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Mice21 [21]
4 years ago
11

An hour before show​ time, only 342 people are seated for a rock fest. According to ticket​ sales, 91​% of the people have yet t

o arrive. How many tickets were sold for the rock fest​? Explain your thinking.
Mathematics
1 answer:
tia_tia [17]4 years ago
5 0

Answer:

3800

Step-by-step explanation:

342 is the 9 percent that have shown up. If you divide that by nine you would get on percent of the total which comes out to be 38. so you multiply 38 by 100 and get 3800 which is 100% of the tickets sold.

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Answer:

18.75 ft

Step-by-step explanation:

The given scenario can be drawn as shown in the following attachment.

Given that,

m\angle DCE=\angle ACB and m\angle DEC=\angle ABC=90^{\circ}

Therefore, \Delta ABC\sim \Delta DEC

\Rightarrow \dfrac{AB}{DE}=\dfrac{BC}{EC}

\Rightarrow \dfrac{AB}{5}=\dfrac{15}{4}

\Rightarrow AB=\dfrac{15\times 5}{4}=18.75 ft

AB is the height of the house.

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What is angle A in the following equation?
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Answer:

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Step-by-step explanation:

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3 years ago
Calculus 3 help please.​
Reptile [31]

I assume each path C is oriented positively/counterclockwise.

(a) Parameterize C by

\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}

with -\frac\pi2\le t\le\frac\pi2. Then the line element is

ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt

and the integral reduces to

\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt

The integrand is symmetric about t=0, so

\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt

Substitute u=\sin(t) and du=\cos(t)\,dt. Then we get

\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}

(b) Parameterize C by

\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{3^2+4^2} \, dt = 5\,dt

and

\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt

Integrate by parts with

u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}

\displaystyle \int u\,dv = uv - \int v\,du

\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}

(c) Parameterize C by

\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt

and

\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}

8 0
1 year ago
A particular psychological test is used to measure need for achievement. The average test score for all university students in O
Angelina_Jolie [31]

Answer:

Only B and C are always true.

Step-by-step explanation:

To analyse which statements are always true, we go through the process of finding confidence intervals for sample means, from the start.

Confidence Interval = (Sample mean) ± (Margin of error)

From this expression, it is evident that the margin of error determines how wide the confidence interval would be.

Sample Mean = 110 (given)

Margin of Error = (Critical value) × (Standard deviation of the distribution of sample means)

Since no information about the population standard deviation is provided, the critical value is obtained using t-distribution.

The critical value usually varies at different confidence levels and degree of freedoms.

The higher the confidence level, the higher the critical value and the higher the margin of error leading to a wider range.

Hence, a confidence interval of 95% will have a higher critical value than a confidence interval of 90%. Hence, statement C is proved once that 'for n = 100, the 95% confidence interval will be wider than the 90% confidence interval'.

After obtaining the critical value, we then obtain the standard deviation of the distribution of sample means or simply the standard error of the mean. This is given as

σₓ = σ/√n

where σ = standard deviation; which isn't given. The standard deviation might be high enough to guarantee that the Margin of error is high too for the confidence interval to contain 115 or low enough to ensure that the Margin of error is very small and the confidence interval will not contain 115.

Or the sample size might be high enough to make the standard error of the mean to be eventually small and lead to a small margin of error and the condidence interval will not contain 115.

The point is, it isn't always sure that the resulting interval.would contain 115. So, statement A isn't always true.

Then from σₓ = σ/√n,

n = sample size, a large sample size means a more narrow confidence interval and a small sample size means a wider sample size. This proves statement C.

The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50.

Hence, Only B and C are always true.

Hope this Helps!!!

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What is your question?

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