You want to find the area left over after the pool is built, so subtract the area of the pool from the area of the yard.
Area of Yard= Base x Height = 14x*19x = 266x^2
Area of Circle= Pi x Radius^2 = (6x)^2*pi = 36x^2*pi
Now subtract the two areas:
266x^2-(36^2*pi)
266x^2-36x^2*pi
Take 2x^2 as a common factor:
2x^2(133-18pi)
D: <span>2x^2(133-18pi)
Hope this helps :)</span>
$365 increased to $428
365/428=1.17
which translates to an 117% increase
<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>
441ft² (i believe this is the answer, idk though, i might be wrong sorry)
Please show a picture so that I can help you
