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s2008m [1.1K]
3 years ago
6

Jeremy's score was 1.75 standard deviations above the mean. which kf thr following is closest to his percentile rank

Mathematics
1 answer:
Gelneren [198K]3 years ago
4 0
The answer is 96%.

Explanation:
It is generally presumed that the scores are normally distributed.

1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75

2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599

3) Multiply the area by 100 in order to find the percentile:
<span>0.9599 </span>× 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.

Hence, the answer is 96%.

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Answer:

x > 3

Step-by-step explanation:

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Need help quick not many points cause I'm low on them will make it brainlies answer if right
adoni [48]
I think the answer is B.
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How much did rosemary pay per pound?
velikii [3]

Answer:

0.89

Step-by-step explanation:

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2 years ago
Please help me answer this :) and explain how you got your answer ! ( will give brainslt )
Bess [88]

Answer: y = - 0.4 x - 4.5

Step-by-step explanation:

Equation y=mx+b

m is the slope

x is *ANY* x coordinate

y is *ANY* y coordinate

b is y intercept or WHERE the point is on the Y AXIS

Lets look at the given:

Slope (m) is -2/5

Passes through point 15, -9/2

As you can see, the point goes through the Y axis at (-9/2)

Our equation is y = - 0.4 x - 4.5

5 0
2 years ago
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

6 0
3 years ago
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