Answer:
x=-3
Step-by-step explanation:
The x values repeat and the y values are different that means its a vertical line
so it must be x=something the x value in both is -3 so
x=-3
Answer:
1. 4(4-9m) 2. 8(x-5) 3. 7(5-3a) 4. 2(2n-1) 5. -10(3n+1) 6. 6(5n+1) 7. 10(3x+7) 8. -3(n+4) 9. 2(9-2a) 10. 8(1-k) 11. 5(8-7n) 12. 5(9x-5)
Step-by-step explanation:
Answer:
x = 20
Step-by-step explanation:
Please see the step-by-step solution in the picture attached below.
I hope this answer will help you. Have a nice day !
<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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