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Anettt [7]
3 years ago
9

Christian randomly selects students from his grade to rate a math test as easy, moderate, or difficult. Of the students he surve

yed, 13 said the test was easy, 11 rated it as moderate, and 3 found it difficult. Assuming that all students took the same test, how many of the 162 total students in Christian’s grade would probably rate the test something other than easy?
A.
27
B.
78
C.
84
D.
126
Mathematics
1 answer:
Paraphin [41]3 years ago
8 0

Answer:

C.

84

Step-by-step explanation:

This question is solved using proportions.

From the sample:

11 + 3 = 14 out of 13 + 11 + 3 = 27 would rate the test something other than easy.

Out of 162:

Applying the rule of three:

14 - 27

x - 162

Applying cross multiplication:

27x = 14*162

x = \frac{14*162}{27}

x = 84

Thus the correct answer is given by option C.

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Answer:

y = \frac{3}{4} x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

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m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

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The line crosses the y- axis at (0, 2 ) ⇒ c = 2

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4 0
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^{
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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.
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