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Basile [38]
3 years ago
6

If f(x) = x2, which of the following describes the graph of f(x - 1)? A. The graph of f(x - 1) is a horizontal shift of f(x) = x

2 one unit to the right. B. The graph of f(x - 1) is a vertical shift of f(x) = x2 one unit down. C. The graph of f(x - 1) is a vertical shift of f(x) = x2 one unit up. D. The graph of f(x - 1) is a horizontal shift of f(x) = x2 one unit to the left.
Mathematics
2 answers:
MrRissso [65]3 years ago
6 0

Answer:

Choice A is correct.

Step-by-step explanation:

We have given that

f(x) = x²

We have to find the description of the graph f(x-1).

f(x-1) = (x-1)²

When we replace x with x-a , the resulting graph is a horizontal shift of f(x) by a units to the right.

The graph of f(x-1) is horizontal shift of f(x) by 1 units to the right.

Hence, Choice A is correct.

pantera1 [17]3 years ago
4 0

Answer:

A

Step-by-step explanation:

To understand this, we can look at the vertical & horizontal translations of a parabola of the form  f(x)=x^2

  • A vertically translated parabola has the form  f(x)=x^2+k, where k is the vertical shift upward when k is positive and vertical shift downward when k is negative.
  • A horizontally translated parabola has the form  f(x)=(x-a)^2, where a is the horizontal shift rightward when a is positive and horizontal shift leftward when a is negative.

When we replace x of the original function with (x-1), we have  f(x)=(x-1)^2. According to the rules, this means that the original function is shifted 1 unit right (horizontal shift).

Correct answer is A.

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Find the probability of winning a lottery by selecting the correct six integers, where the order in which these inte
stich3 [128]

The probability of winning a lottery by selecting the correct six integers, are

\begin{aligned}&(a) 1.68 \times 10^{-6} \\&(b) 5.13 \times 10^{-7} \\& (c) 1.91 \times 10^{-7} \\&(d)8.15 \times 10^{-8}\end{aligned}

<h3>What is binomial distribution?</h3>

The binomial distribution is a type of probability distribution that expresses the probability that, given a certain set of characteristics or assumptions, a value would take one of two distinct values.

Part (a); positive integers not exceeding 30.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other twenty-four.

\frac{\left(\begin{array}{c}6 \\6\end{array}\right)\left(\begin{array}{c}24 \\0\end{array}\right)}{\left(\begin{array}{c}30 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}30 \\6\end{array}\right)}=1.68 \times 10^{-6}

Part (b); positive integers not exceeding 36.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other thirty.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}30 \\0\end{array}\right)}{\left(\begin{array}{c}36 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}36 \\6\end{array}\right)}=5.13 \times 10^{-7}

Part (c); positive integers not exceeding 42.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the 36 other integers.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}36 \\0\end{array}\right)}{\left(\begin{array}{c}42 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}42 \\6\end{array}\right)}=1.91 \times 10^{-7}

Part (d); positive integers not exceeding 48.

To calculate the probability, use binomial coefficients. Choose six of the six accurate integers and none of the other 42.

\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}42 \\0\end{array}\right)}{\left(\begin{array}{c}48 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}48 \\6\end{array}\right)}=8.15 \times 10^{-8}

To know more about binomial probability, here

brainly.com/question/9325204

#SPJ4

The complete question is-

Find the probability of winning a lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding a) 30. b) 36. c) 42. d) 48.

5 0
2 years ago
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