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3 years ago

11

Yotimo has 9 dogs that pull his sled. During winter, he puts 4 boots on each dog so their feet are protected from the snow. How

many boots does he need in all?
5
13
38
36

1 answer:

Temka [501]3 years ago

6 0

Answer:

36

Step-by-step explanation:

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Evaluate (x + y ) 0 for x = –3 and y = 5.

zubka84 [21]

(x+y)0
(-3+5)0
(-15)0
THE ANSWER IS 0 BECAUSE YOU ARE MULTIPLYING BY 0

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3 years ago

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ANSWER AND I WILL DO YOU A FAVOR!!!!

Sonja [21]

T=21+4.5X

21 is the original temp.

4.5 is the constant rate I got when I subtracted 75-21=54 then divided 54/12min=4.5.

75 =21 + 4.5(12)

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3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Can someone help me?

Roman55 [17]

Answer:

The upper quartile is 6.

Step-by-step explanation:

FIrst, I found the median of the data, which is 4. Then, I determined the median of the upper half of the data, which is 6. The median of the upper half of the data is known as the upper quartile.

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3 years ago

Among 420 randomly selected employees at a company, the mean number of hours of overtime worked per month is 10 hours and the st

kari74 [83]

The margin of error of the random selection is 0.20

The given parameters are:

$n = 420$ --- the sample size

$\sigma = 1.6$ --- the standard deviation

$\bar x = 10$ --- the mean

$\alpha = 99\%$ --- the confidence level.

The margin of error (E) is calculated as follows:

$E = z \times \sqrt{\frac{\sigma^2}{n}}$

So, we have:

$E = z \times \sqrt{\frac{1.6^2}{420}}$

$E = z \times \sqrt{\frac{2.56}{420}}$

The z-value for 99% confidence level is 2.576.

Substitute 2.576 for z

$E = 2.576 \times \sqrt{\frac{2.56}{420}}$

$E = 2.576 \times \sqrt{0.006095}$

Take square roots

$E = 2.576 \times 0.0781$

Multiply

$E = 0.2012$

Approximate

$E = 0.20$

Hence, the margin of error is 0.20

Read more about margin of error at:

brainly.com/question/14396648

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2 years ago

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