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2 years ago

Questions

Mathematics

1 answer:

victus00 [196]2 years ago

3 0

Answer:

4-2 to the 3 of 4 is 5

Step-by-step explanation:i took the test

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24 over 32 as an equivelent fraction

andrew-mc [135]

12/16,
6/8,
3/4, and
45/60

are all equivalent to 24/32.

Now that I've listed those four, there are
only an infinite number of others remaining.

4 0

3 years ago

On a test that has a normal distribution, a score of 32 falls three standard

lions [1.4K]

Answer:

Mean = 62

Step-by-step explanation:

X = mean ± sd

82 = mean + 2(s.d) - - (1)

32 = mean - 3sd ___(2)

82 = mean + 2sd

82 - 32= 2sd + 3sd

50 = 5sd

sd = 50/5

sd = 10

From (2):

32 = mean - 3sd

32 = mean - 3(10)

32 = mean - 30

32 + 30 = mean

62 = mean

3 0

3 years ago

What is the mode and median of 1,2,1,3,3,5,3,4,5,4,3,2,5,3,1.

tankabanditka [31]

hope it helps uh ....!!!!

8 0

3 years ago

Eliminate the parameter and obtain the standard form of the rectangular equation. line through (x1, y1) and (x2, y2): x = x1 + t

Helen [10]

The parametric equations for the line passes through the points (X₁ , Y₁) and (X₂ , Y₂) are :

X = X₁ + t (X₂ - X₁) ......................... (1)

Y = Y₁ + t (Y₂ - Y₁) .......................... (2)

For eliminating the parameter (t) , first we will solve any one equation for 't' and then substitute that into another equation.

From the equation (1),

⇒ t = $\frac{X- X1}{(X2 - X1)}$

Substituting this t = $\frac{X- X1}{(X2 - X1)}$ into the equation (2), we will get :

Y = Y₁ + $\frac{(X- X1)}{(X2 - X1)}$ (Y₂ - Y₁)

Y = Y₁ + $\frac{(X- X1)(Y2 - Y1)}{(X2 - X1)}$

So, this is the standard form of rectangular equation.

For two given points (0, 0) and (4, -4)

X₁ = 0 , Y₁ = 0, X₂ = 4 and Y₂ = -4

For finding the parametric equations, we will plug these values into the given parametric equations.

X = X₁ + t (X₂ - X₁)

⇒ X = 0+ t (4 - 0)

⇒ X = 4t

and Y = Y₁ + t (Y₂ - Y₁)

⇒ Y = 0+ t (-4 - 0)

⇒ Y = - 4t

So, the parametric equations for the line passing through (0,0) and (4, -4) are: X = 4t and Y = -4t

6 0

3 years ago

A 5 in by 7in photograph is surrounded by a frame of uniform width the area of the frame equals the area of the photograph find

alexira [117]

Let x be the width of the frame.
Area of the picture is 5*7 = 35 sq in

Area of the frame given as the same, frame = 35 sq in also

Therefore the total area (picture & frame) will be 70 sq in

A simple equation would be made and it will look like this: (7+2x) * (5+2x) = 70

Using the FOIL method
35 + 14x + 10x + 4x^2 = 70
4x^2 + 24x + 35 - 70 = 0; subtract 70 from both sides:
4x^2 + 24x - 35 = 0

unluckily this will not factor easily, so we should use the quadratic equation:
where: a = 4; b = 24; c = -35
 x = -b sqrt b^2 – 4*a*c divided by 2 * a

x = -24 ± sqrt 24^2 -4*4*(-35) divided by 2*4

x = -24 ±sqrt 576 - -(560) divided by 8

x = -24 ± sqrt 576 + 560 divided by 8

x = -24 ± sqrt1136 divided by 8

x = -24 + 33.7046 divided by 8
x = 9.7046 divided by 8
x =1.213 inches is the answer

6 0

3 years ago