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Leno4ka [110]
4 years ago
14

One million people have equal chances of being called by telephone from a broadcasting show. Determine the probability of one pa

rticular person receiving a call.
Mathematics
2 answers:
Softa [21]4 years ago
7 0
One out of one million
Anna [14]4 years ago
4 0
.0001% or a one in a million chance.
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Can someone please solve
pentagon [3]

Answer:

78

Step-by-step explanation:

Evaluate 6 (2 x^2 - 5) where x = -3:

6 (2 x^2 - 5) = 6 (2×(-3)^2 - 5)

Hint: | Evaluate (-3)^2.

(-3)^2 = 9:

6 (2×9 - 5)

Hint: | Multiply 2 and 9 together.

2×9 = 18:

6 (18 - 5)

Hint: | Subtract 5 from 18.

| 1 | 8

- | | 5

| 1 | 3:

6×13

Hint: | Multiply 6 and 13 together.

6×13 = 78:

Answer:  78

7 0
2 years ago
A state vector X X for a four-state Markov chain is such that the system is four times as likely to be in state 2 as in 4, is no
Vesna [10]
All the components in the state vector need to sum to 1. You're given that component corresponding to state 1 is 0.2, and that the component for state 3 is 0.

That leaves states 2 and 4, for which you're told that the component for state 2 is four times as large. If x_i is the component for state i, then you have

x_1+x_2+x_3+x_4=1\iff 0.2+4x_4+0+x_4=1\implies5x_4=0.8\implies x_4=0.16

which means x_2=4x_4=0.64. So the state vector is \mathbf x=(0.2,0.64,0,0.16).
3 0
4 years ago
The​ life, in​ years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o
Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

\bf P(X

So, the probability that a switch fails in the first year is

\bf P(X

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and  

\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}

where  

\bf \binom{100}{k} equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999

3 0
3 years ago
A jacket is on sale for 20%20% off the regular price. The sale price is $50.00$50.00. What is the regular price of the jacket?
algol13
62.50 hope this helps may be wrong :4
5 0
3 years ago
What is f(-2) when f(x) = 3x - 5
babymother [125]
F(-2) = 3(-2)-5 = -11
hope it helps
7 0
3 years ago
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