How many cm are in 6.9km 6.9cm 690,000cm 6,900cm

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. x2 – 6 = x

mezya [45]

Answer:

3, -2

Step-by-step explanation:

8 0

3 years ago

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The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

antoniya [11.8K]

Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

8 0

3 years ago

Marta is making two different charms for necklace one charm has a 1 cm diameter and a 3.14 centimeter circumference A similar ch

dlinn [17]

Both charms are "similar".

And notice one diameter is 1, the other is 4, the diameters are on a ratio of 1:4 then, and so both figures are on a ratio of 1:4.

so, if another unit, like the circumference is at 3.14 on the small one, then the large one will have a circumference 4 times that much, because the circumferences are also on a 1:4 ratio.

4 0

3 years ago

Write the equation of a line in slope intercept form that is perpendicular to the line y = –4x and passes through the point (2,

Gnesinka [82]

1. If the line that we are searching for is perpendicular to the line y = -4x, this means that the gradient of our line and the gradient of the perpendicular line will multiply to give -1. Thus if we call the gradient of our line m, then:

m*(-4) = -1

-4m = -1

m = 1/4

2. Since we know that m = 1/4 and we have a point (2,6) on our line, we can use the formula y - y1 = m(x - x1) to find the equation of our line, where (x1, y1) is the coordinates of a point on the line. Thus:

y - y1 = m(x - x1)

y - 6 = (1/4)(x - 2)

y - 6 = (1/4)x - 2/4 (Expand (1/4)(x - 2))

y = (1/4)x - 1/2 + 6 (Simplify 2/4 and add 6 to each side)

y = (1/4)x + 11/2 (Evaluate -1/2 + 6)

Slope-intercept form is given by y = mx + c. As our equation is already in this form, there is nothing more to do. Thus, the answer is y = (1/4)x + 11/2

6 0

3 years ago

On a time quizzed please help

zhuklara [117]

Answer:

Element Valence electrons

Nitrogen 5

Silicon 4

Oxygen 6

Magnesium 2

Explanation :

Valence electrons : Valence electrons are the electrons which is present on the outermost shell of the electron.

The given element is nitrogen that belongs to group 15 and the atomic number 7.

The number of electrons present on nitrogen element are 7.

The electronic configuration of nitrogen element is,

The number of valence electrons present on nitrogen element are, 5.

The given element is silicon that belongs to group 14 and the atomic number 14.

The number of electrons present on silicon element are 14.

The electronic configuration of silicon element is,

The number of valence electrons present on silicon element are, 4.

The given element is oxygen that belongs to group 16 and the atomic number 8.

The number of electrons present on oxygen element are 8.

The electronic configuration of oxygen element is,

The number of valence electrons present on oxygen element are, 6.

The given element is magnesium that belongs to group 2 and the atomic number 12.

The number of electrons present on magnesium element are 12.

The electronic configuration of magnesium element is,

The number of valence electrons present on magnesium element are, 2.

Step-by-step explanation:

4 0

3 years ago

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How many cm are in 6.9km6.9cm690,000cm6,900cm​

How many cm are in 6.9km6.9cm690,000cm6,900cm