How to write an equation in slope intercept form given a point and a slope?

Malcolm has been watching a roulette-style game at a local charity bazaar. The game has only ten numbers on the wheel, and every

Mariana [72]

Answer:

Malcolm is showing evidence of gambler's fallacy.

This is the tendency to think previous results can affect future performance of an event that is fundamentally random.

Step-by-step explanation:

Since each round of the roulette-style game is independent of each other. The probability that 8 will come up at any time remains the same, equal to the probability of each number from 1 to 10 coming up. That it has not come up in the last 15 minutes does not increase or decrease the probability that it would come up afterwards.

5 0

4 years ago

NEED HELP WITH 7 AND 8<br> I WILL GOVE BRAINEST ONLY IF YOU ACTUALLY DO THE PROBLEMS

mihalych1998 [28]

7:
Parallel lines have the same slope
y = -3x + b
Plug in the x and y values from the point
5 = -3(-4) + b
5 = 12 + b
-7 = b
Answer to 7: y = -3x -7

8:
Perpendicular lines have opposite reciprocal slopes
y = -2x + b
Plug in x and y from the point
-6 = -2(7) + b
-6 = -14 + b
8 = b
Answer to 8: y = -2x + 8

3 0

3 years ago

Evaluate the following limit:

Makovka662 [10]

If we evaluate the function at infinity, we can immediately see that:

$\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}$

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

By comparison of infinities:

We first expand the binomial squared, so we get

$\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}$

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

Dividing numerator and denominator by the term of highest degree:

$\large\displaystyle\text{$\begin{gathered}\sf L = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4 }{x^{3}-5 } \end{gathered}$}\\$

$\ \ = \lim_{x \to \infty\frac{\frac{x^{4} }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} } }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}} } }$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} } }{\frac{1}{x}-\frac{5}{x^{4} } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}$

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0

2 years ago

A square matrix A is said to be idempotent iff A2 = A. (i) Show that if A is idempotent, then so is I − A. (ii) Show that if A i

muminat

Answer:

Step-by-step explanation:

Given that A is a square matrix and A is idempotent

$A^2 = A$

Consider I-A

i) $(I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A$

It follows that I-A is also idempotent

ii) Consider the matrix 2A-I

$(2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I$

So it follows that 2A-I matrix is its own inverse.

7 0

4 years ago

"Ten subtracted from the quotient of a number<br> and 5 is 18."

Anika [276]

Answer: n/5 - 10 = 18: this would be the equation. Your answer would be 140.

Step-by-step explanation: Let consider the number as ‘X’

Quotient of a number and 5 can be written as

X divided by 5

Ten subtracted from the quotient of a number and 5 can be written as

(X divided by 5)-10

Ten subtracted from the quotient of a number and 5 is 18 can be written as

(X divided by 5)-10=18

By solving the above equation, find ‘X’

(X divided by 5) = 18 + 10

X/5=28

X = 28 x 5 = 140

4 0

3 years ago