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AleksAgata [21]

2 years ago

9

What is the probability of rolling a 3 on a 6-sided number cube and then NOT rolling a 3 on a 6-sided number cube?

1 answer:

yarga [219]2 years ago

5 0

Do determine the probability of two events occurring in sequence, multiply the probability of each event against the other.

There's a 1 in 6 chance of rolling a 3 on a 6-sided number cube; there's a 5 in 6 chance of rolling not-a-3 on a 6-sided number cube; thus, the chance of rolling a 3 that is immediately followed by a not-3 is:

(1/6) times (5/6) = (5/36)

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Determine the net pay if gross pay was $546.54 and the following deductions were taken out. Federal tax $66.00, state tax

kirill115 [55]

Answer:

416.87, you subtract all from gross pay.

Step-by-step explanation:

546.54-66-21.86-33.89-7.92=416.87

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2 years ago

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larisa86 [58]

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3 years ago

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Graph a line with a slope of - 3/4 that contains the point (2 3)

Savatey [412]

Answer:

Step-by-step explanation:

1. Graph point (2, 3)

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3 years ago

Write an equation for the nth term of the arithmetic sequence 23, 16, 9, 2, .... Then find a25.

Gelneren [198K]

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3 0

2 years ago

How do you rationalize the numerator in this problem?

maw [93]

To solve this problem, you have to know these two special factorizations:

$x^3-y^3=(x-y)(x^2+xy+y^2)\\ x^3+y^3=(x+y)(x^2-xy+y^2)$

Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:

$\sqrt[3]{x+h}=x\\ \sqrt[3]{x}=y$

That tells us that we have:

$\frac{x-y}{h}$

So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:

$\frac{x-y}{h}*\frac{x^2+xy+y^2}{x^2+xy+y^2}=\frac{x^3-y^3}{h*(x^2+xy+y^2)}$

So, we have:

$\frac{x+h-h}{h(\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2})}=\\ \frac{x}{\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2}}$

That is our rational expression with a rationalized numerator.

Also, you could just mutiply by:

$\frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\ \frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}}$

Either way, our expression is rationalized.

7 0

3 years ago