A b c is the order of answers
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
4*(2*x-6)-(10*x-6)=0
2x - 6 = 2 • (x - 3)
8 • (x - 3) - (10x - 6) = 0 -2x - 18 = -2 • (x + 9)
-2 • (x + 9) = 0 Solve : -2 = 0
<span>This equation has no solution.
</span>A a non-zero constant never equals zero.
Solving a Single Variable Equation :
Solve :
x+9 = 0 <span>
</span>Subtract 9 from both sides of the equation :<span>
</span> x = -9
Answer:
It's a.
Step-by-step explanation:
(a - 2b)^2 + 8ab
= a^2 - 4ab + 4b^2 + 8ab
= a^2 + 4ab + 4b^2.
((a + 2b)^2 = a^2 + 4ab + 4b^2.
Answer:
Which question needs to be answered?
