Help me plz due right now

5x (x - 4) = 0 Solve it in standard form

murzikaleks [220]

Answer:

x=0 or x=4

Step-by-step explanation:

5x(x−4)=0

4 0

3 years ago

What are the ordered pairs of thesolutions for this system of equations? f(x) = x^2 – 2x + 3; f(x) = -6x

babunello [35]

First you subtract the two equations

x^2-2x+3-6x

You simplify that and get

x^2+4x+3 = 0

Now we solve using the quadratic formula.

We get x = -1 and x = -3.

Now we find the y values by plugging the x values into the equation.

f(x) is the same as y.

y = (-1)^2 - 2(-1) + 3

y = 1+2+3

y = 6

Now for the other x value.

y = (-3)^2 - 2(-3) + 3

y = 9+9

y = 18

So the two ordered pairs are (-1,6) and (-3,18)

8 0

2 years ago

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago

I really need help plzzzzz

Sonbull [250]

ANSWER

D. rotation, translation, reflection

EXPLANATION

Dilation is not a rigid motion.

When the triangle is dilated the image triangle may become bigger ( magnification) or smaller (reduction).

When the scale factor is negative 1, a congruent triangle is obtained.

Hence dilation will not always produce a congruent triangle.

Rotation, reflection and translation are rigid motions and hence will always produce a congruent triangle.

8 0

3 years ago

(-6, 1) and 3, 5) M= rise/run=(y2-y1)/(x2-x1)

Sergeeva-Olga [200]

Answer:

4/9

Step-by-step explanation:

The slope of these points would be: (5-1)/(3+6) = 4/9

7 0

3 years ago