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n200080 [17]
2 years ago
11

5.

Mathematics
1 answer:
Marysya12 [62]2 years ago
3 0
The function has a maximum at (1, 3).

_____
You know it is not a minimum because the coefficient of the squared term is negative.

You know the vertex is (1, 3) because you match the pattern to
.. y = a(x -h)^2 +k
which has its vertex at (h, k).

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Researchers for the University of Maryland Department of Civil and Environmental Engineering used stochastic dynamic programming
aalyn [17]

Answer:

Check the explanation

Step-by-step explanation:

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6 0
2 years ago
Order these numbers from least to greatest. 7.60, 7.228, 29/4, 7 2/9
Anarel [89]
7 2/9 < 7.228 < 29/4 < 7.60
8 0
3 years ago
Read 2 more answers
-4.64/(-0.725) how to solve
rusak2 [61]
6.4 is the answer hope this helps
5 0
3 years ago
Read 2 more answers
shirley has $540 un her bank account. She withdraws $35 each week to cover her expenses. write an equation that relates the amou
Veseljchak [2.6K]
540-35x= y

540 is the amount she has minus 35 and x is the weeks she withdrawn the money and y is the answer.
6 0
3 years ago
A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get
Tems11 [23]
An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899  (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
P(x)=C(n,x)p^x(1-p)^{n-x}where
p=probability of success
n=number of trials
x=number of successesC(n,x)=\frac{n!}{x!(n-x)!}

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more) 
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%.  The work is similar to that of n=10.
8 0
3 years ago
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