All categories

3 years ago

Find f (3) +g(1) – g(3) =

Mathematics

1 answer:

pantera1 [17]3 years ago

3 0

Sorry I don’t know the answer hope you find one bye halleluja

You might be interested in

This H.W I dont get it plz help.

steposvetlana [31]

I think it’s D I’m not sure

4 0

4 years ago

The kitchen-tile installer has 20 green, 14 beige, and 16 white tiles in a box. What is the probability of picking a beige tile

Stels [109]

Answer: P = 0.28

Step-by-step explanation:

If we pick it at random, the probability of selecting each tile must be equal. this means that the probability of picking a tile of a giving color is equal to the number of the tiles of that color divided the total number of tiles.

We have 14 beige tiles and 20 + 14 + 16 = 50 tiles in total.

Then the probability of piking a beige tile is:

P = 14/50 = 0.28

8 0

3 years ago

9. You have saved S2500 to spend on vacation. You spend the money at an average rate of $125 per day.

BigorU [14]

Answer: F(x)=125x+2500
Hope this helps let me know if you need me to explain

5 0

3 years ago

See image for question.

kotegsom [21]

Answer:

(a)

<u>Sample space:</u>

1*1, 1*2, 1*3, 1*4,
2*1, 2*2, 2*3, 2*4,
3*1, 3*2, 3*3, 3*4,
4*1, 4*2, 4*3, 4*4

Total 16 outcomes

(b)

(i) P(even) = 12/16 = 3/4
(ii) P(3x) = 7/16
(iii) P(perfect square) = 4/16 = 1/4

(a) <u>Sample space:</u>

1T, 1H, 2T, 2H, 3T, 3H, 4T, 4H

Total 8 outcomes

(b)

(i) 1T, 3T

P(T and odd) = 2/8 = 1/4

(ii) 1H, 4H

P(H and square) = 2/8 = 1/4

(iii) 1H, 2H, 3H, 4H

P(H) = 4/8 = 1/2

6 0

3 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.