<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>
Answer: The two sides are 15m and 10m.
Step-by-step explanation:
Area = l x w
150 = l x w
There are two solutions for l
l 1= P/4+1/4√P2﹣16A
=50/4+1/4·√50^2﹣16·150=15m
l2 = P/4﹣1/4√P2﹣16A
=50/4﹣1/4·√50^2﹣16·150=10m
<span>4.8 (x+4)=2.6
Use distributive property
4.8x + 19.2= 2.6
Subtract 19.2 from both sides
4.8x = -16.6
Divide 4.8 on both sides so that the only thing remaining on the left side is the variable x.
Final Answer: x = -3.458</span>
Answer:
Convenience sampling
Step-by-step explanation:
