Described an event that has a probability of 0 and a probability of 100

1 answer:

3 0

The event with the probability of 0 will definitely not happen.

The event with the probability of 100% will definitely happen.

As human beings we need some safe assumptions. Without those we could not get out of bed in the morning.

We assume that the probability of the Moon falling into the Pacific Ocean tomorrow is 0. We assume that the probability of the Sun rising tomorrow is 100%. We have no way to know either one but we assume both.

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A certian natural number is not divisible by 7. Can it be divisible by 14?

Iteru [2.4K]

Yes, because I think two could work

8 0

3 years ago

Which inequality can be used to explain why these three segments cannot be used to construct a triangle?

ryzh [129]

The answer choice which explains that the three segments cannot be used to construct a triangle is; AC + CB < AB.

<h3>Which inequality explains why the three segments cannot be used to construct a triangle?</h3>

Since, It follows from the triangle inequalities theorem that sum of the side lengths of any two sides of a triangle is greater than the length of the third side.

Hence, since the sum of sides AC + CB is less than AB, it follows that the required inequality is; AC + CB < AB.

Read more on triangle inequalities;

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7 0

1 year ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

<u>Zero Product Property</u>: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$