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Inessa05 [86]
3 years ago
11

One final question, what would be the proper way to write this? I thought it was the last one answer but I’m not sure

Mathematics
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

OMG NO- KEEP IT THAT WAY

You're already right! :)

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A regulation basketball hoop has a rim with an 18-inch diameter. What is the approximate length around the rim of a regulation b
pogonyaev

Answer:

Step-by-step explanation:

4 0
4 years ago
the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a fr
Naily [24]
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}

\bf f=k\sqrt{t}\quad \textit{we also know that }
\begin{cases}
f=300\\
t=8
\end{cases}\implies 300=k\sqrt{8}
\\\\\\
\cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k
\\\\\\
\textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k

\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\
-------------------------------\\\\
\textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}
5 0
4 years ago
Read 2 more answers
PLEASE HELP ME WITH THESE TWO PROBLEMS. QUICK POINTS
uranmaximum [27]

Answer:

can i pls have brainlysit? I really need them. In return I will try to answer this question

Step-by-step explanation:

length(cm) × width(cm) × height(cm) = cubic centimeters(cm³)

use this formula as a first step.

then substitute your values in

length = 3 cm

width = 4 cm

height = 5 cm

so 3 * 4 * 5 = 60 cm^3

7 0
4 years ago
Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo
ycow [4]

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

n = 201, p = 0.45

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

3 0
3 years ago
Show that solving the equation 3^2x=4 by taking common logarithms of both sides is equivalent to solving it by taking logarithms
Sophie [7]

Answer:

Step-by-step explanation:

Case I:  use common logs:

 2x log 3 = log 4, or 2x(0.47712) = 0.60206

 Solving for x, we get 0.95424x = 0.60206, and then x = 0.60206/0.95424.

 x is then x = 0.631

Case II:  use logs to the base 3:

 2x (log to the base 3 of 3) = (log to the base 3 of 4)

 This simplifies to 2x(1) = 2x = (log 4)/log 3 = 1.262.  Finally, we divide this

 result by 2, obtaining x = 0.631

5 0
3 years ago
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