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3 years ago

When rolling two number cubes or a set of six-sided dice, what is the probability of rolling a 7?

Mathematics

2 answers:

egoroff_w [7]3 years ago

7 0

Impossible because it is a 6 sided dice and anything higher than 6 is impossible

Elina [12.6K]3 years ago

5 0

P(7) =0.

Since the maximum probability is one (certainty) & the minimum is Zero, impossibility.

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What is the slope of the line graphed on the coordinate plane?

Alekssandra [29.7K]

Answer:

-1/12

Step-by-step explanation:

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-1/12

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3 years ago

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Semenov [28]

Answer:

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Step-by-step explanation:

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3 years ago

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The difference of twice a number and 7 is 9

UNO [17]

2x-7=9

Add 7 to both sides
2x=16

Divide both sides by 2
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Final answer: The number is 8

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3 years ago

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Katen [24]

Answer:

1. 2x can be seen as +2x therefore when you put it as +2x-5x the answer will be - 3x

2. +3a+2a also equals +5a

3. - 5 is left alone and there is no other similar expression

Put them together the answer becomes

-3x+5a-5

It doesn't have to be the same order of format but the numbers must be as such

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2 years ago

Plz help me solve this pre calc problem, I have to condense using the properties of log

Ahat [919]

Given:

$\log \left(x^{2}\right)+\log \sqrt{x^{3}+1}$

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The condense of the logarithmic function.

Solution:

$\log \left(x^{2}\right)+\log \sqrt{x^{3}+1}$

Apply log rule:

$\log _{c}(a)+\log _{c}(b)=\log _{c}(a b)$

Using the log rule:

$\log _{10}\left(x^{2}\right)+\log _{10}(\sqrt{x^{3}+1})=\log _{10}\left(x^{2} \sqrt{x^{3}+1}\right)$

$\log \left(x^{2}\right)+\log \sqrt{x^{3}+1}=\log \left(x^{2} \sqrt{x^{3}+1}\right)$

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4 years ago