Answer:
It's a simple question. This swimming pool is basically a rectangular cuboid. Length x Width x Height = how much water it will hold.
Water = 1000000 feet^(3)
Step-by-step explanation:
1000 x 100 x 10 = 1000000 feet^(3)
or you could just converted to meters cube which should be 28316.85 (ish).
(2/3) - (7/12)
= .0834
or in fraction form: 1/12
Answer:
XY = 18
Step-by-step explanation:
The figures are similar so the ratios of corresponding sides are equal, that is
= 
, substitute values
= 
( cross- multiply )
10 XY = 180 ( divide both sides by 10 )
XY = 18
Answer:
3 7/8
Step-by-step explanation:
_____ + 5 3/4 = 9 5/8
x + 5 3/4 = 9 5/8
Subtract 5 3/4 from each side
x + 5 3/4 - 5 3/4 = 9 5/8 - 5 3/4
x = 9 5/8 - 5 3/4
Get a common denominator of 8
5 3/4 *2/2 = 5 6/8
x = 9 5/8 - 5 6/8
We need to borrow 1 from the 9 in the form of 8/8
x= 8 + 8/8+ 5/8 - 5 6/8
x = 8 13/8 - 5 6/8
= 3 7/8
<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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