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Mrac [35]
2 years ago
6

Which equation describes how the parent function, y = x cubed, is vertically stretched by a factor of 4?

Mathematics
1 answer:
storchak [24]2 years ago
4 0

The equation that describes how the parent function, y = x³, is vertically stretched by a factor of 4 is y = 4x³.

<h3>We can find how the equation is vertically stretched below:</h3>

When an equation is vertically stretched, it means that the parent equation has been multiplied by a number that is more than one.

This means that the equation is multiplied by some factor.

It is given that we should describe how the parent function is vertically stretched by a factor of 4.

The parent function is given as y = x³.

Since the equation is said to be vertically stretched by a factor of 4, we must multiply the parent equation by 4 to stretch it.

Therefore, we have found the equation that describes how the parent function, y = x³, is vertically stretched by a factor of 4 to be y = 4x³.

Learn more about vertically stretching equations here: brainly.com/question/13671886

#SPJ4

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Step-by-step explanation:

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Here n = 6 , then

sum = 180° × 4 = 720°

let x be the sixth angle, then sum and equate to 720°

150 + 100 + 80 + 165 + 150 + x = 720

645 + x = 720 ( subtract 645 from both sides )

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A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect
Vladimir79 [104]

The probability that the first two electric toothbrushes sold are defective is 0.016.

The probability of an event, say E occurring is:

P(E)=\frac{n(E)}{N}

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = the number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is n = 20.

The number of defective electric toothbrushes is x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

(\left {{20} \atop {2}} \right. )=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

(\left {{3} \atop {2}} \right. )=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =3

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

\frac{3}{190}\\ =0.01579\\=0.016

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

Learn more about probability here brainly.com/question/27474070

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