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Genrish500 [490]
3 years ago
13

I need to do all of them.

Mathematics
1 answer:
OLEGan [10]3 years ago
5 0

uh, solve for x? Its simple, but first, since it is a prep for a test, you should study how to solve for x, which is easy. So To solve for x, bring the variable to one side, and bring all the remaining values to the other side by applying arithmetic operations on both sides of the equation. Simplify the values to find the result.

Let’s start with a simple equation as, x + 2 = 7

How do you get x by itself?

Subtract 2 from both sides

⇒ x + 2 - 2 = 7 - 2

⇒ x = 5

Now, check the answer, x = 5 by substituting it back into the equation. We get 5 + 2= 7.

L.H.S = R.H.S

And then for a triangle:

Solve for x" the unknown side or angle in atriangle we can use properties of triangle or thePythagorean theorem.

Let us understand solve for x in a triangle with the help of an example.

△ ABC is right-angled at B with two of its legs measuring 7 units and 24 units. Find the hypotenuse x.

then in △ABC by using the Pythagorean theorem,

we get AC2 = AB2 + BC2

⇒ x2 = 72 + 242

⇒ x2 = 49 + 576

⇒ x2 = 625

⇒ x = √625

⇒ x = 25 units

get it? Good

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If a fair 5-sided die is rolled 5 times, what is the probability that each possible outcome (1, 2, 3, 4, and 5) will occur exact
yarga [219]

Answer:

<em>0.0384</em>

Step-by-step explanation:

Given a fair 5 sided die rolled 5 times.

Numbers on it are 1, 2, 3, 4, 5.

To find:

The probability that each number will occur exactly once.

Solution:

<em>Formula for probability of an event E:</em>

<em></em>P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}<em></em>

At the first roll of die, any number can occur.

So number of possible outcomes = 5

Total number of possible outcomes = 5

P(1^{st}\ roll) = \dfrac{5}{5} = 1

At the second roll of die, any number can occur other than that occurred in first roll.

So number of possible outcomes = 4

Total number of possible outcomes = 5

P(2^{nd}\ roll) = \dfrac{4}{5}

At the third roll of die, any number can occur other than that occurred in first and second roll.

So number of possible outcomes = 3

Total number of possible outcomes = 5

P(3^{rd}\ roll) = \dfrac{3}{5}

At the fourth roll of die, any number can occur other than that occurred in first, second and third roll.

So number of possible outcomes = 2

Total number of possible outcomes = 5

P(4^{th}\ roll) = \dfrac{2}{5}

At the fifth roll of die, any number can occur other than that occurred in first, second, third and fourth roll.

So number of possible outcomes = 1

Total number of possible outcomes = 5

P(5^{th}\ roll) = \dfrac{1}{5}

The required probability will be multiplication of all the five probabilities.

1 \times 0.8 \times 0.6 \times 0.4 \times 0.2 = \bold{0.0384}

5 0
3 years ago
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