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Lostsunrise [7]
3 years ago
5

O triplo de um mumero somado a 9 e igual a 30 .qual e esse numero

Mathematics
1 answer:
-Dominant- [34]3 years ago
7 0
Do you think than i ve understood it right 

3x+9=30

3x=21

x = 21/3 = 7

x = 7

hope helped 
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Sindrei [870]
H^3+3h^2x+3hx^2+x^3 should be the answer
7 0
2 years ago
Elizabeth attempts a field goal by kicking a football from the ground with an initial vertical
sammy [17]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t\implies \stackrel{\textit{hits the ground}~\hfill }{0=-16t^2+64t} \\\\\\ 0=-16t(t-4)\implies t= \begin{cases} 0\\ \boxed{4} \end{cases}

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.

5 0
3 years ago
Find the annual percentage rate of change for the population of Oregon. In 2000, there were 4.8
MrRa [10]

Answer:

The annual percentage rate of change for the population of Oregon is of 8.125%.

Step-by-step explanation:

Total percentage change:

Change multiplied by 100 and divided by the initial value.

Change: 8.7 - 4.8 = 3.9 million

Initial value: 4.8

Percentage change: 3.9*100/4.8 = 81.25%

Annual percentage rate of change

81.25% during 10 years(from 2000 to 2010).

So, per year

81.25%/10 = 8.125%

The annual percentage rate of change for the population of Oregon is of 8.125%.

5 0
2 years ago
A research team at Cornell University conducted a study showing that approximately 10% of all businessmen who wear ties wear the
LenKa [72]

Answer:

a) The probability that at least 5 ties are too tight is P=0.0432.

b) The probability that at most 12 ties are too tight is P=1.

Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432

The probability that at least 5 ties are too tight is P=0.0432.

a) We can express the probability that at most 12 ties are too tight as:

P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1

The probability that at most 12 ties are too tight is P=1.

5 0
2 years ago
8. The 1st student flipped a fair coin 7 times
REY [17]
My guess is he gets heads about half the time
4 0
3 years ago
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