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3 years ago

8

A stadium is shaped like a hemisphere with a radius of 150 feet. Last month, the owners of the stadium paid $0.05 per cubic foot

to cover the cost of utilities. What was the total cost for utilities last month?

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

353,249 feet cubed

Step-by-step explanation:

V=(4/3) $\pi (3.14)rx^{3}$ /2 times 0.05

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A sample size of n = 64 is drawn from a population whose standard deviation is o = 5.6.

AlekseyPX

Answer:

The margin of error for a 99% confidence interval for the population mean is 1.8025.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this problem:

$\sigma = 5.6$

So

$M = 2.575*\frac{5.6}{\sqrt{64}} = 1.8025$

The margin of error for a 99% confidence interval for the population mean is 1.8025.

5 0

3 years ago

By how much does 5 exceed -4?<br>(a) 1<br>(b)-1<br>(c) 9<br>(d) -9

Elden [556K]

Answer:

C

Step-by-step explanation:

-4+9=5

6 0

4 years ago

What percent of 64 is 8

steposvetlana [31]

<span>n/100 * 64 = 8 </span><span>n/100 = 8/64<span>
n = 8*100/64 = 25/2 = 12.5</span><span>
8 is 12.5% of 64 hope this helps:)</span></span>

8 0

3 years ago

Solve the initial value problem y′+y=f(t),y(0)=0 where f(t)={1,−1, if t<4 if t≥4 Use h(t−a) for the Heaviside function shifte

Anton [14]

Looks like the function on the right hand side is

$f(t)=\begin{cases}1&\text{for }t$

We can write it in terms of the Heaviside function,

$h(t-a)=\begin{cases}1&\text{for }t\ge a\\0&\text{for }t>a\end{cases}$

as

$f(t)=h(t)-2h(t-4)$

Now for the ODE: take the Laplace transform of both sides:

$y'(t)+y(t)=f(t)$

$\implies s Y(s)-y(0)+Y(s)=\dfrac{1-2e^{-4s}}s$

Solve for <em>Y</em>(<em>s</em>), then take the inverse transform to solve for <em>y</em>(<em>t</em>):

$(s+1)Y(s)=\dfrac{1-e^{-4s}}s$

$Y(s)=\dfrac{1-e^{-4s}}{s(s+1)}$

$Y(s)=(1-e^{-4s})\left(\dfrac1s-\dfrac1{s+1}\right)$

$Y(s)=\dfrac1s-\dfrac{e^{-4s}}s-\dfrac1{s+1}+\dfrac{e^{-4s}}{s+1}$

$\implies y(t)=1-h(t-4)-e^{-t}+e^{-(t-4)}h(t-4)$

$\boxed{y(t)=1-e^{-t}-h(t-4)(1-e^{-(t-4)})}$

4 0

3 years ago

Help me pls i keep getting 3/2 but thats not the answer

Gemiola [76]

$\huge\text{Hey there!}$

$\large\boxed{\mathsf{\dfrac{6x^2}{y^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4^2)}{4^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4\times4)}{4\times4\times4}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{16\times4}}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{64}}}\\\large\boxed{\approx \mathsf{\dfrac{96\div32}{64\div32}}}\\\large\boxed{\approx \mathsf{\dfrac{3}{2} \rightarrow \bf \dfrac{96}{64}}}$

$\huge\text{Thus, your answer is: \boxed{\mathsf{Option\ A. } \ \mathsf{\dfrac{96}{64}}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy}\\\huge\text{your day!}$

~ $\frak{Amphitrite1040:)}$

3 0

3 years ago

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