Answer:
0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3
Step-by-step explanation:
We have the mean, which means that the poisson distribution is used to solve this question.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
The auto parts department of an automotive dealership sends out a mean of 6.3 special orders daily.
This means that
What is the probability that, for any day, the number of special orders sent out will be exactly 3?
This is P(X = 3). So
0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3
Answer:
x = -9
Step-by-step explanation:
Distribute the -7 to (x + 9) and distribute 9 to (x - 5), creating the equation, -7x - 63 = 9x - 45 -14x. Then you add the like terms, 9x - 14x, which equals -5x. So the equation is now -7x - 63 = -5x -45. Add the -5 to both sides which makes -2x -63 = -45. Then add the -63 to both sides, which equals 18, then divide that by -2 and you get -9 as your answer.
Answer:
The expression x^3 + ax^2 - 15x + b has a factor of x – 2 and leaves a remainder 75 when divided
by x + 3. Find the value of a and of b.
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Sorry I just need points so I can have it on the phone later after work and I will have to go get it bruh was a day and a half day