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Assoli18 [71]
3 years ago
7

I need help pls 11+4b=8b-13 b=?

Mathematics
1 answer:
spin [16.1K]3 years ago
4 0
With this, you need side A to equal side B. Your first step is to move the variable to one side. (I've also worked in out, and color coordinated the steps, so you can see better).

11 + 4b = 8b - 13 < let's subtract "4b" from the left side to get rid of it (we're subtracting because it is added on to left side). This will give us:

11 = 8b - 13 < we've subtracted "4b" from the left side, however what you do to one side of the equal sign, you must do to the other, so subtract "4b" from the right side as well. Once done, you should have this"

11 = 4b - 13 < we have "4b-13" because you can only work with like terms. 8b-4b=4b, and the 13 stays untouched.
 
Now let's work on isolating the variable. See how we have a -13 on the side with the variable? We don't want that, so we're going to add 13 to that side, to give us 0 and cancel it out. Remember, what you do to one side, you must do to the other, so add 13 to the 11 on the other side as well. This will give you 24, and your equation should now look like this:

24 = 4b < we're almost done, but the variable is not quite by itself yet. We need to get rid of the "4" in "4b". 4 in being multiplied to "b" so we need to do the opposite operation to cancel it out. We need that 4 to be a 1 because 1x=x. So, divide 4b by 4 to get 1b or just plain b. 

Remember, what you do to one side, you must do to the other, so divide 24 by 4 to get 6, and you're left with your answer.

Answer 6 = b 

Hope this helps!
      

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An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6
SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6  times.

The above situation can be represented through binomial distribution;

P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......

where, n = number trials (samples) taken = 6 trials

            r = number of success = at least 4

           p = probability of success which in our question is probability of

                 rolling a “six", i.e; p = 0.20

<u><em>Let X = Number of sixes on a die</em></u>

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X \geq 4)

P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)

=  \binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}

=  15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}

=  0.0154 + 0.00154 + 0.000064

=  0.01696

<em />

Therefore, probability of rolling at least 4 sixes is 0.01696.

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