What is the f(x)=(x-2)(x-4)

What is the value in millions of dollars of a downtown office building that cost 12 million dollars to build 20 years ago and de

Fudgin [204]

Answer:

The value in millions of dollars of a downtown office building that cost 12 million dollars to build 20 years ago and depreciated at 9% per year is $1819738.95

Step-by-step explanation:

Cost of building 20 years ago = 12000000

We are given that the cost depreciated at 9% per year

Formula : $N(t)=N_0(1-r)^t$

Where N(t)= Population after t years

$N_0$ =Initial population=12000000

r= rate of depreciation=0.09

t = 20 years

Substitute the values in the formula:

$N(t)=N_0(1-r)^t\\N(t)=12000000(1-0.09)^{20}$

N(t)=1819738.95

Hence the value in millions of dollars of a downtown office building that cost 12 million dollars to build 20 years ago and depreciated at 9% per year is $1819738.95

4 0

3 years ago

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

Note that: 1 is subtracted because it is a probability without replacement

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

3 years ago

The map shows the route for a race.You are at X, 6000 ft from the first checkpoint C.The second checkpoint D is located at the m

Alex

Answer:

Distance between the two check point (C and D) is 4920 ft.

Step-by-step explanation:

From the question given above, we obtained the following information:

X is 6000 ft from C (1st check point)

D (2nd check point) is the midpoint point between C and Y ( the end of the race)

Total race = 3 miles

1 mile = 5280 ft

Next, we shall express the total race in feet (ft).

This is illustrated below:

1 mile = 5280 ft

Therefore,

3 miles = 3 × 5280 = 15840 ft

Next, we shall determine the distance between the 1st check point (C) and the end of the race (Y)

This can be obtained as follow:

Total distance = 15840 ft

Distance between X and C = 6000 ft

Distance between C and Y =?

Distance between C and Y = (Total distance) – (Distance between X and C)

Distance between C and Y = 15840 – 6000

Distance between C and Y = 9840 ft

Finally, we shall determine the distance between the two check point (C and D) as follow:

Since D is midpoint between C and Y, therefore, we shall we shall divide the distance between C and Y by 2 in order to obtain the distance between the two check point (C and D).

This is illustrated below:

Distance between C and Y = 9840 ft

Distance between the two check point (C and D) = 9840/2 = 4920 ft

Therefore, the distance between the two check point (C and D) is 4920 ft

5 0

3 years ago

Help me pls, do and work it out number 6 plss help

grigory [225]

28

25/715

-50

215

-200

28 15/25 equals to 28 3/5

7 0

3 years ago

A. 2x4 + 21x2 +3x -2 B. 2x4 + 9x2 -2 C. 3x2 + 6x+ 1 D. 2x4 + 9x3 +12x2 + 3x -2

Inga [223]

You need to foil, The answer is D

6 0

3 years ago

What is the f(x)=(x-2)(x-4)​

What is the f(x)=(x-2)(x-4)