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3 years ago

Which equation would shift the parabola down 3 units? A) y = x2 B) y = x2 - 3 C) y = x2 + 3 D) y = (x + 3)2

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

5 0

Answer

x^2-3

Step-by-step explanation:

olga nikolaevna [1]3 years ago

4 0

Answer:

B) y = x2 - 3

Step-by-step explanation:

Suppose you have a function y = f(x), you can do these following operations on the function:

Shift up a units: $y = f(x) + a$

Shift down a units: $y = f(x) - a$

Shift left a units: $y = f(x + a)$

Shift right a units: $y = f(x - a)$

So the correct answer for this question is:

B) y = x2 - 3

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Answer: 2z + 6

Step-by-step explanation: Distribute the 2 to z and 3.

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Step-by-step explanation:

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3 years ago

Read 2 more answers

In one town, the number of burglaries in a week has a poisson distribution with a mean of 1.9. find the probability that in a ra

tester [92]

Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9

We have to find the probability that in a randomly selected week the number of burglaries is at least three.

P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........

= 1 - P(X < 3)

= 1 - [ P(X=2) + P(X=1) + P(X=0)]

The Poisson probability at X=k is given by

P(X=k) = $\frac{e^{-mean} mean^{x}}{x!}$

Using this formula probability of X=2,1,0 with mean = 1.9 is

P(X=2) = $\frac{e^{-1.9} 1.9^{2}}{2!}$

P(X=2) = $\frac{0.1495 * 3.61}{2}$

P(X=2) = 0.2698

P(X=1) = $\frac{e^{-1.9} 1.9^{1}}{1!}$

P(X=1) = $\frac{0.1495 * 1.9}{1}$

P(X=1) = 0.2841

P(X=0) = $\frac{e^{-1.9} 1.9^{0}}{0!}$

P(X=0) = $\frac{0.1495 * 1}{1}$

P(X=0) = 0.1495

The probability that at least three will become

P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]

= 1 - [0.2698 + 0.2841 + 0.1495]

= 1 - 0.7034

P(X ≥ 3 ) = 0.2966

The probability that in a randomly selected week the number of burglaries is at least three is 0.2966

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3 years ago

How is the the graph of y=

kati45 [8]

Answer:

did u ever find out the answer ?

Step-by-step explanation:

7 0

3 years ago

Let f(x)=x^2+17x+72 What are the zeros of the function?

densk [106]

Answer:

The zeros of the given function $f(x)=x^2+17x+72$ is $\left(x+8\right)\left(x+9\right)$ are -8 and -9.

Step-by-step explanation:

Given : Function $f(x)=x^2+17x+72$

We have to find the zeros of the functions.

Consider the given Function $f(x)=x^2+17x+72$

Since, we have to find the zeros of the given functions.

Put f(x) = 0

Thus, $x^2+17x+72=0$

We can solve the given equation using middle term splitting method.

17x can be written as 8x + 9x

$x^2+17x+72=0$ can be written as $x^2+8x+9x+72=0$

Taking x common from first two term and 9 common from last two terms, we have,

$x(x+8)+9(x+8)=0$

Simplify, we have,

$\left(x+8\right)=0$ or $(x+9\right)=0$

x = -8 and x = -9

Apply zero product rule, $a\cdot b= 0 \Rightarrow a=0 \ or \ b=0$

$\left(x+8\right)\left(x+9\right)=0$

Thus, The zero of the given function $f(x)=x^2+17x+72$ is $\left(x+8\right)\left(x+9\right)$ are -8 and -9.

3 0

3 years ago