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alexdok [17]
3 years ago
13

What is the definition of mean???

Mathematics
1 answer:
Scilla [17]3 years ago
3 0
Mean, in terms of math, is the total added values of all the data in a set divided by the number of data <em>in</em> the set. Make sense? If not, here' an example...

Let's say this is my data set:
1, 2, 5, 4, 3, 8, 7, 4, 6,10

To find the mean...
Step 1: Add all of them together.
1+2+5+4+3+8+7+4+6+10 is what? 50. Now that you have this number...
Step 2: Divide by the amount there are. Basically, count up all of the numbers. How many are there? There are 10. Finally...
Step 3: Divide. 50/10 is 5, so the mean of this data set would be 5. Get it? I sure hoped this helped :)
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An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
2 years ago
Sound intensity of certain Vacuum Cleaner is a random variable having a normal distribution with a mean of 75 dB and a standard
fomenos

Answer:

10 thousand times loud pls mark me as the brainliset hope it helps you

8 0
3 years ago
What is the value of 4g + 5h when g=1 and h=4
erma4kov [3.2K]

Answer:

24

Step-by-step explanation:

4(1)+5(4)

4+20

24

8 0
3 years ago
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NEED HELP ASAP 100PT
Mnenie [13.5K]

Answer:

It is 3 square root of 7

Step-by-step explanation:

:)

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2 years ago
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I will mark you brainliest if you answer this
zhenek [66]

\huge\color{purple}\boxed{\colorbox{black}{♡Solution}}

8.

7 {x}^{3}  + 3 {x}^{2}  + 4x + 10 - 10 - 8x - 3 {x}^{3}  \\  = 7 {x}^{3}  - 3 {x}^{3}  + 3 {x}^{2}  + 4x - 8x + 10 - 10 \\  = 4 {x}^{3}  + 3 {x}^{2}- 4x

9.

5 {x}^{4}  - 4 {x}^{3}  - 3x - 4 - 2x + 6 {x}^{3}  + 2 {x}^{4}  \\  = 5 {x}^{4}  + 2 {x}^{4}  - 4 {x}^{3}  + 6 {x}^{3}  - 3x - 2x - 4 \\  = 7 {x}^{4}  + 2 {x}^{3}  - 5x - 4

10.

7 {x}^{3}  - 9 {x}^{2}  - 7x - 8 - 8 + 4 {x}^{2}  + 6 {x}^{3}  \\  = 7 {x}^{3} + 6 {x}^{3}   - 9 {x}^{2}  + 4 {x}^{2}  - 7x - 8 - 8 \\  = 13 {x}^{3}  - 5 {x}^{2}  - 7x - 16

11.

4a + 20 - 5 {a}^{2}  + 20a - 35 \\  =  - 5 {a}^{2} +  24a - 15

12.

8y + 48 - 6 {y}^{2}  + 36y - 24 \\  =  - 6 {y}^{2}  + 44y + 24 \\  = 2( - 3 {y}^{2}  + 22y + 12)

13.

3c - 12 - 5 {c}^{2}  - 20c + 40 \\  =  - 5 {c}^{2}  - 17c + 28

\large\mathfrak{{\pmb{\underline{\orange{Mystique }}{\orange{♡}}}}}

7 0
3 years ago
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