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Margarita [4]
3 years ago
8

Side DJ is congruent to which side of CWM

Mathematics
2 answers:
dmitriy555 [2]3 years ago
8 0
The lines on the triangle tell you which one it’s congruent to. In the first triangle side DJ has 2 tally marks in between them so that would mean CW is the exact same, because it also has 2 tally marks in the middle. So DJ mad CW are congruent
NeTakaya3 years ago
6 0

Side DJ is congruent to side CW

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Sever21 [200]
That would be C, D, E and F.
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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.

8 0
3 years ago
How many solutions does the system of equations have?
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3 years ago
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erastovalidia [21]

Answer:

The correct option is 4.

Step-by-step explanation:

The given expression is

(2x^2y^6z^5)(5x^4y^5z^3)

Simplify the given expression.

(2x^2y^6z^5)\times (5x^4y^5z^3)

(2\times 5)(x^2x^4)(y^6y^5)(z^5z^3)

10x^{2+4}y^{6+5}z^{5+3}                  [\because x^mx^n=x^{m+n}]

10x^6y^{11}z^8

Therefore correct option is 4.

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a point is chosen at random on AK. what is the probability that the point will be on CD. don't forget to reduce
Novay_Z [31]

Answer:

your answer is 1/10. hope this helps!

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