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2 years ago

What is 3 root of 27

1 answer:

4 0

It is 3.

You can check this by seeing 3^3 ie. 3x3x3, which comes out as 27.
You can check the cube root on a calculator.

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PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! Graph g(x) = 3x^2 - 12x - 3

Paul [167]

Answer: Vertex = (2, -15) 2nd point = (0, -3)

Step-by-step explanation:

g(x) = 3x² - 12x - 3

= 3(x² - 4x - 1)

a=1 b=-4 c=-1

Find the x-value of the vertex by using the formula for the axis of symmetry: $x = \dfrac{-b}{2a}$

$x = \dfrac{-(-4)}{2(1)}$

$= \dfrac{4}{2}$

= 2

Find the y-value of the vertex by plugging the x-value (above) into the given equation: g(x) = 3x² - 12x - 3

g(2) = 3(2)² - 12(2) - 3

= 12 - 24 - 3

= -15

So, the vertex is (2, -15) ← PLOT THIS COORDINATE

Now, choose a different x-value. Plug it into the equation and solve for y. I chose x = 0

g(0) = 3(0)² - 12(0) - 3

= 0 - 0 - 3

= -3

So, an additional point is (0, -3) ← PLOT THIS COORDINATE

5 0

2 years ago

X2-5x+10=0 is it quadratic equation? how can i identify the value of a,b,c? help

natali 33 [55]

A= 1
B= -5
C= 10

In order to figure it out, you can think about it like this..

4 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$