Can anyone help me with this

Mathematics

2 answers:

Kazeer [188]3 years ago

8 0

Answer:

very simple algebric question...hope its help you

stiv31 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

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Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 195,

Semenov [28]

Given:
n = 195, sample size.
x = 162, successes in the sample

The proportion is
p = x/n = 162/195 = 0.8308

n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.

The proportion mean is
μ = 0.8308
The proportion standard deviation is
$\sigma = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.8308(1-0.8308)}{195} } =0.0269$

σ/√n = 0.0269/√195 = 0.00192

At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)

Answer: The 95% confidence interval is (0.827, 0.835)

5 0

4 years ago

assume that tge volume,v, of a sphere is expanding at a rate of 100 inches cubex per minute, the volume of a sphere is given by

Zielflug [23.3K]

$\bf \textit{volume of a sphere}\\\\
V=\cfrac{4}{3}\pi r^3\\\\
-----------------------------\\\\
\cfrac{dv}{dt}=\cfrac{4\pi }{3}\cdot 3r^2\cfrac{dr}{dt}\implies \cfrac{dv}{dt}=4\pi r^2\cfrac{dr}{dt}
\\\\\\
\cfrac{\frac{dv}{dt}}{4\pi r^2}=\cfrac{dr}{dt}\qquad 
\begin{cases}
\frac{dv}{dt}=100\\\\
r=8
\end{cases}\implies \cfrac{100}{4\pi 8^2}=\cfrac{dr}{dt}$