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Taya2010 [7]
3 years ago
12

In a calculus​ class, Jack Hartig scored 4 on a quiz for which the class mean and standard deviation were 2.9 and 2.1​, respecti

vely. Norm Alpina scored 8 on another quiz for which the class mean and standard deviation were 6.5 and 1.9​, respectively. Relatively​ speaking, which student did​ better? Make use of​ z-scores.
Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer: Norm Alpina did better with z-score 0.79

Step-by-step explanation:

Z score formula = (raw score - mean) / standard deviation

For Jack Hartig,

score = 4; mean = 2.9; standard deviation = 2.1

Hence, Z score = (4 - 2.9) /2.1

= 1.1/2.1

= 0.52

For Norm Alpina,

score = 8; mean = 6.5; standard deviation = 1.9

Hence, Z score = (8 - 6.5) /1.9

= 1.5/1.9

= 0.79

Relatively, Norm Alpina did better for having Z score 0.79

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<h3>Find the power series representation for the function:</h3>

In the question the given function is,

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We know that series representation of sin x and eˣ are:

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Substituting the series representation in the function we get,

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Hence f(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2(2n+2)} + e^{2} \sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}  is the power series representation for the function f(x) = x³sin(x) + e³ˣ⁺².

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