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4 years ago

5

A company that manufactures toothpaste is studying five different package designs.Assuming that one design is just as likely to

be selected by a consumer as any other design, what selection probability would you assign to each of the package designs (to 2 decimals)

1 answer:

otez555 [7]4 years ago

6 0

Answer:

The selection probability to be assigned to each of the package designs is 0.20

Step-by-step explanation:

Firstly, we need to assume that one design is just as likely to be selected by a consumer as any other design

so the probability of selecting any of the design is same and that is 1/5 = 0.20

Thus, what we are trying to say is that each of the package designs have an equal selection probability of 0.20

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Write the percent as a fraction in simplest form and as a decimal 36%

Darya [45]

Answer:

0.36, 9/25

Step-by-step explanation:

36% converts to the decimal fraction 0.36.

36% can also be expressed in the reduced fractional form 36/100= 9/25.

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3 years ago

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Completely simplify the rational expression LaTeX: \frac{7x}{3y} + \frac{12x}{9y}

irina [24]

Answer:

$\frac{11x}{3y}$

Step-by-step explanation:

$\frac{7x}{3y} + \frac{12x}{9y}$

To add the two fractions, you need a common denominator. In this case, the lowest common denominator (LCD) is 9y. That's because 9y is divisible by both 3y and 9y.

The first fraction must be changed so that its denominator is 9y. Do this by multiplying both numerator and denominator by 3.

$\frac{7x}{3y}\cdot\frac{3}{3}+\frac{21x}{9y}=\frac{21x}{9y}+\frac{12x}{9y}\\=\frac{21x+12x}{9y}\\=\frac{33x}{9y}$

Finally, this answer can be simplified ("reduced") by dividing both numerator and denominator by 3, their greatest common factor.

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3 years ago

Find all the zeros of the equation. -x^3-3x^2=6x+4

k0ka [10]

Answer:

The zeros of given expression $-x^3-3x^2=6x+4$ is -1, $-1+i\sqrt{3}$ and $-1-i\sqrt{3}$

Step-by-step explanation:

Given expresssion is $-x^3-3x^2=6x+4$

To find zeros of given expression we have to equate the expression to zero.

ie., $-x^3-3^2-6x-4=0$

$-(x^3+3x^2+6x+4)=0$

$x^3+3x^2+6x+4=0$

By using synthetic division

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| 0 -1 -2 -4

|________________

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Therefore (x+1) is a zero

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For quadratic equation $ax^2+bx+c=0$ we have

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Here a=1 ,b=2 and c=4 now substitute the values

$x=\frac{-2\pm \sqrt{2^2-4(1)(4)}}{2(1)}$

$x=\frac{-2\pm \sqrt{4-16}}{2}$

$x=\frac{-2\pm \sqrt{-12}}{2}$

$x=\frac{-2\pm \sqrt{12i^2}}{2}$ where $i^2=-1$

$x=\frac{-2\pm i \sqrt{4\times 3}}{2}$

$x=\frac{-2\pm i \sqrt{4}\sqrt{3}}{2}$

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Step-by-step explanation:

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