Answer:
Your answers are A,B,D, and F
Step-by-step explanation:
When dealing with negatives the answer is always going to be negatives under -8 like -5, or whole numbers like 3
1. your leading coefficient has to be 1 (nothing before the x^2). If there is you have to divide that out before you start.
2. Move your constant (the number without any x attached) to the other side of the equation
3. take 1/2 of the b term (the one with the x attached) and then square it and then add it to both sides
4. Factor the left side
5. Set each factor equal to 0 and solve
Here is an example:
4x^2-24x+20=0
The first term is not a 1 so we have to divide it out by 4 first
x^2-6x+5=0
Move the 5 to the other side. It becomes negative.
x^2-6x=-5
Take 1/2 of 6 (3) then square it (9) and add it to both sides.
x^2-6x+9=-5+9
Factor the left side
(x-3)(x-3)=4
(x-3)^2=4
To solve you need to square root both sides
x-3=+/-
x-3=+/-2
x=3+2=5
x=3-2=1
Those would be your two answers.
<span>Hope that helps</span>
Answer:
Given that an article suggests
that a Poisson process can be used to represent the occurrence of
structural loads over time. Suppose the mean time between occurrences of
loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a
4-year period? c). How long must a time period be so that the probability of no loads
occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a
4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads
occurring during that period is at most 0.3 is given by: 3.3 years
Step-by-step explanation:
Answer:
64%
Step-by-step explanation:
:)
The basketball coach spent 80 dollars more than the baseball coach
