Which angles are congruent?

Which of the following equations is non-linear?

lakkis [162]

2x + 6y = 14y - 19x^2 + 12 is a non-linear equation

Step-by-step explanation:

Lets define a linear equation first.

A linear equation is an equation in which there is no variable with exponent greater than 1 or the degree of the equation is 1.

So,

The equation is a linear equation because the degree of the equation is 1.

The equation is a linear equation because the degree of the equation is 1.

The equation involve a term with exponent 2 which makes the degree of the equation 2 making it a quadratic equation

The equation is a linear equation because the degree of the equation is 1.

Hence,

2x + 6y = 14y - 19x^2 + 12 is a non-linear equation

Keywords: Linear, quadratic

Learn more about equations at:

#LearnwithBrainly

5 0

2 years ago

6. What is the sum of the solutions?

vlada-n [284]

x * x = x^2

x * .7 = .7x

-6 * x = -6x

-6 * .7 = -4.2

x^2 + .7x - 6x -4.2

x^2 - 5.3 - 4.2

x = 6

8 0

2 years ago

R is in the middle of pqs btw

Vlad [161]

Answer:

46

Step-by-step explanation:

5x = 3x - 2 + 34

5x = 3x + 32

2x = 32

x = 16

plug back in

3( 16 ) - 2

48 - 2

46

5 0

2 years ago

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,00

timama [110]

Answer: the probability that a randomly selected tire will have a life of exactly 47,500 miles is 0.067

Step-by-step explanation:

Since the life expectancy of a particular brand of tire is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = life expectancy of the brand of tire in miles.

µ = mean

σ = standard deviation

From the information given,

µ = 40000 miles

σ = 5000 miles

The probability that a randomly selected tire will have a life of exactly 47,500 miles

P(x = 47500)

For x = 47500,

z = (40000 - 47500)/5000 = - 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.067

6 0

2 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago