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Anuta_ua [19.1K]
2 years ago
8

Find the volume of a sphere with a surface area of 1296 in.

Mathematics
1 answer:
seraphim [82]2 years ago
8 0

Answer:

V≈4387.14

Step-by-step explanation:

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\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad  \begin{cases} I=\textit{interest earned}\dotfill&\$540\\ P=\textit{original amount deposited}\dotfill & \$4800\\ r=rate\to 4.5\%\to \frac{4.5}{100}\dotfill &0.045\\ t=years \end{cases} \\\\\\ 540=(4800)(0.045)t\implies \cfrac{540}{(4800)(0.045)}=t\implies 2.5=t

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1. 2x

2. 22

Step-by-step explanation:

question 1

f(x)-g(x)

5x-3-(3x-3)

5x-3-3x+3

5x-3x-3+3

2x+0

2x

question 2

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3 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
3 years ago
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