_1 + x_2 = 0$

$x_2 =-5x_1$

Now substitute -5x1 on second equation

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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