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miv72 [106K]
3 years ago
5

What is the probability of getting either a sum of 6 or at least one 5 in the roll of a pair of​ dice

Mathematics
1 answer:
Gekata [30.6K]3 years ago
3 0
In percentage you mean?
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creativ13 [48]

Answer:

answer is b jejejejsjsjwj

7 0
2 years ago
100 points!!! need help asap
ExtremeBDS [4]

Answer:

Step-by-step explanation:

1)

res = (10.6+8.5+6.1)/2 = 12.6 kg

2)

The biggest  = 6.5 kg

mid = 4.1 kg

the samllest = 2 kg

7 0
3 years ago
Read 2 more answers
Solve for w 4w - 9 = 7
bagirrra123 [75]

Answer:

w = 4

Step-by-step explanation:

1. Isolate the variable by adding 9 to both sides

--> 4w = 16

2. Divide each side by 4

--> w = 4

-If this answer helps you, please remember to vote it as 'brainliest answer' :)

Thank you!

6 0
3 years ago
What is the answer to the problem
Tanya [424]

Answer:

a - 8

b - 20

c - 40

answer - 40% of 20 is 8

Step-by-step explanation:

20 x .4 = 8

.4 is found by the 40 / 100

8 0
3 years ago
Read 2 more answers
*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts
Nataly [62]

Answer: Assuming the function is g(x)=\frac{2x+1}{x-3}:

The x-intercept is (\frac{-1}{2},0).

The y-intercept is (0,\frac{-1}{3}).

The horizontal asymptote is y=2.

The vertical asymptote is x=3.

Step-by-step explanation:

I'm going to assume the function is: g(x)=\frac{2x+1}{x-3} and not g(x)=2x+\frac{1}{x}-3.

So we are looking at g(x)=\frac{2x+1}{x-3}.

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

0=\frac{2x+1}{x-3}

A fraction is only 0 when it's numerator is 0.  You are really just solving:

0=2x+1

Subtract 1 on both sides:

-1=2x

Divide both sides by 2:

\frac{-1}{2}=x

The x-intercept is (\frac{-1}{2},0).

The y-intercept is when x is 0.

Replace x with 0.

g(0)=\frac{2(0)+1}{0-3}

y=\frac{2(0)+1}{0-3}  

y=\frac{0+1}{-3}

y=\frac{1}{-3}

y=-\frac{1}{3}.

The y-intercept is (0,\frac{-1}{3}).

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is y=\frac{2}{1}.  After simplifying you could just say the horizontal asymptote is y=2.

Or!

I could do some division to make it more clear.  The way I'm going to do this certain division is rewriting the top in terms of (x-3).

y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}

y=2+\frac{7}{x-3}

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0
3 years ago
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