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WINSTONCH [101]
3 years ago
12

Which set of numbers CANNOT represent the lengths of the sides of a triangle?

Mathematics
1 answer:
posledela3 years ago
4 0

Answer:

C

Step-by-step explanation:

The general rule is that any 2 must exceed the third in order for a triangle to form. If any one is an exception, it won't work.

A will form a triangle. Any 2 exceeds the third.

B also forms a triangle. Same rule.

C  does not work. 7 + 11 = 18 which is the value of the third leg.

D words Any two exceeds the 3rd.

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Find the number of possible outcomes in the sample space.<br> You flip a coin once.
igor_vitrenko [27]
2 it will either land on heads or tails if you flip it once, hence 2 possible outcomes.
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2 years ago
Write these values in order starting with the smallest . <br> 0.15 12% 1/10 1/20
Vikki [24]

Answer:

1/10, 1/20, 12%, 0.15

Step-by-step explanation:

1/20=1÷20=0.05

12%= 12÷100=0.12

1/10= 1÷10= 0.01

6 0
3 years ago
For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
pshichka [43]

Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables X_i with the following distribution:

X_i Bin (1,p) = Be(p) bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

Z = \sum_{i=1}^N X_i

From the info given we know that N \sim Bin (M,q)

We need to proof that Z \sim Bin (M, pq) by the definition of binomial random variable then we need to show that:

E(Z) = Mpq

Var (Z) = Mpq(1-pq)

The deduction is based on the definition of independent random variables, we can do this:

E(Z) = E(N) E(X) = Mq (p)= Mpq

And for the variance of Z we can do this:

Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2

Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2

And if we take common factor Mpq we got:

Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]

And as we can see then we can conclude that   Z \sim Bin (M, pq)

8 0
3 years ago
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stellarik [79]

Answer:

a dry-erase maker

Step-by-step explanation:

7 0
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I NEED HELP ASAP!! will be willing to mark you brainliest!
Softa [21]

Answer:

8/75

Step-by-step explanation:

5 0
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