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3 years ago

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A principal of 5500 is invested in an account paying the amount of an annual rate of 5%. Find the amount in the account after 4

years if the account is compounded semi-annually, quarterly, and monthly.

1 answer:

kondor19780726 [428]3 years ago

7 0

Answer:

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Which polynomial represents (6 x 2 −x−9)(4x−7)?

nydimaria [60]

Answer:

The answer to your question is 24x³ - 46x²- 28x + 63

Step-by-step explanation:

Data

(6x² - x - 9)(4x - 7)

Process

1.- Multiply 4x by the three terms of the first polynomial

(6x² - x - 9)(4x) = 24x³ - 4x² - 36x

2.- Multiply -7 by the three terms of the first polynomial

(6x² - x - 9)(-7) = -42x² + 7x + 63

3.- Add both result

24x³ - 4x² - 36x -42x² + 7x + 63 = 24x³ - 46x²- 28x + 63

6 0

3 years ago

The Temperature at 5:00 A.M. Is -38 F. The temperature rises 20 by 11:00 A.M. Use addition to find the temperature at 11:00 A.M.

GarryVolchara [31]

-38F + 20F equals -18F

8 0

3 years ago

A bus went halfway from a village to a city at a certain speed. Then it stopped in traffic for an hour. In order to catch up the

k0ka [10]

We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.

Since 200 km is "halfway", the total distance must be 400 km.

time = distance / speed
total time = (time for first half) + (delay) + (time for second half)

400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40

The speed of the bus before the traffic holdup was 40 km/h.

4 0

3 years ago

determinar la longitud de uno de los lados de un hexágono regular en donde l medida de su perímetro es 48x+90

RUDIKE [14]

P = 48x + 90
P = 6L
L = (48x + 90)/6
L = 8x + 15 (comprimento de cada lado)

3 0

3 years ago

An adult can lose or gain two pounds of water in the course of a day. Assume that the changes in water weight are uniformly dist

Levart [38]

Answer:

Standard deviation of the weight over a day is 1.15

Step-by-step explanation:

Since it is given that water weight is uniformly distributed between -2 to 2.

So applying formula for variance of uniformly distribution as follows,

$variance=\dfrac{\left (b-a\right )^{2}}{12}$

From the given data, values of a and b is , $a=-2$ , $b=2$ ..

Substituting the values,

$variance=\dfrac{\left (2-\left (-2 \right )\right )^{2}}{12}$

Now, $-\left ( -2 \right )=2$

$variance=\dfrac{\left (2+2\right )^{2}}{12}$

$variance=\dfrac{\left (4\right )^{2}}{12}$

$variance=\dfrac{16}{12}$

Dividing the fraction by 4,

$variance=\dfrac{4}{3}$

So, the value of variance is $\dfrac{4}{3}$

The formula for standard deviation is given as,

$Standard\:deviation=\sqrt{variance}$

Substituting the value,

$Standard\:deviation=\sqrt{\dfrac{4}{3}}$

$Standard\:deviation=1.15470$

Rounding to 2 decimal places,

$Standard\:deviation=1.15$

So, the value of standard deviation is 1.15.

4 0

3 years ago

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