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3 years ago

I need help with number 3

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

79200 feet in 15 miles and 1 hour there is 60 seconds

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Parking at a large university has become a very big problem. University administrators are interested in determining the average

dimaraw [331]

Answer:

Step-by-step explanation:

8 0

3 years ago

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Calculate how many different sequences can be formed that use the letters of the given word. Leave your answer as a product of t

sdas [7]

Answer: E) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)

Step-by-step explanation:

According to the combinations: Number of ways to choose r things out of n things = C(n,r)

Given word: "georgianna"

It is a sequence of 10 letters with 2 a's , 2 g's , 2 n's , and one of each e, o,r, i.

If we think 10 blank spaces, then in a sequence we need 2 spaces for each of g.

Number of ways = C(10,2)

Similarly,

1 space for 'e' → C(8,1)

1 space for 'o' → C(7,1)

1 space for 'r' → C(6,1)

1 space for 'i' → C(5,1)

1 space for 'a' → C(4,2)

1 space for 'n' → C(2,2)

Required number of different sequences = C(10,2) ×C(8,1)× C(7,1)× C(6,1)×C(5,1)×C(2,2).

Hence, the correct option is E) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)

7 0

3 years ago

Relative density was determined for one sample of second-growth Douglas fir 2 x 3 4s with a low percentage of juvenile wood and

Citrus2011 [14]

Answer:

0.03865

Step-by-step explanation:

Since the true density for the two types of trees are given, we would evaluate the average for both of them

Low percentage of juvenile wood density = 0.523 and 0.0543

Average density = 0.523 + 0.0543/2

Average density = 0.55015

Moderate percentage of juvenile wood density = 0.489 and 0.0450

Average density = 0.489 + 0.0450/2

Average density = 0.5115

Therefore, difference in true average density for the two woods are = 0.55015 - 0.51150

= 0.03865

5 0

3 years ago

Find the volume of the composite figure. Explain your thought, explain how your arrived at your answer. SHOW YOUR WORK FOR FULL

Bond [772]

At the bottom of the composite figure we have half a sphere, of radius 10 in.

The volume of this hemisphere would be half the volume of the full sphere, or:

(1/2)(4/3)π(10 in)^3, or (2/3)π(1000 in^3), or (2000/3)π in^3.

On top is the cone of radius 10 and slant height 15 in. To find the volume of this cone-shaped solid, we'll need the height of the cone. This can be found using the Pyth. Thm. as follows:

15^2 = 10^2 + h^2, where h is the height of the cone.

225 = 100 + h^2, so that h= √125, or 5√5. The height of the cone is 5√5 in.

Then the volume of the cone is V = (1/3)(base)(height)

= (1/3)(π)(100 in^2)(5√5 in)

= 500√5/3(π) in^3

The total volume of the composite solid is then

(2000/3)(π in^3) + ( 500√5/3(π) ) in^3), or

(π/3)(4+√5) in^3. This comes out to 6.53 in^3, to the nearest hundredth.

7 0

3 years ago

Please Help I dont Know what to do with this question!!!!

Anestetic [448]

Answer:

1. Certain

2. Impossible

3. Likely

4. Unlikely

Step-by-step explanation:

I believe this is correct. Hope this helps!

6 0

2 years ago

Read 2 more answers