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jonny [76]
3 years ago
12

HELP WITH BOTH PARTSSSSSSSSS

Mathematics
2 answers:
const2013 [10]3 years ago
4 0

Answer:

Wait... that's my name... whatever, here is the answer XD

Step-by-step explanation:

c+9.75=20.75 is the answer :D

Cost of his ticket is $20.75

Arturiano [62]3 years ago
3 0

Answer:

c+9.75=20.75  c=11

Step-by-step explanation:

c+9.75=20.75

c=20.75-9.75

c=11

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Factor the expression. r^2 – 49 a. (r – 7)(r 9) b. (r 7)(r 7) c. (r – 7)(r 7) d. (r – 7)(r – 7)
gayaneshka [121]
r^2-49=r^2-7^2\\\\/use\ a^2-b^2=(a-b)(a+b)/\\\\=\boxed{(r-7)(r+7)}
6 0
3 years ago
5.69÷7 you can round to nearest hundredth
pogonyaev
The answer would be .81 becuase you can't round the 1 up to 2 becuase the number in the thousandths place is not above 5. 
Hope this helps!!
5 0
3 years ago
If the club is sending 3 members to a convention, how many different groups of 3 members are possible?
Natasha2012 [34]
There are 120 different possible groups.
(10 * 9 * 8)/(3 * 2 * 1) + 120
6 0
3 years ago
Im really bad at the functions stuff so- yk help
atroni [7]

Answer:Isolate the variable by dividing each side by factors that don't contain the variable.

w =-3u/2+2

4 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number o
Ksenya-84 [330]

Answer:

Step-by-step explanation:

Step1:

We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t

Step2:

Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “”

The probability mass function of poisson distribution is given by

P(X) = , x = 0,1,2,3,...,n.

Where, μ(mean of the poisson distribution)

a).

Given that time period t = 1hr.

Then,μ = 8t

             = 8(1)

             = 8

Now,

The probability that exactly 6 small aircraft arrive during a 1-hour period is given by

P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6)

Consider,

P(X = 6) =  

              =  

              =  

              = 0.1219.

Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.

1).P(At least 6) = P(X 6)

Consider,

P(X 6) = 1 - P(X5)

                = 1 - {+++++}

                = 1 - (){+++++}

                = 1 - (0.000335){+++++}

                = 1 - (0.000335){1+8+32+85.34+170.67+273.07}

                = 1 - (0.000335){570.08}

                = 1 - 0.1909

                = 0.8090.

Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090.

2).P(At least 10) = P(X 10)

Consider,

P(X 10) = 1 - P(X9)

                 = 1 - {+++++

5 0
3 years ago
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