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3 years ago

I need help with this question

Mathematics

1 answer:

densk [106]3 years ago

3 0

Answer:

So we have an original figure with a value of 50 and then a scaled version or dilation

Hence we have a scale factor of

$\frac{1}{5}$

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Given x = wavelength and y = frequency are inversely related, solve for the constant for the given type of

Natasha_Volkova [10]

Answer:

given

x = k 1/ y

k = x y

= 300,000 × 40

= 12,000,000

4 0

3 years ago

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Evaluate the expression a/b for a = 80 and b = 8.

tangare [24]

A / B

= 80 / 8

= 10

Therefore the answer is 10

7 0

3 years ago

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NEED HELP ASAP!!

Solnce55 [7]

Answer:

Only Cory is correct

Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

$F =G\times \dfrac{M \cdot m}{r^{2}}$

Where;

G = The universal gravitational constant

M = The mass of one object

m = The mass of the other object

r = The distance between the centers of the two objects

For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km

Therefore, for an astronaut in the international Space Station, r = 6,800 km

The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

$\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2} = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx 1.14$

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity

Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

6 0

3 years ago

Which value of y below is the solution to the system shown below. A) y=6 B) y=-1 C) y=-4 D) y=8

Grace [21]

Answer:

Step-by-step explanation:

given the 2 equations

x + 2y = 27 → (1)

2x + 3y = 46 → (2)

Rearrange (1) expressing x in terms of y by subtracting 2y from both sides

x = 27 - 2y ← substitute into (2)

2(27 - 2y) + 3y = 46

54 - 4y + 3y = 46

54 - y = 46 ( subtract 54 from both sides )

- y = - 8 ( multiply both sides by - 1 )

y = 8 → D

4 0

3 years ago

The height, in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84. (

Schach [20]

Answer:

Step-by-step explanation:

Given that the height in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84.

X is N(64, 2.8)

Or Z = $\frac{x-64}{2.8}$

a) the probability that the height of a randomly chosen woman is between 59.8 and 68.2 inches.

$=P(59.8$

b) $P(X\geq 59)\\= P(X\geq -1.78)\\ \\=0.9625$

c) For 4 women to be height 260 inches is equivalent to

4x will be normal with mean (64*4) and std dev (2.8*4)

4x is N(266, 11.2)

$P(4x>260)= \\P(Z\geq -0.53571)\\=0.7054$

d) Z is N(0,1)

E(Z19) = $P(Z>19)\\= 0.000$

since normal distribution is maximum only between 3 std deviations form the mean on either side.

3 0

3 years ago