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3 years ago

Given that f(x) = x2 − 3x + 3 and g(x) = the quantity of x minus one, over four , solve for f(g(x)) when x = 5. (1 point)

Mathematics

1 answer:

Anna007 [38]3 years ago

6 0

Rewrite g(x) as x-1

------

and then substitute this result for x in f(x) = x^2 - 3x + 3:

f(g(x)) = (x-1)^2 / 4^2 - 3(x-1)/4 + 3.

At this point we can substitute the value 5 for x:

f(g(5)) = (5-1)^2 / 4^2 - 3(5-1)/4 + 3

= 16/16 - 3(4/4) + 3 = 1 - 3 + 3 = 1

Therefore, f(g(5)) = 1.

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The point of concurrency of the medians of a triangle is called the

tangare [24]

Answer:

centroid of the triangle

Step-by-step explanation:

centroid of the triangle

8 0

3 years ago

Bas_tet [7]

Answer:

a) We reject H₀ We find difference between the mean of the two groups

b) we accept H₀ : There is no statistical difference ( in the scale of pain) when the difference in samples means is 1

Step-by-step explanation:

Group 1: Only with educational material

Sample size n₁ = 42

Sample mean x₁ = 3,2

Sample standard deviation s₁ = 2,3

Group 2: With exercises with the treatment of individuals

Sample size n₂ = 42

Sample mean x₂ = 5,2

Sample standard deviation s₂ = 2,3

Test Hypothesis

Null Hypothesis H₀ x₂ - x₁ = 0 or x₁ = x₂

Alternative Hypothesis Hₐ x₂ - x ₁ > 0 or x₂ > x₁

Significance level α = 0,01

Sample sizes are n₁ = n₂ > 30

We use z-test

for α = 0,01 z(c) ≈ 2,32

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √ (2,3)²/42 + (2,3)²/42

z(s) = ( 5,2 - 3,2 )/ √0,2519

z(s) = 2 / 0,5

z(s) = 4

Comparing z(s) and z(c)

z(s) > z(c)

z(s) is in the rejection region. We reject H₀ the true average pain level for the control condition exceeds that for the treatment condition.

b) Does the true average pain level for the control condition exceed that for the treatment condition by more than 1.

In this case

z( s) = x₂ - x₁ / 0,5

z(s) = 1 /0,5

z(s) = 2

Comparing z(s) and z(c) now

z(s) < z(c) 2 < 2,32

And z(s) is in the acceptance region ( for the same significance level) and we should accept H₀ equivalent to say that the two groups statistical have the same mean

5 0

3 years ago

If ADEF is reflected over the x-axis, what would be the coordinates of point F?

Brut [27]

Answer:

Please check the explanation.

Step-by-step explanation:

Let the coordinates of the point F be (x, y).

When a point F(x, y) is reflected over the x-axis, the x-coordinate of the point F remains the same, and the y-coordinate of the point reverses the sign.

Thus, the rule of reflection over the x-axis:

F(x, y) → F'(x, -y)

Here,

F'(x, -y) would be coordinates of point F after the reflection over the x-axis.

Let say, the point F(1, 2).

The coordinate of the point F after the reflection over the x-axis would be:

F(1, 2) → F'(1, -2)

Thus, F'(1, -2) would be the coordinates of point F after the reflection over the x-axis.

6 0

3 years ago

There are 30 students in Mrs. Taylor's kindergarten class. If there are twice as many students with blond hair as with blue eyes

weqwewe [10]

Answer:

Step-by-step explanation:

Let e represent the number of students with blue eyes. Then the number of students with blond hair is 2e. The total number of students is ...

3 + e + 2e -6 = 30

We subtracted 6 because the expressions e and 2e cause the 6 students with both blond hair and blue eyes to be counted twice.

3e = 33 . . . . add 3 and simplify

e = 11

11 students have blue eyes.

5 0

3 years ago

the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?

miss Akunina [59]

The equation of a circle in standard form is

$(x - h)^2 + (y - k)^2 = r^2$

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

$d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)}$

$d = \sqrt{144 + 64}$

$d = \sqrt{208}$

$r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52}$

$(x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2$

$x^2 + (y - 2)^2 = 52$

7 0

3 years ago