Answer:
Mei is correct because -9 is to the left of -3 on the number line.
Step-by-step explanation:
This is because -9 is LESS than -3, when you add or subtract using negative numbers, such as 7-9, you get -2. However when you subtract 7-13, you get -6. Because you subtracted a greater number from 7, your answer is -6 is LESS than -2. This is the same with inequalities, greater negative numbers will always be smaller and smaller negative numbers will always be greater.
Hope this helps!
Answer:
4
Step-by-step explanation:
set
constrain:
Partial derivatives:
Lagrange multiplier:
![\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2x%5C%5C2y%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3x%5E2%5C%5C3y%5E2%5Cend%7Barray%7D%5Cright%5D)
4 equations:
By solving:
Second mathod:
Solve for x^2+y^2 = 7, x^3+y^3=10 first:
The maximum is 4
That answer your looking for is 80
1. 3/1
2. Will be positive. Aka always rational.
3. 3/1 + 3/1 = 6/2 = 3/1
<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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