(6m^5+3-m^3-4m)-(-m^5+2m^3-4m+6)=

Mathematics

2 answers:

Helen [10]4 years ago

8 0

Answer:

$7m^5-3m^3-3$

Step-by-step explanation:

$(6m^5+3-m^3-4m)-(-m^5+2m^3-4m+6)= \\\\(6m^5+m^5)+(-m^3-2m^3)+(-4m+4m)+3-6= \\\\7m^5-3m^3-3$

Hope this helps!

likoan [24]4 years ago

6 0

Answer:

Step-by-step explanation:

6m^5-2m^3-m^3-4m+4m+3-6

7m^5-3m^3-3

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3 years ago

Read 2 more answers

General Hospital has noted that they admit an average of 9 patients per hour.

serious [3.7K]

Answer:

(a) The probability that during the next hour less than 3 patients will be admitted is 0.00623.

(b) The probability that during the next two hours exactly 8 patients will be admitted is 0.00416.

Step-by-step explanation:

The complete question is: General Hospital has noted that they admit an average of 8 patients per hour.

(a) What is the probability that during the next hour less than 3 patients will be admitted?

(b) What is the probability that during the next two hours exactly 8 patients will be admitted?

The above situation can be represented through Poisson distribution as it includes the arrival rate of the pattern. So, the probability distribution of the Poisson distribution is given by;

$P(X = x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; x = 0,1,2,......$

Here X = Number of patients admitted in the hospital

$\lambda$ = arrival rate of patients per hour = 9 patients

So, X ~ Poisson( $\lambda$ = 9)

(a) The probability that during the next hour less than 3 patients will be admitted is given by = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\frac{e^{-9} \times 9^{0} }{0!} + \frac{e^{-9} \times 9^{1} }{1!} + \frac{e^{-9} \times 9^{2} }{2!}$

= $e^{-9} +(e^{-9} \times 9)+ \frac{e^{-9} \times 81}{2}$

= 0.00623

(b) Here, $\lambda = 9 \times 2$ = 18 because we have to find the probability for the next two hours and we are given in the question of per hour.