Answer: 15 boxes
Step-by-step explanation:
From the question, we are informed that Denise wants to cover a 21.75 square foot wall with weathered wood and that each box of weathered wood contains 1.5 square feet of wood.
The number of boxes of weathered wood that she need will be calculated by dividing 21.75 square feet by 1.5 square feet. This will be:
= 21.75/1.5
= 14.5
= 15 boxes approximately
Answer:
If I’m seeing it right I think it is 160units^3
Answer:
-1 is your answer for this mate
3/5 does nor belong, as the other three fractions are just different versions of the same fraction :) 4/10 can be simplified into 2/5. If you multiply this fraction by three, 2/5 would become 6/15 :) Hope this helped!!
<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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