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tensa zangetsu [6.8K]
2 years ago
7

This scatter diagram shows the marks scored in a class test in English and Science

Mathematics
1 answer:
kiruha [24]2 years ago
6 0
U didn’t show the scatter diagram lol
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The probability that Hank Aaron hits a home run on any given at-bat is 0.14, and
Liula [17]

Answer:

Step-by-step explanation:

3 0
3 years ago
Taiga wants to make a circular loop that she can twirl around her body for exercise. He will use a tube that is 2.5 meters long.
Travka [436]

Answer: 0.796 meters

Step-by-step explanation:

Circumference of circle = 2\pi r , where r=radius

Here, Circumference of loop = 2.5 meters

i.e. 2\pi r=2.5

\Rightarrow\ r=\dfrac{2.5}{2\pi}

\Rightarrow\ r=\dfrac{2.5}{2\times\dfrac{22}{7}}\\\\\Rightarrow\ r=\dfrac{2.5\times7}{2\times22}\\\\\Righttarrow\ r=0.398\ m

Diameter = 2r = 2(0.398) = 0.796 meters

Hence, The diameter of Taiga's exercise hoop = 0.796 meters

7 0
3 years ago
Write an algebraic expression to represent each scenario.<br><br> Part A: “the sum of 6 and y”
Nostrana [21]

Answer:

6 + y

Step-by-step explanation:

Here, we want to write an algebraic expression to represent the situation

Mathematically, we want to add 6 and y

The algebraic expression will be;

6 + y

3 0
3 years ago
A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
Allushta [10]

Answer:

  • 4968.6 m from where it was fired
  • 221.33 m/s

Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

  (x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

To find the range, we can solve the equation y=0 for t, and use this value of t to find x.

Using the quadratic formula, we find t at the time of landing to be ...

  t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

  t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

  t ≈ 31.939 seconds

Then the horizontal travel is

  x = 220·cos(45°)·31.939 ≈ 4968.6 . . . . meters

__

As it happens, the value under the radical in the above expression for time, when multiplied by s, is the vertical speed at landing. The horizontal speed remains s·cos(θ), so the resultant speed is the Pythagorean sum of these:

  landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

  ≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

6 0
2 years ago
HELP MEEE!!!!!!!!! PLEASE!
Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

b

8 0
2 years ago
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