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14 days ago

XYZ Inc. produces two types of paper towels?regular and super-soaker. Regular uses 2 units of recycled paper per unit of product

ion, and super-soaker uses 3 units of recycled paper per unit of production. The total amount of recycled paper available per month is 10,000. Let X1 be the number of units of regular produced per month and X2 the number of units of super-soaker produced per month. The appropriate constraint/s will be:_________
A) 2X1 + 3X2 = 10000

B) 180X1 + 80X2 geq 360

C) 2X1 + 3X2 geq 10000

D) 2X1 = 3X2

E) A, B, and C

Mathematics

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Let c1(t) = eti + (sin(t))j + t3k and c2(t) = e−ti + (cos(t))j − 6t3k. Find the stated derivatives in two different ways to veri

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Answer:

$i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

$\frac{d}{dx}(e^x) = e^x$

Step-by-step explanation:

Step 1:-

We have c1(t) = e^ t i + (sin(t))j + t³k

and c2(t) = e^−t i + (cos(t))j − 6t³k.

By adding c1(t) and c2(t):

c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k

Now, employing the derivative formula:

$\frac{d}{dx}(e^x) = e^x$

$\frac{d}{dx}(sinx) = cosx\\\frac{d}{dx}(cosx) = -sinx$

Next, differentiate with respect to 't'

$\frac{d}{dt}c_{1}+ c_{2} } = e^ t i +cost j +3t^2 k - e^-t i - sintj -18t^2 k$

By factoring out i, j, and k terms, we arrive at:

$\frac{d}{dt}(C_{1} +C_{2} ) = i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

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2 months ago

A bag contains chips of which 27.5 percent are blue. A random sample of 5 chips will be selected one at a time and with replacem

lawyer [12517]

Answer:

$\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997$

Step-by-step explanation:

It is known that the mean and standard deviation of the sampling distribution of the sample proportion( $\hat{p}$ ) are represented as follows:-

$\mu _{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}$

, where p= Population proportion and n = sample size.

Let p denote the proportion of blue chips.

According to the information provided, we have

p= 0.275

n= 5

Thus, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be:

$\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{ 0.275(1- 0.275)}{5}}\\\\=0.19968725547\approx0.1997$

Therefore, you will have the mean and standard deviation for the sample proportion of blue chips for samples of size 5:

$\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997$

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