Express the ratio as a fraction in the lowest term.3600s:2hours

Compute the probability of the event E1 that Bob wins in a duel against Eve alone, assuming he shoots first. (Hint: Let x be the

skad [1K]

Answer: Hello your question is incomplete attached below is the missing

n ( 1 + n )

Step-by-step explanation:

P( Bob hits target ) = 1/3

P( Eve hits target ) = 2/3

P( Carol hits target ) = 1

<u>Compute the P that Bob wins in a duel against Eve alone</u>

P(Bob hits the target in first shot ) = n = 1/3

P(Bob hits the target in second shot ) = n^2 = ( 1/3 * 1/3 ) = 1/9

hence the probability of Bob winning( i.e. P( Bob wins Event E1 ) = n + n^2 = n ( 1 + n )

6 0

2 years ago

The graph of an absolute value function opens up and has a vertex of (0, -3).

Ad libitum [116K]

Absolute value function domain is all real numbers

absolute value has vertex of(0,-3)
and it opens up
means that the lowest y value is -3
highest is infinity

domain=-inifnity to infinity
range=-3 to positive initnify

the answer is C for the domain and B for the range

4 0

2 years ago

Read 2 more answers

Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul

IgorLugansk [536]

Answer:

(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: $P(k) = \frac{\lambda^ke^{-k}}{k!}$ where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and $\lambda = 7$ into the formula:

$P(k=7) = \frac{7^7e^{-7}}{7!}$

$P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%$

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

$P(k\leq2) = P(k=0)+P(k=1)+P(k=2)$

$P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}$

$P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}$

$P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%$

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

$P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%$