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3 years ago

So I need help finding the cross section of all the shapes, so any cross section experts mind helping?

Mathematics

1 answer:

Kisachek [45]3 years ago

3 0

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Step-by-step explanation:

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Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was

Zolol [24]

Answer: 10 customers.

Step-by-step explanation:

The formula to find the required sample size :

$n=p(1-p)(\dfrac z^*}{E})^2$ (1)

, where p= prior population proportion.

n= sample size.

$\sigma$ = Population standard deviation from previous studies.

Let p be prior population proportion of the customers who click on ads on their smartphones .

As per given , we have

p=0.62

E= 0.27

For 92% confidence , significance level : $\alpha=1-0.92=0.08$

The critical value of z for 92% confidence interval from z-table would be

$z_{\alpha/2}=z_{0.04}=1.75$

Put theses values in the formula (1), we will get

$n=(0.62)(1-0.62)(\dfrac{(1.75)}{0.27})^2$

$n=(0.2356)(6.48148)^2= 9.89745775254\approx 10$

Therefore , the company should survey 10 customers .

6 0

3 years ago

The intensity of light with wavelength λ traveling through a diffraction grating with N slits at an angle θ is given by I(θ) = N

Ymorist [56]

Answer:

0.007502795

Step-by-step explanation:

We have

N = 10,000

$\bf d=10^{-4}$

$\bf \lambda = 632.8*10^{-9}$

Replacing these values in the expression for k:

$\bf k=\frac{\pi Ndsin\theta}{\lambda}=\frac{\pi10^4*10^{-4}sin\theta}{632.8*10^{-9}}=\frac{\pi 10^9sin\theta}{632.8}$

So, the intensity is given by the function

$\bf I(\theta)=\frac{N^2sin^2(k)}{k^2}=\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}$

The total light intensity is then

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=\int_{-10^{-6}}^{10{-6}}\frac{10^8sin^2(\frac{\pi 10^9sin\theta}{632.8})}{(\frac{\pi 10^9sin\theta}{632.8})^2}d\theta$

Since $\bf I(\theta)$ is an even function

$\bf \int_{-10^{-6}}^{10^{-6}} I(\theta)d\theta=2\int_{0}^{10^{-6}}I(\theta)d\theta$

and we only have to divide the interval $\bf [0,10^{-6}]$ in five equal sub-intervals $\bf I_1,I_2,I_3,I_4,I_5$ with midpoints $\bf m_1,m_2,m_3,m_4,m_5$

The sub-intervals and their midpoints are

$\bf I_1=[0,\frac{10^{-6}}{5}]\;,m_1=10^{-5}\\I_2=[\frac{10^{-6}}{5},2\frac{10^{-6}}{5}]\;,m_2=3*10^{-5}\\I_3=[2\frac{10^{-6}}{5},3\frac{10^{-6}}{5}]\;,m_3=5*10^{-5}\\I_4=[3\frac{10^{-6}}{5},4\frac{10^{-6}}{5}]\;,m_4=7*10^{-5}\\I_5=[4\frac{10^{-6}}{5},10^{-6}]\;,m_5=9*10^{-5}$