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stepan [7]
3 years ago
10

What is the equation of the line in slope-intercept form?

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:y = 3x/4 + 4

Step-by-step explanation:

The equation of a line in the slope-intercept form is represented by

y = mx + c

Where c = y-intercept

Slope, m = change in the value of y on the vertical axis / change in the value of x in the horizontal axis.

change in the value if y is y2 -y1 and change in the value of x is x2 - x1

Where

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the graph,

y2 = 4

y1=0

x2=0

x1 = -3

we can determine the slope.

Slope, m = (4-0) / (0- -3)

= 4/3

Let us pick points (0,4) to determine the intercept

Substituting x = 0 , y = 4 and m = 4/3 into the slope-intercept equation,

y = mx + c

4 = 3/4×0 + c

c = 4

Equation of the line in slope-intercept form is

y = 3x/4 + 4

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The compound interest formula can help with that.

  A = P(1 +r/n)^(nt) . . . . value of principal P at rate r for t years, compounded n times per year.

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Momba needs to deposit $6307.95 today to have $8000 in 6 years.

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Prove A-(BnC) = (A-B)U(A-C), explain with an example​
NikAS [45]

Answer:

Prove set equality by showing that for any element x, x \in (A \backslash (B \cap C)) if and only if x \in ((A \backslash B) \cup (A \backslash C)).

Example:

A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace.

B = \lbrace0,\, 1 \rbrace.

C = \lbrace0,\, 2 \rbrace.

\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}.

\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}.

Step-by-step explanation:

Proof for [x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))] for any element x:

Assume that x \in (A \backslash (B \cap C)). Thus, x \in A and x \not \in (B \cap C).

Since x \not \in (B \cap C), either x \not \in B or x \not \in C (or both.)

  • If x \not \in B, then combined with x \in A, x \in (A \backslash B).
  • Similarly, if x \not \in C, then combined with x \in A, x \in (A \backslash C).

Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) as required.

Proof for [x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]:

Assume that x \in ((A \backslash B) \cup (A \backslash C)). Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

  • If x \in (A \backslash B), then x \in A and x \not \in B. Notice that (x \not \in B) \implies (x \not \in (B \cap C)) since the contrapositive of that statement, (x \in (B \cap C)) \implies (x \in B), is true. Therefore, x \not \in (B \cap C) and thus x \in A \backslash (B \cap C).
  • Otherwise, if x \in A \backslash C, then x \in A and x \not \in C. Similarly, x \not \in C \! implies x \not \in (B \cap C). Therefore, x \in A \backslash (B \cap C).

Either way, x \in A \backslash (B \cap C).

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) implies x \in A \backslash (B \cap C), as required.

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Answer:

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c) The exponential distribution is memory-less, so it is independent of past events.

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nata0808 [166]

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7 0
3 years ago
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