Answer:
The nonpermissible replacement of the variable y is -3
Step-by-step explanation:
The nonpermissible replacement of the variable in a certain expression is the value of the variable that will make the denominator of the expression zero.
The expression 
has the denominator 
This denominator cannot be equal zero, thus,
Okay here we go!
The answer is D.
Because when you solve the equation you get 2 and that means the arrow has to be or end at 2.
HOPE THIS HELPED! :)
Answer:
6³ = 216
9² = 81
3⁴ = 81
18² = 324
Step-by-step explanation:
6³ = 6 · 6 · 6 = 36 · 6 = 216
9² = 9 · 9 = 81
3⁴ = 3 · 3 · 3 · 3 = 9 · 3 · 3 = 27 · 3 = 81
18² = 18 · 18 = 324
<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.
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One way to carry out this problem is to simply take the value that you already know, which is that 1/5 is 20 % and multiply it by 2 for 2/5 or 2 (1/5ths) and do this for all of them. There is a pattern that occurs for every increase in one out of denominator of 5.
